smps
TRANSCRIPT
Switched Mode Power Suppliesby Dr.Ing. Heinz Schmidt-Walter
Switched mode power supplies (SMPS) are used in nearly all electronic systems. Everytelevision set and computer is powered by an SMPS as are many state-of-the-art industrialequipment. Battery powered equipment also uses SMPSs to provide a constant internal supplyoperating voltage independent of the state of charge of the battery. SMPSs are also used toachieve a higher supply voltage compared to the powering battery voltage. This is normallythe requirement for tape recorders, CD players, note books mobile phones and cameras.
SMPS have remarkable advantages in compared to linear regulated power supplies.Theoretically SMPS work loss-free and in practice efficiencies of about 70% to 95% areachieved, this results in low temperature operation and consequently high reliability. Theother major advantage is that SMPS operate at a high frequency which results in small lowweight components. Compared to linear power supplies SMPS are therefore inherently moreefficient, smaller, lighter and cheaper to manufacture.
In general all SMPS operate by the same principle whereby packets of energy are taken froman input voltage by an electronic switch (transistor) which switches at a high frequency.Switching frequencies are normally in the range of 20kHz to 300kHz, depending on therequired performance. The conditions between turn-on and turn-off time of the switch determines the average energy flow. A low pass filter is placed at the output of all SMPSs tosmooth the discontinuous energy flow. The high efficiency of SMPSs is a direct result of thetheoretically loss free switching component and low pass filter.
There are a number of different types of switch SMPSs which are described below.
SMPSs can be configured as secondary or as primary switched power supplies. Secondaryswitched power supplies have no isolation between input and output. They are used inapplications where isolation in respect to the mains already exists or where isolation is notrequired, for example in battery supplied devices. Primary switched power supplies offer anisolation between input and output, their switching transistors operate on the primary side of atransformer. The energy is be transfered to the secondary side at a high frequency via a highfrequency transformer. The transformer can be relatively small, because of its high operatingfrequency.
There are three basic SMPS configurations which are: flyback, forward andresonant-converters. Flyback converters transfer their energy during the off-time of thetransistors. Forward converters transfer their energy during the on-time of the transistors.Resonant converters use a resonant circuit for switching the transistors when they are at thezero current or zero voltage point, this reduces the stress on the switching transistors.
A power factor pre-regulator is also a SMPS and is used to ensure that the mains current issubstantially sinusoidal.
Overview: Switch mode power supplies
Buck converterVout ≤ V in
short-circuit and no load proof simply achievable has to floatVGS
Usage: Repacement of analoge voltage regulators
Boost converterVout ≥ V in
Not short cicuit proofNot no load proof if not operating in a closed loopUsage: Battery supplied devices as notebooks,mobilphones, camera flashesPhotoblitze
Inverting converterVout < 0Vshort-circuit proof simply achievableNot no load proof if not operating in a closed loopusage: Achieve of a negative voltage out of a positive
Flyback converterSeveral, isolated output voltages, regulated by one controlcircuit, achievable Power up to some 100WWide range for input and output voltage (mains voltage85...270VAC achievable)Transistor breakdown voltage VDS ≥ 2Vin
Very good magnetic coupling necessaryBig core with air gap necessary
Single transistor forward converterOnly one output voltageOutput power up to several 100WTransistor breakdown voltage VDS ≥ 2Vin
Duty cycle ton
T≤ 0, 5
Very good magnetic coupling necessarySmall core without an air gap
Two transistor forward converterOnly one output voltageOutput power up to some kWTransistor breakdown voltage VDS = Vin
Duty cycle ton
T≤ 0, 5
Small core without an air gapNot an extaordinary magnetic coupling necessary
Page 2
Vin Vout
Vin Vout
Vin Vout
Vin
Vout1
Vout2
Vin
Vout
VinVout
Full-bridge push-pull converterOnly one output voltageOutput power up to many kW Transistor breakdown voltage VDS = V in
Small core without an air gapNot an extaordinary magnetic coupling necessaryBalancing problems
Half-bridge push-pull converterOnly one output voltageOutput power up to some kW Transistor breakdown voltage VDS = V in
Small core without an air gapNot an extaordinary magnetic coupling necessaryBalancing problems
ZCS push-pull resonant converterSeveral, isolated output voltages achievableOutput power up to several kWTransistor breakdown voltage VDS = V in
Small core without an air gapNot an extaordinary magnetic coupling necessaryControl with fixed pulse duration and variable frequencyIf the output power is low compared with the rated power,the frequency can be audible
Page 3
Vin Vout
Vin Vout
Vin Vout
C
L
Buck-converter
The Buck-converter converts an input voltage into a lower output voltage, it is also calledstep-down converter.
Figure1.1.1: Buck-converter
Figure 1.1.1 shows the circuit diagram of a Buck-converter. The transistor operates as theTswitch, which is turned on and off by a pulse width modulated control voltage withVPWM
high frequency. The ratio between on-time to the period time is called the dutyt1
Tt1 T
cycle.
Figure 1.1.2: voltages and currents of the Buck-converter
In the following analysis it will be assumed that the conducting voltage drop of the transistorand the diode is zero.
During the on-time of the transistor the voltage is equal to . When the transistorV1 V in
switches off (blocking phase) the inductor continues to drive the current through the load inLparallel with and the diode, consequently the voltage becomes zero. The voltage Cout V1 V1
stays at zero during the off-time of the transistor provided that the current does not reduceIL
to zero. This mode of operation is called continous mode. In this mode is a voltage whichV1
changes between and zero, corresponding to the duty cycle of ,(see Figure 1.1.2).V in Vcont
L
CoutLoad
T I L
V1Vin
Iin
VoutCin
VL
D
Iout+ ++
VPWM
V1
Vcont
t
t
V
t
Vout = V1
Vin
IL Iout = IL
t1 T
L (Vin-Vout)
-Vout
t
∆ IL
Seite 4
The low-pass filter, formed by und , produces an average value of , i.e. ,L Cout V1 Vout = V1
therefore for continuous mode:
Vout = t1
TVin
For the continuous mode the output voltage is a function of the duty cycle and the inputvoltage, it is independent on the load.
The inductor current has a triangle shape, its average value is determined by the load. TheIL
peak-to-peak current ripple is dependent on and can be calculated with the help of∆IL LFaraday's Law:
V = L didt
→ ∆i = 1L
⋅ V ⋅ ∆t → ∆IL = 1L
(Vin − Vout) ⋅ t1 = 1L
Vout (T − t1)
For and a switching frequency it follows that for the continous mode:Vout = t1
TV in f
∆IL = 1L
(V in − Vout) ⋅ Vout
Vin⋅ 1
f
The current ripple is independent of the load. ∆IL
The average of the current is equal to the output current .IL Iout
At low load current, in case that , the current becomes zero in every switchingIout ≤ ∆IL
2IL
cycle. This mode is called discontinous mode and for this mode the calculations above arenot valid.
Calculation of and :L Cout
To the calculate the value of a realistic value of has to be selected. The problem is asL ∆IL
follows: If is selected at a very low value, the value of has to be relatively high and∆IL Lthis would require ae very heavy and expensive inductor. If is selected at very high level∆IL
the switch-off current of the transistor would be very high (this would result in high losses inthe transistor). A good and usual compromise between these effects is: ∆IL ≈ 0.2Iout
For it follows: L
L = 1∆IL
(V in − Vout) ⋅ Vout
Vin⋅ 1
f
The maximum value of the inductor current is:
Seite 5
IL = Iout + 12∆IL
Assuming that the inductor ripple current is small compared to its dc current the RMS valueof the current flowing through the inductor is given by:
IL(RMS) ≈ Iout
The capacitor is chosen usually for a cut-off frequency of the -low-pass filter,Cout LCout
which is approximately 100 to 1000 times lower than the switching frequency. An exactcalculation of the capacitor depends on its maximum rating of the AC current and its serialequivalent impedance , both can be verified from the relevant data sheet.Zmax
The current ripple causes a voltage ripple at the output capacitor . For normal∆IL ∆Vout Cout
switching frequencies this voltage ripple is determined by the equivalent impedance .Zmax
The output voltage ripple is given by Ohm's law:
∆Vout ≈ ∆IL ⋅ Zmax
The choice of the output capacitor depends not on its capacitance, but on its series equivalentimpedance at the switching frequency which can be verified from the capacitor dataZmax
sheet.
Seite 6
Boost converter
The boost converter converts an input voltage to a higher output voltage. The boostconverter is also called a step-up converter.Boost converters are used in battery powered devices, where the electronic circuit requires ahigher operating voltage than the battery can supply, e.g. notebooks, mobile phones andcamera-flashes.
Figure 1.2.1: Boost converter
Figure 1.2.1 shows the basic circuit diagram of the boost converter. The transistor operatesTas a switch, which is turned on and off by a pulse-width-modulated control voltage . Vcont
Figure 1.2.2: voltages and currents of the boost converter
In the following analysis it will be assumed that the conducting voltage drop of the transistorand the diode is zero.
During the on-time of the transistor, the voltage across is equal to and the current L V in IL
increases linearly. When the transistor is turned off, the current flows through the diodeIL
and charges the output capacitor. The function of the boost converter can also be described in terms of energy balance. Duringthe on-time of the transistor the inductance is charged with energy and during the off-time of
Cin Cout
L
LoadTV1Vin
I
VoutVL
D
Vcont
IoutIin L+ + +
VoutV1
Vcont
t
t
V
t
IL
t1 T
L
-(Vout-Vin)
Vin
t
∆ ILIin = IL
Iout = IDt
ID
page 7
the transistor this energy is transfered from the inductor through the diode to the outputcapacitor.
If the transistor switch is not turned on and off by the clock the output capacitor charges via L and to the level . When the transistor is switched the output voltage will increaseD Vout = V in
to higher levels than the input voltage.
As with the buck converter (see chapter 1.1: "buck converter") discontinous and continousmode is dependent on whether the inductor current reduces to zero during the off-time of IL
the transistor.
With the help of Faraday's Law the continous mode and steady state conditions (see alsofigure1.2.2) can be established:
.∆IL = 1LVin ⋅ t1 = 1
L(Vout − V in) ⋅ (T − t1)
From which:
Vout = V inT
T − t1
For the continous mode the output voltage is a function of the duty cycle and the inputvoltage, it is independent of the load.
The boost converter is not short circuit proof, because there is inherently no switch-offdevice in the short-circuit path.
NOTE:If the boost converter is not regulated in a closed loop but is controlled by a fixedduty cycle of a pulse generator (this could be the case for a laboratory set up), theboost converter is not no-load proof. This is because each switching cycle results in energy in the choking coil being tranfered to the output capacitor. This will result inthe output voltage continously increasing until the devices are eventually destoyed.
Calculation of and : L Cout
As with the buck converter the starting point of calculating is to select a current ripple L ∆IL
of about 20% of the input current: . The input current can be calculated by∆IL ≈ 0.2I in I in
assuming zero losses (input power = output power), therefore:
V in ⋅ I in = Vout ⋅ Iout → I in = IoutVout
Vin
can be calculated as follows:L
L = 1∆IL
(Vout − V in) Vin
Vout⋅ 1
f
The peak value of the inductor current is (Ref: Fig. 1.2.2):
page 8
.IL = Iin + 12∆IL
Assuming that the inductor ripple current is small compared to its dc current the RMS valueof the current flowing through the inductor is given by:
IL(RMS) ≈ I in
The output capacitor is charged by pulses (Ref Fig. 1.2.2). The ripple of the output∆Vout
voltage results from the pulsating charge current and is mainly determined by theID
impedance at the switching frequency of capacitor . can be verified from the Zmax Cout Zmax
the capacitor data sheet.
The output voltage ripple is given by Ohm's law:
∆Vout ≈ ID ⋅ Zmax
page 9
Buck-Boost Converter
The buck-boost converter converts a positive input voltage to a negative output voltage.
Figure 1.3.1: Buck-Boost converter
Fig 1.3.1 shows the basic circuit of the buck-boost converter. The transistor works as aTswitch, which is turned on and off by the pulse-width-modulated voltage . During theVcont
on-time of the transistor, the inductor current increases linearly. During the off-time theIL
current continous and charges the output capacitor . Note the polarity of the outputIL Cout
voltage in Fig1.3.1.
For the continous mode and steady state conditions the output voltage is given by:
Vout = V int1
T − t1
The inductor current is given by:IL
IL = IoutT
T − t1= Iout
Vout
V in+ 1
and ∆IL = 1L
V in t1 = 1L
⋅ Vin Vout
Vin + Vout⋅ 1
f
Figure 1.3.2: Voltages and currents of the buck-boost-converter
LCout
Load
T
I L
Vin
Iin
VoutCin
VL
D
Vcont
Iout+
+
Vcont
t
V
t
IL
t1 T
L
-Vout
Vin
t
∆ IL IL
Iout = IDt
ID
page 10
Flyback converter
The Flyback converter belongs to the primary switched converter family, which means thereis isolation between in and output. Flyback converters are used in nearly all mains suppliedelectronic equipment for low power consumption, up to approximately 300W. Examples ofwhich are televisions, personal computers, printers, etc..
Flyback converters have a remarkably low number of components compared to other SMPSs,they also have the advantage that several isolated output voltages can be regulated by onecontrol circuit.
Figure 2.1.1: Flyback converter
Fig. 2.1.1 shows the basic circuit of a flyback converter. The transistor works as a switch,which is turned on and off by the pulse-width-modulated control voltage . During theVcont
on-time of the transistor the primary voltage of the transformer is equal to the inputV1
voltage which results in the current increasing linearly. During this phase, energy isV in I1
stored in the transformer core. During the on-phase the secondary current is zero, because thediode is blocking. When the transistor is turned off the primary current is interrupted andI1
the voltages at the transformer invert due to Faraday's Law ( ), the diode conducts andv = Ldidt
the energy moves from the transformer core via the diode to the output capacitor . Cout
During the on-phase of the transistor the drain-source voltage is equal to zero. DuringVDS
the off-time of the transistor, the output voltage will be transformed back to the primaryVout
side and the drain-source voltage theoretically steps up to . If a mainsVDS = Vin + Vout ⋅ N1
N2
voltage of 230V/50Hz is used will jump up to approximately 700V. In practice thisVDS
voltage will be even higher due to the self induction of the leakage inductance of thetransformer. To allow for this effect the minimum rated drain-source breakdown voltage ofthe transistor must be 800V.
The transformer is not a "normal" transformer, because its function is to store energy duringthe on-time of the transistor and to deliver this energy during the off-time via the diode to theoutput capacitor. In effect the transformer is a storage inductor (often called a choke) with a
primary and secondary winding. To store energy the transformer core needs an air gap(normal transformers do not have an air gap). An important consideration for this transformeris, that primary and secondary windings are closely coupled to achieve a minimum leakageinductance. It should be noted that the energy of leakage inductance cannot be transfered to
the secondary side and is therefore disipated as heat on the primary side.
V 1
Vin
VoutV2
I1
Vcont
N1 N2
VDS
Cout
L1, L2,
Cin
Load++
page 11
Figure 2.1.2: Voltages and currents at the flyback converter
Design of the flyback converter:
For the primary voltage of the transformer the average must be equal to zero for steadyV1 V1
state conditions (if not, the current will increase to infinity).
This leads to: and:V in ⋅ t1 = Vout ⋅ N1
N2⋅ (T − t1)
Vout = V in ⋅ N2
N1⋅ t1
T − t1
The turns ratio of the transformer should be choosen so that for the rated output power theon-time (energy charge time) is equal to the off-time (energy discharge time) . Thist1 T − t1
leads to the turns ratio:
N1
N2= V in
Vout
V
Vcont
t
t
IL
t1 T
t
∆ IL
DS Vin+Vout( )N1N2
I1
t
∆
I2
t
I1
∆ I2
secondary current calculated to the primary side:
I1 I2 N2N1 I = I1+I2L
N2N1
t
VinV1
N1N2
-Vout
Î2
page 12
The breakdown voltage of the transistor and the reverse voltage of the diode must be for thiscase:
Transistor: VDS = V in + Vout ⋅ N1
N2≈ 2V in
Diode: VR = Vout + V in ⋅ N2
N1≈ 2Vout
It should be noted that the rated breakdown voltage of the transistor must be chosensignificantly higher, because at the turn-off instant the energy of the leakage inductance Ls
will not be taken over by the secondary winding. To keep the overvoltage in an acceptablerange a snubber circuit is required, see Fig 2.1.3. At the instant of turn-off the current of theleakage inductance is diverted through by the diode and charges the capacitor . TheLs D Cpower is dissipated in resistor .RIf and are required to operate at 230VAC,, a value of R has to be determinedR Cexperimentally to ensure that the dc voltage across falls within the region of 350V toC400V.
Figure 2.1.3: Snubber circuit to limit the peak voltage across the transistor
To design the transformer the primary inductance has to be calculated first.L1
has to store energy during the on-time of the transistor, which is the energy required at theL1
output. This energy is given by: , where is the periodic time of the switching W = Pout ⋅ T Tfrequency and is the rated power. This energy is stored in the primary inductance duringPout
the first half of the period time and is transfered to the output capacitor during the second halfof the switching period. As before the switching period is divided into two equal parts, onepart to store the energy and the other part to transfer the energy.
During the on-time of the transistor the voltage across the primary inductance is equal to V in
and the current is a ramp waveform. For every cycle of the input energy it follows that :I1
(see Fig. 2.1.4)W = V inI1
2T2
This energy is stored in and can be calculated as:L1
W = 12
L1I12
For the size of the primary inductance this leads to:
. L1 ≈V in
2
8 Pout ⋅ f
VinVout
Ls
DRC
+ +
page 13
The calculation above assumes an efficiency of 100 %. If we consider an efficiency of , itηmeans that we have to store more energy in and not all of this energy is delivered to theL1
output, then can be calculated as follows:L1
L1 ≈Vin
2
8 Pout ⋅ f⋅ η
has to be estimated because its value is not known at this point of calculation. ( isη η ≈ 0.75normally a good estimate.)
Fig. 2.1.4: Shape of the input current for rated powerI 1
The peak value of the current is: I 1 I 1 = 4 ⋅ Pout
Vin ⋅ η
The RMS-value of the current is: I 1 I 1RMS = I 1
6The core of the transformer and the windings can now be calculated with the help of Chapter5: "Calculation of inductors and high frequency transformers"
The output capacitor is charged by pulses (Ref Fig. 2.1.2). The ripple of the outputCout ∆Vout
voltage results from the pulsating charge current and is mainly determined by theI 2
impedance of the capacitor . can be verified from the capacitor data sheet. Zmax Zmax
The magnitude of the ripple voltage is given as follows:
∆Vout ≈ I 2 ⋅ Zmax
The input capacitor can be calculated for 230V/50Hz-mains as follows:Cin
Cin ≈ 1µFW
⋅ Pin
A special feature of the flyback converter is the possilbility of controlling several isolatedoutput voltages with only one control circuit (Fig. 2.1.5).
T/2 T
I1Î1
t
page 14
Fig. 2.1.5: Flyback converter for several output voltages
One output voltage is regulated (in Fig. 2.1.5 ). Voltage is coupled to via theVout 3 Vout 2 Vout 3
turns ratio: . The energy which is stored in during the on-time of theVout2
Vout3= N2
N3L1 (N1)
transistor moves during the off-time to the outputs. These output voltages maintain theirvalues in relationship to the turns ratio. The output voltages in relation to the the turns ratiofrom the primary side appear to be in parallel. Therefore the energy from the primary sidetransfers to the output where the lowest voltage appears.
VinN1 Vout2
Vout3
control circuit
N2
N3
+
+
+
page 15
Single Transistor Forward Converter
The single transistor forward converter belongs to the primary switched converter familysince there is isolation between input and output. It is suitable for output powers up to 1kW.The single transistor forward converter is also called a single ended forward converter.
Figure 2.2.1: Single transistor forward converter
The forward converter transfers the energy during the on-time of the transistor. During thistime the voltage is equal to the input voltage. The winding is in the same direction as V1 N2
. When the transistor is on voltage at is given by . The voltage N1 V2 N2 V2 = VinN2
N1V2
drives the current through the diode which during this time is equal to through I 2 D2 I 3 Lwhich charges the output capacitor . Cout
Fig. 2.2.2: voltages and currents at the single transistor forward converter
During the off-time of the transitor, and are without current. The inductor drawes itsN1 N2 Lcurrent through the diode . The value of the voltage is equal to zero (neglecting theD3 V3
forward voltage drop of ).D3
Vin
Vout
I1
Vcont
N1 N1' N2
CoutV3
LD2
D1
D3V2V1
I1' I2
I3
Load
Ia
Cin
+ +
V1
Vcont
t
t
V3
t
Vin
I3Iout = I3
t1 T
Vin
t
∆ I3ID3
-VinN2N1
t
I1 }}
∆ I1
∆ I3
I1'∆ I1 }
ID2
t
N2N1
I1: magnetizing-current
∆
Page 17
During the off-time of the transistor, the magnetic flux of the transformer has to reduce tozero. The core is demagnetized with via to . has the same number of turns asN1 D1 V in N1
therefore the demagnetization needs an equal time intervall as the on-time. For thisN1
reason the minimum off-time has to be as long as the on-time. This causes a maximum dutycycle of 0.5 for the single transistor forward converter. t1/T
During the off-time, the voltage at is equal to the input voltage . This voltage will beN1 V in
transformed back to the primary winding and for follows: . Due to this theN1 V1 V1 = −V in
drain source voltage steps up to when the transistor is turned off.VDS ≥ 2Vin
In comparison to the transformer of the flyback converter the transformer in this forwardconverter is a "normal" transformer. Its job is not to store energy but to transfer energy. Forthis reason the core has no air gap.
The breakdown voltage of the transistor has to be .VDS > 2V in
The windings and must be closely coupled. However a snubber circuit as shownN1 N1
in Fig. 2.1.3, chapter 2.1. "flyback converter" is necessary.
In comparison to the flyback converter, the forward converter can only have oneregulated output voltage.
The maximum duty cycle is .t1
T= 0.5
Design of the single transistor forward converter:
The output voltage is equal to average of . The maximum duty cycle is 0.5.This leadsVout V3
to (see also Chap. 1.1: "buck converter"):
Vout = V in ⋅ N2
N1⋅ t1
T
For the turns ratio follows:
N2
N1= 2 ⋅ Vout
V inand N1 = N1
For further calculation of the transformer see Chapter 5: "Inductors and high frequencytransformers"
To calculate the method used for the buck converter is appropriate. Initially the currentLripple of the inductor current has to be selected. A value for it is 20% of the output∆I3 I3
current is normally acceptable: . Assuming a maximum duty cycle of 0.5, this∆I3 ≈ 0, 2 ⋅ Iout
leads to:
Page 18
L = Vout ⋅ T/2∆I3
The value of depends on the acceptable voltage ripple of the output voltage. ThisCout ∆Vout
voltage ripple is mainly determined by the impedance of the output capacitor :Zmax Cout
∆Vout ≈ ∆IL ⋅ Zmax
can be verified from the datasheet of .Zmax Cout
The input capacitor for 230V/50Hz-mains should be:C in
C in ≈ 1µFW
⋅ P in
2.2.1 Two-Transistor forward converter:
The two-transistor forward converter is a variant of the single transistor forward converter.
Fig. 2.2.3: two-transistor forward converter
The transistors and are switching at the same time. During the on-time of theT1 T2
transistors, the voltage at the primary winding is equal to the input voltage . During theV in
off-time of the transistors the transformer will be demagnetized via the diodes and D1 D2
into the input voltage . In comparison to the single transistor forward converter thisV in
converter has the advantage that its transistors only have to block the input voltage and thewinding is not required. In addition to this the coupling of the transformer windings is noN1
longer critical. These advantages make this converter type suitable for significantly higheroutput powers compared to the single transistor converter.
The calculation of the components is equivalent to the single transistor forward converter.
For the two transistor forward converter the breakdown voltage of the transistors is onlyrequired to be .VDS = Vin
The two transistor forward converter can be used for powers up to a few kWs. It is asimple converter, which is not critical in regard to its physical design and its electricaloperation.
VinVout
N1 N2 L
Load
IoutT1
T2D1
D2+
+
Page 19
Push-Pull Converters
The push-pull converter can be designed for high power..
Fig. 2.3.1: Push-pull converter, here: full-bridge typ
Fig. 2.3.2: Voltages and currents at the push-pull converter
The push-pull converter drives the high frequency transformer with an AC-voltage, where thenegative as well as the positive half swing transferes energy. The primary voltage can beV1
depending on which pair of transistors ( or ) are turned on or+V in, −Vin or zero T1, T4 T2, T3
off. At the secondary side the AC-voltage is rectified and smoothed by und .L Cout
For continious mode follows (see also Chapter 1.1 "buck converter"):
Vin Vout
N1 N2 L
Load
Iout
T1
T4
T3
T2
V3V2V1
I3
Cin Cout
++
V1
V
t
t
V3
t
Vin
I3Iout = I3
Vin
t
∆ I3
-VinN2N1
t
I1
control of T1 and T4
V
tt1 T
control of T2 and T3
GS
GS
Vout = V3
Page 20
Vout = V in ⋅ N2
N1⋅ t1
T
The duty cycle may theoretically increase to100%. This is not possible in practice becauset1
T
the serial connected transistors or have to be switched with a time difference toT1, T2 T3, T4
avoid a short-circuit of the input supply. The turns ratio of the transformer has to be:
N2
N1≥ Vout
V in
The transistors of the push-pull converter can be switched with the maximum duty
cycle of 0.5. This leads to the maximum duty cycle of after rectification.t1
T= 1
The calculation of and follows those of the buck converter (chapter 1.1).L Cout
2.3.1 Half-Bridge Push-Pull Converter:
Fig. 2.3.3: Half-bridge push-pull converter with full-wave rectifier
A variant of the push-pull converter is the half-bridge push-pull converter. The capacitors and divide the input voltage into two. Therfore the magnitude of the primaryC1 C2 V in
voltage is . In comparison to the full-bridge push-pull converter follows for the±V in/2
half-bridge typ the turns ratio of the transformer to: .N2
N1≥ 2Vout
V in
NOTE:In Fig. 2.3.3 a two diode full-wave rectifier is used instead of a full-wave bridgerectifier. The choice of rectifier type is dependent on the output voltage and current. Thedifference between these two rectifier types is, that the current has to pass through twodiodes in the bridge type and only one diode in the full-wave type. Consequently thefull-wave type is used for high current to reduce the rectifier losses and the bridge typeis used for high voltage purpose to save one secondary winding of the transformer.
Vin Vout
N1 N2
Load
Iout
T1
T2
V2
V1
C1
C2
V2
L
Cout
++
Page 21
Resonant Converters
Resonant converters use a resonant circuit for switching the transistors when they are at thezero current or zero voltage point, this reduces the stress on the switching transistors and theradio interference. We distinguish between ZVS- and ZCS-resonant converters (ZVS: ZeroVoltage Switching, ZCS: Zero Current Switching).
To control the output voltage, resonant converters are driven with a constant pulse duration ata variable frequency. The pulse duration is required to be equal to half of the resonant periodtime for switching at the zero-crossing points of current or voltage.
There are many different types of resonant converters. For example the resonant circuit can beplaced at the primary or secondary side of the transformer. Another alternative is that a serialor parallel resonant circuit can be used, depending on whether it is required to turn off thetransistor, when the current is zero or the voltage is zero.
The technique of resonant converters is described below giving the ZCS-push-pull resonantconverter as an example.
2.4.1 ZCS-Push-Pull Resonant Converter:
Fig. 2.4.1: The ZCS-push-pull resonant converter
The converter works as follows:The resonant circuit is formed by and . Assume an initial condition of the voltage L C VC
across equal to zero. If now the transistor is turned on, a sinusoidal current half-swingC T1
starts through . This half-swing charges the capacitor from zero to .T1, L, Tr, C and Cin C Vin
If this first half sinusoidal swing is finished, can be switched off without losses and after aT1
LoadVout
VinCin
T1
T2
L
Tr
D1
D2 C
Cout
control circuit: constant pulse duration, variable frequency
V'outI
Vc
+
+
Page 22
short delay can be switched and a next half sinusoidal swing starts, this discharges T2 Cfrom back to zero Volts. V in
Every half sinusoidal swing transfers a certain amount of energy from the primary to thesecondary side of the transformer. The transformer operates on its primary side as aTrvoltage source. For the duration of the current swing through the primary winding, the outputvoltage will be transformed to the primary side: . The energy which isVout Vout = Vout
N2
N1
transfered by every half-swing is equal to . This energy will be transferedW = Vout ⋅ ∫ i(t) dttwice in each resonant period. This leads to the output power being given by
( : frequency of the converter). Fig. 2.4.2 shows an equivalentPout = W ⋅ 2 fswitching fswitching
circuit for one half-swing.
Fig. 2.4.2: Equivalent circuit for one half sinusoidal swing of the ZCS-push-pull resonant converter
The resonant frequency is:
f0 = 12π LC
This leads to the minimum on-time of the transistors. The on-time should be a little higherthan half of the resonant period time to ensure that the current reduces to zero. For maximum energy transfer, must be half of . This leads to the turns ratio of theVout Vin
transformer:
Vout = 12
Vin ⇒ N1
N2= 1
2⋅ Vin
Vout
The maximum output power is achieved if one half current swing instantly follows the next.
The transfered energy of each half-swing further depends on the value of and . The higherC Lthe value of and the lower the value of , to maintain a certain resonant frequency, theC Lhigher the amount of energy transfer. (see also the peak value of the current in Figs. 2.4.2 and2.4.3).For a certain output power , considering , it can be shown that for and :Pout Vout = Vin/2 L C
I
tton
To/2
(Vin -V'out)CL
Î =
C Vin
T1 =Vout N1N2
VcVin
t
Vc
IS L V'out
toff
+
Page 23
LC
=
Vin
2
2
⋅ 2π ⋅
fSwitching
f0
Pout⇒ C = 1
2π ⋅ LC
⋅ f0
und L =
LC
2
⋅ C
Figure 2.4.3: Voltages and currents at the ZCS-push-pull resonant converter
I addition to the general advantages of resonant converters, having lower switching losses andlower radio interference, this particular resonant converter has two more additionaladvantages:
The ZCS-push-pull resonant converter can regulate several output voltages using onecontrol circuit, as per the flyback converter. This because several output voltages seemto be parallel connected from the view of the primary side. Due to this the energyalways passes to that output, having the lowest voltage value taking into considerationthe turns ratio.
The ZCS-push-pull resonant converter is no-load and short-circuit proof without anyelectronic precaution. The output voltage cannot reach more than twice of the nominalvalue, because then is . The current cannot reach more than twice of theVout = Vin
nominal output current, because then is and .Vout = 0 I = Vin C/L
CL
t
t
V
V
I
t
GS1
GS2
Vc
t
V'out
t
Vin
VinN1N2
-VoutN1N2
12Vin
Page 24
Control of switch mode power supplies
The output voltage of a switch mode power supply is kept constant with the help of closedloop control. The value of the output voltage (actual value) is compared with a referencevoltage (nominal voltage). The difference between actual and nominal value controls the dutycycle of the transistor drive. The function of the control loop is to regulate the variation of themains and of the change of the output current. This is called line regulation and loadregulation.
There are two different methods of regulation: voltage-mode and current-mode control.The voltage-mode control is the "traditional" method of regulation. Most modern systems use current-mode control which is the basis of nearly all IC current-mode controllers.
Both controller types can be explained using a boost converter shown in fig 4.1: Voltage-mode control:
Figure 4.1: voltage-mode-control for a boost converter
The output voltage is compared to the reference voltage via a voltage divider Vout Vref R1, R2
and amplified by the PI-regulator. A pulse width modulator (PWM, see Fig.4.1a) converts theoutput voltage of the PI-regulator into a pulse width modulated voltage . The outputV2 t1/Tof the pulse width modulator (PWM) controls the transistor of the boost converter (see alsoChapter 1.2: "boost converter").
Figure 4.1a:Pulse width modulator
LoadVin Vout
-
+-+
saw-toothgenerator
VrefV'out
C1 R4
R1
R2
R3PWM
PI-regulator
V2t1/T
L
Cout
saw-tooth-signal -
+V2
t1 T
t
V2V Vs
V3
Vs
Vs
peak
Pulse With Modulator
t
V3
Page 26
The closed loop operates as follows: If the output voltage is to low, the voltage willVout Vout
be lower than the reference voltage , this will cause the output voltage of theVref V2
PI-regulator to increase. In the PWM circuit is compared with a saw tooth signal and as itV2
increases the duty cycle also increases, this causes the output voltage to increase untilt1/T. V out = Vref
Current-mode control:
Figure 4.2: current-mode control for a boost converter
The output voltage is compared to a reference voltage via the voltage divider Vout Vref R1, R2
and amplified by the PI-regulator. The output voltage of the PI-regulator is compared withV2
ramp voltage across the current measuring resistor . When the voltage across Ri exceeds V2Ri
the output of the comparator resets a RS-flip-flop and turns the transistor off. TheRS-flip-flop is set before by the clock. The transistor is turned on by the clock and turned off,when the ramp voltage (which means the inductor current) reaches a certain value. In this waythe PI-regulator directly controls the inductor current.
The closed loop operates as follows: If the output voltage is too low, the voltage Vout Vout
will be lower than the reference voltage . This causes the output voltage of theVref
PI-regulator increases. The comparator compares the voltage with the ramp voltageV2 V2
across . In this way determines the value to which the ramp voltage across increasesRi V2 Ri
(which means the value to which the inductor current increases) until the transistor isI L
turned off. If increases because the is lower than , the inductor current willV2 Vout Vref
increase until is exactly equal to the reference voltage.Vout
Comparision of voltage-mode to current-mode control:
The PI-regulator of the current-mode control regulates the inductor current directly. Thiscurrent feeds the output capacitor and the load resistance . and form a firstCout RL Cout RL
order system and the step response is an an exponential function.
The voltage-mode control regulates the duty cycle , which means that the voltage acrosst1/T is controlled. This voltage operates on a second order system, formed by .L L, Cout and RL
The step response of such a system is a sinusoidal transient approaching a fixed value.
LoadVin Vout
-
+-+
VrefV'out
C1 R4
R1
R2
R3
PI-regulator
V2
Flip-Flop
Ri
ClockS
R
Q
ramp
comparator
ILVLID
Cout
L
Page 27
The current-mode control has therefore a better control response, for this reason mostcontrollers are current-mode types.
regulator1+sR CaL
RLVref VoutID
R2R1+R2
V'out
regulatorVref VoutID
R2R1+R2
V'out
1+sR CaL
RLt1/T ksL
+
-
+
-
current-mode voltage-mode
Fig 4.3: Block-diagramms for current-mode- and voltage-mode control
Design of the PI-regulator:
The PI regulated system tends to oscillate, if the capacitance is selected at too small aC1
value and if the resistor is too high a value. To allieviate this problem should initiallyR4 C1
be selected high ( A 1 F foil capacitor is normal in most control circuits). should beR4
selected so that the cut-off frequency of the PI-regulator stays well below the cut-offfrequency of and :L Cout
12π LCa
≥ 10 12πR4C1
The controller should now operate in a stable mode (if not, internal interference or an unfitarchitecture of the board might could be a problem). To improve the reaction of the closedloop, can be decreased step by step with a parallel increase of . If the loop starts toC1 R4
oscillate, can be increased by the factor of ten and decreased. Using these designC1 R4
guides the loop will operate in a stable mode with sufficient regulation speed for mostapplications.
HINT:In many control circuits the operational amplifier (normally called the error amplifier)is a transconductance amplifier. It supplies an output current (very high outputimpedance), which is proportional to the input voltage. In this case and areR4 C1
connected from the output to ground to achieve the PI-characteristic of the regulator.
Page 28
Design of Inductors and High Frequency Transformers
Inductors store energy, transformers transfer energy. This is the prime difference. Themagnetic cores are significantly different for inductors and high frequency transformers:Inductors need an air gap for storing energy, transformers do not. Transformers for flybackconverters have to store energy which means they are not a high frequency transformer butthey are in fact an inductor with primary and secondary windings. The material of the cores isnormally ferrite. In addition to this other marterials with high permeability and with a highsaturation point are used.
Calculation of Inductors:
An inductor with certain inductance and certain peak current can be determined by theL Ifollowing calculation:
Inductors should store energy. The stored energy of an inductor is: . This energy isW = 12LI 2
stored as magnetic field energy, within the ferrite core and within the air gap (see Fig.5.1.1).The higher the required stored the energy the larger the required core.
The size of a inductor is approximately proportional to the stored energy.
Fig. 5.1.1: inductor with its magnetic and mechanical sizes
The field energy in the inductor is:
(1)W = 12 ∫ H ⋅ B dV ≈
energy in the ferrite
12
HFe ⋅ BFe ⋅ VFe +
energy in the air gap
12
Hδ ⋅ Bδ ⋅ Vδ
The magnetic field density is continuous and within the air gap and the ferrite isBapproximately equal, i.e. . The magnetic field strength is not continuous,B ≈ BFe ≈ Bδ Hwithin the air gap it is increased by a factor compared to that than within the ferrite. If thisµr
is substituted into equation (1) and considering this leads to:B = µ0µr ⋅ H, VFe = lFe ⋅ A and Vδ = δ ⋅ A
I
N
Φ=ΒΑA
Hfe
Hδδ
lfe
A: cross-section area of the corel : magnetic length of the core
Φ: magnetic fluxB: magnetic flux density
δ : air gap
H : magnetic field strength within the ferritH : magnetic field strength within the air gap
fe
δ
fe
N: number of turnsI : inductor current
Page 29
W ≈ 12
B2
µ0
lFeµr
+ δ ⋅ A
of the ferrite amounts to 1000...4000. It should be noted that the magnetic length of theµr
ferrite is reduced by in the above equation. Therefore it can be seen that the energy isµr
mainly stored within the air gap.
This leads to: W ≈ 12
B2 ⋅ A ⋅ δµ0
Inductors require an air gap to store energy.
Because the energy is stored within the air gap, an inductor requires a certain volume for theair gap to store a certain amount of energy. The energy is given by . The core material1
2LI 2
has a limit for the maximum magnetic flux density , this limit is about forB Bmax = 0, 3 Tusual ferrite materials. This leads to a minimum required volume of the air gap:Vδ
Vδ = A ⋅ δ ≥L I 2 ⋅ µ0
Bmax2
where Bmax = 0, 3 T
Knowing the required volume of the air gap, a core can be selected from a databook of ferritecores.
The number of turns can be calculated with help of the magnetic conductance (oftenN AL
simply called the -value): AL
N = LAL
AL :magnetic conductance
The -value can be verified from the databook of the ferrite cores. AL
The maximum flux density should not be higher than 0.3 Tesla. The maximum flux densitywithin the ferrite can be calculated using the data of the core datasheet.
B = L ⋅ IN ⋅ Amin
= N ⋅ AL ⋅ IAmin
!≤ 0, 3 T
: Minimum cross-cut of the core. The flux density has its maximum. at . can beAmin Amin Amin
verified from the datasheet.
Page 30
Calculation of the wire: The current density of the wire can be chosen between 2 und 5 A/mm² (depending on theSsize and the isolation, which determines the heat transport out of the inductor). This leads tothe diameter of the wire :d
d = 4 ⋅ IRMS
π ⋅ Swith S = 2…3…5 A
mm2
Calculation of High Frequency Transformers
A high frequency transformers transfer electric power. Its mechanical size depends on thepower to be transfered and on the operating frequency. The higher the frequency the smallerthe mechanical size. Usually frequencies are from 20 to 100kHz. The material of the core isferrite.
Databooks for appropriate cores provide information about the possible tranfer power forvarious cores.
The first step to calculate a high frequency transformer is to choose an appropriate core withthe help of the databook, the size of the core is dependent on the transfer power and thefrequency. The second step is to calculate the number of primary turns. This numberdetermines the magnetic flux density within the core. The number of secondary turns is theratio of primary to secondary voltage. Following this the diameters of the primary andsecondary conductors can be calculated depending on the RMS-values of the currents.
Calculation of the minimum number of primary turns:
Figure 5.2.1: Voltages and currents at a transformer
V2
V1
T/2 T
∆
t
V1 N2N1
t
I2 V1R
N2N1
IM IMV1R
N2N1
2I1
t
tIM : magnifiing current
V1 V2
I1 I2
N1 N2
R
simple equivalent circuit:
V1 V2'
I1 I2'
R
IM
L1N1N2
2
Page 31
The voltage at the primary side of the transformers has a rectangle shape. This causes anV1
input current , which is the addition of the back transformed secondary current and theI1 I2
magnetising current (see figure 5.2.1). To keep the magnetising current low, aIM IM
magnetic core without an air gap is used.
The rectangle voltage causes a triangle shape for the magnetising current . TheV1 IM
magnetising current is approximately independent of the secondary current (see the simpleI2
equivalent circuit in figure 5.2.1). The magnifiing current is approximately proportional to themagnetic flux or flux density. The input voltage determines the magnetic flux. TheV1
physical correlations are given by Faradays law of induction: . V = N ⋅ dΦdt
Figure 5.2.2: input voltage and magnetic flux density at the transformer
For the transformer in figure 5.2.1 follows:
∆B = V1 ⋅ T/2N1 ⋅ A
The change of flux density depends on the frequency and the number of∆B f = 1/Tturns . The higher the frequency and the number of turns the lower the change ofN1
flux density.
The minimum number of turns can be calculated to ensure that a certain change of fluxN1 min
density is not exceeded. The saturation flux density of about (which means∆B B ≈ 0, 3 T) cannot be used in high frequency transformers. In push-pull converters going∆B ≈ 0, 6 T
around the hysteresis loop with every clock would cause unacceptable losses i.e. heatgeneration. If no further information on core losses and termal resistance are available, ∆Bshould be limited to for operating frequencies from 20 to 100 kHz. The∆B ≈ 0, 3…0, 2 Tlower the lower the core losses. ∆B
This leads to a minimum number of turns for :N1
N1 min ≥ V1 ⋅ T/2∆B ⋅ Amin
where ∆B ≈ 0, 2…0, 3 T
: minimum cross-section area of the core. This is where the flux density is at aAmin
maximum. can be checked from the datasheet.Amin
HINT:In single ended forward converters the core is magnetised into one polarity only. Iinpush-pull converters the core is magnetised alternating into both polarities.
V1
T/2 T
∆
t
BV1
BB
Page 32
The calculation of the minimum number of turns is equal for these differentN1 min
types of switch mode power supplies.
Calculation of the winding conductors:The diameter of the conductors depends on the RMS-value of the current. The current can becalculated with the power.
For the push-pull converter follows:
I1RMS ≈ Pout
V inand I2RMS = Pout
Vout
+Pout/Vin
-Pout/Vin
I1+Pout/Vout
-Pout/Vout
I2
For the single ended forward converter follows:
I1RMS ≈ 2 Pout
Vinand I2RMS = 2 Pout
VoutPout/Vin
I1
Pout/Vout
I2
The magnetising current can be neglected in this calculation. The current density can bechosen in a range of 2 to 5 A/mm, depending on the termal resistance of the choke. Thecross-section and the diameter can be calculated as follows: Awire dwire
Awire = IS
and dwiret = I ⋅ 4S ⋅ π
where S = 2… 3 …5 Amm2
HINT:If good coupling is important, the primary and secondary winding should be placedon top of each other. Improved coupling is achieved if the windings are interlocked.The coupling is bad in a) good in b) and in c) about four times better than in b).
B
t
B∆B
t
B∆
single ended forward converters push-pull converters
primary
primary
second.
second.
primary
primary
secondary
secondary
secondary
secondary
primary
primary
c)b)a)
Page 33
HINT:The primary number of turns should not be chosen significantly higher than ,N1 min
otherwise the copper losses of the wire would increase needlessly due to the longerconductor.
HINT:For high frequencies and large diameter of the wire the skin effect should beconsidered. For operating frequencies of more than 20kHz and diameters of morethan 1mm litz wire or copper foil should be used.
Page 34
Power-Factor Control
The European standards EN61000-3-2 define limits for the harmonics of the line current.This concerns appliances, which may be sold to public customers and have an input power of (special regulations see EN61000-3-2). Some limit values from this standard are≥ 75 Wgiven in table 6.1. In practice this standard means that for many applications a mains rectifierwith smoothing is not allowed because of the amount of harmonics (see figure 6.1).
input power75 to 600W input power >600W
harmonic-order
n
maximum value ofharmonic current
per Watt (mA/W) / maximum (A)
maximum value ofharmonic current
(A)
3 3,4 / 2,30 2,30
5 1,9 / 1,14 1,14
7 1,0 / 0,77 0,77
9 0,5 / 0,4 0,40
11 0,35 / 0,33 0,33
table 6.1: RMS limits for the harmonics of the line current
To keep the line current approximately sinusoidal, a boost converter can be used (see figure6.2). In this case the boost converter is called Power Factor Pre-regulator (PFC). Incomparison to the boost converter the PFC is controled in a different way: The output voltageis higher than the input voltage as for the boost converter, but the transistor is turned on andoff in a way that a sinusoidal input current is achieved instead of a exact constant outputvoltage. The transistor is driven in such a way, that the inductor current follows theIL(t)shape of the rectified mains . The output voltage of the PFC is controled toV in(t)approximately .Ua ≈ 380 V
Figure 6.1: Usual rectifieing and smoothing of the mains and its mains current
I
mains
UC
LoadVout
U
I
Vout
t
t
m
m
m
m
Page 35
Figure 6.2: boost converter as a power-factor preregulator
Currents, Voltages and Power of the PFC:
Figure 6.3: Currents, voltages and Power of the PFC
For the following calculations, it is assumed that the output power is constant:
Pout = Vout ⋅ Iout = const.
The input current should be controled to a sinusoidal shape and should be in phase with theinput voltage. The input power is now pulsating and can be calculated as follows:
P in(t) = V in ⋅ I in
2⋅ (1 − cos 2ωt)
The input power consists of a DC-part and of an AC-part .P in = = Vin⋅Iin
2 P in ∼ = Vin⋅Iin
2 ⋅ cos 2ωtThe DC-part is equal to the output power , providing a loss-free PFC . Pout
VoutEMI-Filtermains
DL
TCout
Vin
t t
Vout
Vin
Vout
Iin
t
t
t
VinIe
∆Vout
Vout
VinIin
Pin PoutPinPout
Iin
t
∆IL
a) b)
in
Page 36
P in = V in ⋅ I in
2= Vout ⋅ Iout = Pout
In practice an efficiency of about is realistic which means that .η = 95% P in ≈ Pout
0.95
The output capacitor is charged by the pulsating input power and discharged by theCout P in
constant output power . This causes a voltage ripple at , which depends on thePout ∆Vout Cout
value of . For the 230V/50Hz-mains, providing and , Cout Vout = 380 V ∆Vout/Vout = 10% Cout
can be calculated to:
Cout ≈ 0, 5;µFW
The choke determines the high frequency ripple of the input current (figure 6.3b). TheL ∆IL
higher the inductance and the higher the clock frequency , the lower this current ripple. Iff of the peak value of the input current and assuming that the ∆IL = 20% I in
230V/50Hz-mains voltage is a minimum of , it follows that: V in min = 200V
L ≈ 50 ⋅ 103
f ⋅ P in; L (H), f (Hz), P (W)
and for the maximum inductor current:
ILmax = I in max + 12
∆IL = 1, 1 ⋅ 2P in
U in min
Controlling the PFC:Two feedback circuits are required:
One to control the input current to be sinusoidal (input current control)and
one to keep the average output voltage constant, which means keeping it independentfrom the load power consumption (output voltage control)
The input current control loop is lead by the input voltage. In this case the input currentaquires the same shape as the input voltage and consequently the power factor of the mainscurrent will be unity.
The output voltage is controled by comparing it to a constant reference voltage.
The multiplier links the two loops. The output of the multiplier is sinusoidal and itsmagnitude depends on the output voltage control loop. If the output voltage decreases from itsnominal value, the output voltage of the voltage control amplifier increases which causes themagnitude of the multiplier output to increase and consequently the RMS-value of the inputcurrent also increases.
Page 37
Figure 6.4: The control loops of the PFC
The RMS value of the input current is controled by the output voltage control loopwhile the input current control loop drives the input current to be sinusoidal.
Vref
RM
R1
R2
R3
R4
R5
C5C6
PWM
input current
output voltage
R7
C7
R8R9
R10
multiplier
VinL
Cout
IinVout
V2
V'in
control
control
Page 38
Radio Interference Suppression of Switch Mode Power Supplies
Switch mode power supplies generate radio interference due to the high frequency switching.This interference propagates through space by means of the electromagnetic fields or via themains supply in the form of currents and voltages.
The legislation created limits for the levels of interference. These limits are published in theEuropean Standards. Table 7.1 gives some of the most important limits for mobile highfrequency equipment (interference class B). High frequency equipment is that which operatesat a frequency in excess of 9kHz.
measurand frequency range limits standard
electromagnetic interferenceat 10m distance
30 to 230 MHz230 to 1000 MHz
30 dB(µV/m)37 dB(µV/m)
EN55022class B
current harmonics in the mains
0 to 2 kHz see table 6.1(PFC)
EN61000
conducted-mode interferencevoltages at the mains wires inrepect to earth potential
0.15 to 0.5 MHz**
0.5 to 5 MHz
5 to 30 MHz
66 to 56 dB(µV) Q*56 to 46 dB(µV) M*56 dB(µV) Q*46 dB(µV) M*60 dB(µV) Q*50 dB(µV) M*
EN55022class B
* Q: Measured by quasi-peak detector M: Measured by average-detector** Linear decrease to the logarithm of the frequency
Table 7.1: Limits for mobile high frequency equipment class B
Radio interference radiation:High frequency equipment emission radio interference is measured as radio noise fieldstrength (µV/m) . The amount of radio interference radiation depends on the rise time of theswitched currents and voltages and significantly on the layout of the printed circuit board. Tokeep the radio interference radiation low, three principles should be adhered to:
Meshes, in which a switched current flows, should be as small as possible in theirsurrounded area to keep their electromagnetic field low.Nodes whose potential are in respect to earth step up and down with switching, shouldbe as small as possilble in their volumetric space, to keep their parasitic capacitance toearth low. The switch mode power supply should have a metal housing.
HINT:In addition to the reduction of the interference radiation, the first two principles arealso good for keeping the conducted interferences low, which leave the power supplyvia the mains. It should also be noted that a high interference level results ininaccurate switching of the transistors and problems with the closed loop controlcircuit. This often causes audible noise.
Page 39
Mains input conducted-mode interference:Switch mode power supplies take high frequency currents out of the mains. These currentscause a voltage drop at the source impedance of the mains which can be measured at themains terminals. According to the European Standards the interference voltages have to bemeasured between the mains terminals and earth. For this measurement specific radiointerference test equipment is needed which includes a radio interference meter and anartificial mains network. This equipment is required to define a specific mains impedance forcomparable measurments.
We distinguish between three different radio interference voltages (see Fig.7.1):Unsymmetric radio interference voltage: This is the high frequency voltage betweenearth and each mains terminal. Only this voltage is measured corresponding to thestandards. The limits in table 7.1 are valid or this voltage only.Common-mode radio interference voltage (asymmetric radio interference voltage):This is the sum of all unsymmetric interference voltages in respect to earth. Differential-mode radio interference voltage (symmetric radio interferencevoltage): This is the high frequency voltage between the mains terminals.
Figure 7.1: radio interference voltages at the single phase mains
Although the legislation requires only the measurement of the unsymmetric radiointerference voltages, the common-mode and differential-mode interferences are decisive forthe radio interference suppression. The respective suppression of common-mode anddifferential-mode interference needs different designs and components.
Suppression of common-mode radio interference:
Common-mode radio interference voltages at the mains terminals and (for three phaseL1 Nmains ) are common mode voltages in respect to earth potential , whichL1, L2, L3 and N PEmeans they are equal in magnitude and phase. The interference currents , which are drivenI≈∼
by this common-mode voltage, are also common-mode currents. These flow via earth (earthconductor) and back through the parasitic capacitance . is very low. Due to this,Cearth Cearth
the common-mode interference voltage has a very high impedance, which means that thisinterference source acts like a current source. A low-pass filter to suppress the interferencevoltages at the mains terminals must therefore be arranged as in figure 7.2. Looking from theswitch mode power supply the required low pass filter must have a shunt capacitor (Cy) and a
radio interference
(switch mode com.diff.
radio int.
unsym.
L1
N
PE
Mains
~~~voltage
voltage
voltage
radio int.
radio int.
source
power supply )
Page 40
current compensated inductor. Current compensated chokes are wound so that no magneticfield is generated by the operating current (50- or 60Hz), see figure 7.3. Due to this the chokeonly acts against the common-mode interference current and does not effect the operatingcurrent.
Figure 7.2: Suppression of the asymmetric (common-mode) radio interference voltages
The capacitors are called y-capacitors. Y-capacitors have to fulfil special safety requirements,because they would connect the mains phase to ground in the case of a fault. Y-capacitorsmay not exceed a certain capacity to ensure that the permitted maximum earth leakagecurrent is not exceeded. The earth leakage current is a 50Hz-current (or 60Hz in certaincountries). The maximum earth leakage current is 3.5mA (in medical equipment it is amaximum of 0.5mA). According to the standards for the measurement of earth leakagecurrent, terminals and have to be connected and the maximum mains voltage has to beL1 Napplied between and . This means that the y-capacitors are in parallel. For theL1 & N PEEuropean 230V/50Hz-mains it follows that for the maximum y-capacitor:
Cy ≤ 12
× 230 V + 10%2 π 50 Hz × 3.5 mA
≈ 22 nF
Figure 7.3: left: current compensated choke for common-mode interfernces, right: not current compensatedchoke (in this case a ring core double choke with powder core) for differential-mode interferences
Suppression of the differential-mode radio interference:
Differential-mode radio interference voltages are high frequency voltages between the mainsterminals and . To reduce the interference level, a LC-low-pass filter has to be insertedL1 Nbetween the mains conductors (figure 7.4). The differential-mode interference voltageL1, N
radio interference
(switch mode
L1
N
PE
Mains
~~~
I~~~
I~~~
2I~~~CearthCy Cy
current-compensatedchoke
I50Hz
I50Hz
earth leakage current
source
power supply)
I50HzI50Hz
I~~~
I~~~
I~~~
I~~~
I50Hz I50Hz
current compensated choke ring core double choke with powder core
Page 41
results mainly from the pulsed current, which is taken from the switch mode power supplyfrom the mains rectifier smoothing capacitor. Due to the impedance of the smoothingcapacitor a high frequency voltage is generated between und . This is a low impedanceL1 Nwhich means that the interference source acts as a voltage source. Looking from the switchmode power supply the interference filter must be arranged using a series choke followed by ashunt capacitor (see figure 7.4). The choke must not be a compensated choke, becausedifferential-mode interference current and 50Hz-operating current (which is also a differentialtype) cause a mgnetic field within the core (see figure 7.3). To avoid saturation these chokesrequire an air gap. With a ring core the air gap is not visible, because the air gap is achieveddue to the amount of glue used in the iron powder. Open cores are also used. With this typethe magnetic field loop closes through space. Ring cores are prefered because they have a lowmagnetic field outside the core.
Figure 7.4: Suppression of the differential-mode interferences
The capacitors for this purpose are called x-capacitors. They have a lower test-voltage thany-capacitors and are not limited in their value. Foil type capacitors up to are normally1µFused.
HINT: Sometimes the impedance of the differential-mode interference source isapproximately equal to the mains impedance. In that case a π-low-pass filter usingtwo x-capacitors are appropriate (in figure 7.4 dotted lined).
Complete radio interference filter:
Figure 7.5: radio interference filter for common-mode and differential-mode filtering
Figure 7.5 shows a complete radio interference filter. The component values can be founditeratively and with the help of experience. With the radio interference meter only theunsymmetric interference voltages can be measured. Due to this it is not possible todifferentiate between common-mode and differential-mode interference. In practice theoperating frequency and several harmonics are differential-mode interference and all highfrequencies, say above 5MHz, are common-mode. Often a powder core choke is not required.
radio interference
(switch mode
L1
N
mains
~~~
I~~~
I~~~choke
I50Hz
I50HzCx
source
power supply)
radio interference L1
N
PE
mains
~~~
Cy Cy
I50Hz
I50HzCx
housing
Cx
current-comp.choke
powder-corechoke
source(switch modepower supply)
Page 42
Page 43