appendix 01
DESCRIPTION
RMS Values of Commonly Observed Converter WaveformsTRANSCRIPT
Fundamentals of Power Electronics 1 Appendix A: RMS Values
Appendix 1RMS Values of Commonly Observed Converter Waveforms
A.1 Results for some common waveformsSee text for results
A.2 General piecewise waveform
How to compute the rms value of a waveform that can be broken intosmaller piecewise segments
Example: transistor current waveform, including effects of shortturn-on current spike
Fundamentals of Power Electronics 2 Appendix A: RMS Values
A.2 General piecewise waveform
(rms value) = 1T v2(t)dt
0
TBasic expression for rms value of waveformv(t) having period T:
i(t)
tTsD1Ts D2Ts D3Ts
Triangular
segm
ent
Constant
segm
ent
Trapezoidal
segm
ent
etc.
Suppose the waveform can berepresented as a series ofsegments, with the kth segmenthaving length DkTs and with Tsequal to the switching period:
rms = DkukΣk = 1
n
Then the rms value can beexpressed as:
where uk is the contribution of the kth
segment — see following slides
Fundamentals of Power Electronics 3 Appendix A: RMS Values
Some basic segment shapes
i(t)
t
I1
i(t)
t
I1
0
i(t)
t
I1 I2
i(t)
t
Ipk
i(t)
ωt
Ipk
θ1
θ2
uk = I 12
uk = 13I 1
2
uk = 13I 1
2 + I1I2 + I 22
uk = 12I pk
2
uk = 12I pk
2 1 –sin θ2 – θ1 cos θ2 + θ1
θ2 – θ1
Constant segment
Triangular segment
Trapezoidal segment
Sinusoidal segment(half or full period)
Sinusoidal segment(partial period)
Fundamentals of Power Electronics 4 Appendix A: RMS Values
Example
Transistor current waveform, including turn-on current spike induced bydiode reverse recovery
The turn-on current spike is short but of high magnitude. Does itsignificantly increase the rms current?
The observed current waveform is approximated by piecewise linearsegments as shown below
i(t)
tTs0.2 µs
I1 = 20 A
0.2 µs
I2 = 2 A
0.1 µs10 µs
0.2 µs5 µs
0 A
1 2 3 4 5 6
Six segments:
1-3 are from diodereverse recovery
4 is transistor on time
5 is transistor turn-offtransition
6 is transistor off time
Fundamentals of Power Electronics 5 Appendix A: RMS Values
Calculation
i(t)
tTs0.2 µs
I1 = 20 A
0.2 µs
I2 = 2 A
0.1 µs10 µs
0.2 µs5 µs
0 A
1 2 3 4 5 6
rms = DkukΣk = 1
6
= 3.76 A
Result:
Without the current spike, the rmsvalue is approximately 1.4 A. So inthis example, the diode reverserecovery significantly increases thetransistor rms current.