introduction introduction basic structure of a solid-state relay inputoptical isolation trigger...

www.crouzet-usa.com

Introduction

www.crouzet-usa.com

Introduction

Basic Structure of a Solid-State Relay

Input Optical Isolation

Trigger Circuit

Switching Circuit

Protection

Load

AC Mains

Solid-state relays are available in a variety of packages with various mounting styles. Rating may vary from a few amps to 150Arms, at voltages up to 1600Vpeak.

The basic structure of a SSR is relatively constant between the different manufacturers. Only minor differences differentiate the various solid-state relays available on the market.

www.crouzet-usa.com

Introduction

Input Circuit; Commonly referred to as the ‘primary’, the input of a SSR may consist of a simple resistor in series with the optical-isolator, or of a more complex circuit with current regulation, reverse polarity protection, EMC filtering, etc. In either case, they both serve the same basic function, which is to sense the application of a control signal and to ‘tell’ the SSR that it must turn on.

Basic Structure of a Solid-State Relay

Optical Isolation; The optical isolator in a SSR provides isolation between the input circuitry / control system, and the output circuit connected to the AC mains. The type of optical isolator used in the relay may also determine whether it is a zero-crossing or random-fire output.

Trigger Circuit; This processes the input signal and switches the output state of the SSR. The trigger circuit may be internal or external to the optical-isolator.

Switching Circuit; This is the portion of the SSR that switches the power to the load. It usually consists of a transistor, SCR, or FET in a DC application, or a triac, alternistor, or back-to-back SCR in an AC application.

Protection Circuit; Many applications require some form of electrical protection to prevent the SSR from being damaged in the application, or from mis-firing due to environmental conditions. The protective device(s) may be incorporated into the design of an SSR, or mounted external to the relay.

www.crouzet-usa.com

So in general, a solid-state relay (SSR) is simply an electronic component that serves as an interface and provides electrical isolation between a control circuit (usually at low-voltage) and a power circuit (usually with high power ratings).

Control System; PLC, Temp. Controller,

Pressure Switch, etc.

SSR+

-

Load

AC Line (24-575Vac)

The control system may also supply AC voltage to the SSR. In that case, an AC input SSR is required, which does not have a polarity sensitive input.

Motor / Pump, Heating Element, Solenoid Valve, Transformer, Ballast, etc.

Introduction

The control system may control a single SSR, or a bank of multiple SSR’s.

www.crouzet-usa.com



SSR+

-

Load

AC Line (24-575Vac)

Introduction

In the off-state (0 volts on the input), the SSR prevents load current from flowing through the load.In the on-state (specified voltage on the input), the SSR allows load current to flow through the load.

0 volts

0 volts



SSR+

-

Load

AC Line (24-575Vac)5 volts

Line Vac

OFF

ONAC Current Flow

Introduction

SSR LED

Current regulator maintains constant current level

Full-Wave Bridge and input capacitors convert the AC input into a DC signal.

Regulated AC Input Circuit Example:Basic Structure Explained

Series Resistors

Input current flows through the input LED embedded in the coupler, which gates on the utput device in the coupler.

Introduction

No LED

Full-Wave Bridge (some AC input SSR’s might utilize a half-wave bridge, which is essentially a diode, resistor, and capacitor.

Unregulated AC Input Circuit Example:

Series Resistors

No Current Regulator

Basic Structure Explained

Introduction

LED

Regulated DC Input Circuit Example:

Current Regulator Circuit


( + )

( - )

Introduction

No LED

Unregulated DC Input Circuit Example:

No Current Regulator


IntroductionBasic Structure Explained

Optical Isolator and Trigger Circuit:

The input circuit supplies voltage to the LED internal to the optical isolator.

The trigger circuit is gated on by the light emitted from the internal LED. With no signal on the input, current will not flow through the output.

Isolation is provided by the clearance between the LED and the trigger circuit. Typical isolation strength is 4,000+Vrms.


Optical Isolator and Trigger Circuit:The two most common optical isolators (couplers) used in a solid-state relay are the triac driver (left), and the transistor coupler (right). The triac driver is used in our GN series SSR’s, while many of our competitors still use the transistor couplers. There are advantages and disadvantages to each;

Triac Driver Transistor CouplerPros; •Trigger circuit built into

coupler•Less immune to

noise/transients

•Slightly more expensive than

transistor couplers•Requires more input current

to gate on the output

Cons;

Pros; •Slightly less expensive than

transistor couplers•Requires less input current to

gate on the output

•External Trigger circuit

required (more components)•Less immune to

noise/transients

Cons;


Optical Isolator and Trigger Circuit:

Triac coupler output circuit Transistor coupler output circuit


Output & Trigger Circuit:

Optical Isolator Output

Inductor

Transient Suppressor

Current Limiting Resistor

Back-to-Back SCR Assembly (mounted directly to the base plate)

GN SSR Output Circuit

www.crouzet-usa.com

A

KG

Output & Trigger Circuit:Commonly referred to as the ‘secondary’ of the relay, the output is the part of the relay that directly powers the load. This is done via the trigger circuit and the power device installed in the particular relay. The most common power device used in Crouzet’s relays is the SCR (Silicon Controlled Rectifier).The SCR consists of an anode (A), a cathode (K), and a gate (G). Current flows from the anode to the cathode when voltage is applied to the gate of the SCR. When there is no voltage on the gate, the SCR prevents current from flowing through the load.

SCR

+

-LOAD

Direction of Current Flow with voltage applied to the gate.

Trigger Circuit

If a DC load is connected to an SCR, current will continue to flow even after the trigger circuit removes voltage from the gate. In this case, the anode or cathode must be disconnected to shut off the load.


Output & Trigger Circuit:

A

KG

Since current can only flow from the anode to the cathode (direction of the arrow), an AC load will ‘half-wave’ when controlled by an SCR.

LOAD

AC line

1) + 2) -

1) - 2) +

The SCR prevents current from flowing when A is negative with respect to B

A

B

1 2 1 2Current Flow

No gate voltage Voltage applied to gate


AC Line

Load Voltage / Current

Gate SignalSince current only flows in one direction through an SCR, the power to the load is effectively reduced by 50%.

Single SCR controllers are quite common in household applications, such as speed-controllers for ceiling fans or light dimmers. They are also extremely common in industrial applications where they may serve a similar function.

A

A

K

KG

G

SCR 1 SCR 2

LOAD

Output & Trigger Circuit:In order for current to flow through the load in both the positive and negative swing of the AC sine-wave, two SCR’s are needed in the output. These are connect in inverse-parallel (“back-to-back”), with the cathode of one being tied to the anode of the second, and visa-versa.


Line or Neutral

Line or Neutral

Terminal #1

Terminal #2

SCR #2

SCR #2

The gate of each SCR is tied together via the optical isolator(s). When the coupler turns on, the SCR’s conduct current in both directions.

SCR #1

SCR #1

One Operation Cycle:Basic Structure Explained

1) Input power turns on coupler

2) Terminal 1 swings ‘positive’ with reference to terminal 2

3) As the potential increases on terminal 1, a small amount of current flows through the cathode-gate junction of SCR2, through the coupler, and into the gate off SCR1. When enough current is flowing into the gate of SCR1 (100ma to 400ma), the SCR turns on and energizes the load for the remainder of the half-cycle.

4) As the AC line crosses zero, SCR1 turns off and the process repeats with SCR2.

(gate current)

(load current)

SCR2

SCR1

GN SSR Output Circuit


One Operation Cycle:Voltage across the SSR output terminals during normal conduction of a zero-cross solid-state relay.

Trigger Voltage

Switching ‘Window’

Forward Voltage Drop

Voltage across the terminals

Line Voltage

Protection:Basic Structure Explained

Protection for a solid-state relay may take many forms and serve many purposes.

RC network across output (dv/dt attenuation)

Inductor(dv/dt attenuation)

Internal transient suppressor (protects against transients that might exceed the rating of the coupler(s) and/or SCR die.)

GN SCR Assembly:Basic Structure Explained

SCR Die

Cathode Lead (K)

Gate Lead (G)

A

A

K

KG

G

SCR 1 SCR 2

Anode (A)(trace)

Ceramic Insulator

Base plate

Output Terminal Lead-Frames

Output Terminal

Output Terminal

GN PCB (480Vac) Assembly:Basic Structure Explained

Gate Connection to B-B Assembly

Snubber Capacitors (no longer used in GN)

Internal Transient Protection (across each coupler)

Optical Isolators (2 are used to achieve the 1200V rating)

Input Current Limiting Transistors

LED Input Status Indicator

Input Terminals

www.crouzet-usa.com

Thermal Properties

Thermal Properties

www.crouzet-usa.com

Thermal Properties

All solid-state relays dissipate power in the form of heat. The amount of power dissipated is a product of the load current and the forward-voltage drop of the power device in the output.

Forward Voltage Drop

To calculate the amount of power (P) being dissipated, multiply the load current (I) by the voltage drop (E).

www.crouzet-usa.com

A

KG24 Ohms

Current Flow

Thermal Properties

To calculate the load current, we divide the voltage across the load by the impedance of the load (I=E/R). As stated in the previous slide, once we have this information, we can calculate the power dissipated by the SCR in the circuit.

+24Vdc

-VCC

Vf = 1.1V

The typical voltage drop of an SCR in a GN solid-state relay is 1.1Vpk. In the example below, we can calculate the amount of power being dissipated in a simple SCR control circuit.

22.9V Drop (24V - 1.1V)

1) Load current = 22.9V / 24 Ohms = 0.95 amps

2) SCR power dissipation = 1.1V x 0.95 amps = 1.05 Watts

2) Load power dissipation = 22.9V x 0.95 amps = 21.76 Watts

www.crouzet-usa.com

In many applications the impedance of the load is not known in Ohms. Heating elements are usually rated in Watts, which is the amount of power that the element, not the SSR, will dissipate. To calculate the load current in such an application, you simply divide the wattage of the load by the line voltage.

Thermal Properties

A

A

K

KG

G

SCR 1 SCR 2

10kW

240Vac

1) Load current = 10kW / 240Vac = 41.7 amps

2) SCR power dissipation = 41.7 amps x 1.1Vpk = 45.87 Watts

Thermal Properties

Once the total power dissipation of the relay is known, we can use provided specifications to calculate the actual SCR die temperature of the relay in the application. However, to better understand the concept, we must first be familiar with the thermal characteristics of a solid-state relay.

Typical Back-to-Back SCR Assembly:

Base plate

Copper tracing(fused to substrate and soldered to base plate)

Ceramic substrate

Copper tracing(fused to substrate)

Two SCR Die(Soldered to copper traces)

Thermal Properties

The heat generated in the SCR die propagates through each component in the SCR assembly. This is not a ‘pure’ process, as each component, including the solder between the components, has an impedance based on it’s construction and material type. Thus, each device in the system retains some level of heat based on it’s total thermal impedance.

SCR Junction(core)

Solder Joints

Substrate Base Plate

Heat Sink

Thermal Properties

How well a SSR transfers heat from it’s SCR die to it’s base plate, or it’s “thermal efficiency”, is determined by it’s Rjb rating. Rjb simply means; thermal

impedance (R) between the SCR junction (j) and the SSR base plate (b). This is measured in degrees Celsius per Watt of power being dissipated by the SCR die.

Tdie = Tamb + (Rs-a x (I x Vf)) + (Rjb x (I x Vf))Tdie = Tamb + (Rs-a x (I x Vf)) + (Rjb x (I x Vf))

Using this formula we can calculate the temperature of the SCR die in an SSR for a specific application (within a reasonable level of certainty).

Tdie = SCR TemperatureTamb = Ambient TemperatureRs-a = Heat Sink Rating

Rjb = Specified SSR Thermal Impedance

I x Vf = Power Dissipated by the SSR(as discussed previously in this presentation)

Always corroborate your calculations by evaluating the relay within the application. Minor fluctuations in the variables can result in a significant difference between calculated and actual results

Thermal Properties


If we take the formula in steps, we can calculate the temperature of the two “critical” points of the assembly. The first half of the formula, “Tamb + (Rs-a x (I x Vf))”, gives the actual temperature of the base plate of the SSR. The second half of the formula, “(Rjb x (I x Vf))”, gives the temperature differential between the base plate and the SCR die.

Tamb + (Rs-a x (I x Vf))(Rjb x (I x Vf))

Thermal Properties

The thermal impedance of the heat sink determines how much the temperature will vary between ambient and the base plate, relative to how much power the SSR is dissipating. Heat sink efficiency is also measured in degrees Celsius per Watt of power, but will change depending upon the ambient temperature and the availability of forced airflow. For applications where the device is to be cooled through convection airflow, the heat sink must be mounted in a manner that will allow air flow to move up through the fins.

To estimate the base plate temperature of the SSR, simply multiply the heat sink impedance by the total power being dissipated, then add the sum to ambient.

Base Plate Temperature

Convection AirflowTamb = 40ºC

bp = Tamb + (Rs-a x (I x Vf))bp= 40ºC + (1.0ºC/W x (50A x 1.1Vf))bp= 95.0ºC

Assume the SSR is carrying 50 amps of load current and has a forward voltage drop of 1.1Vpk. The thermal impedance of the heat sink is 1.0ºC/W and Tamb is 40ºC

Thermal Properties


Tamb = 40ºCConvection Airflow

1.0ºC/W Heat Sink

Thermal Heat Sink Compound or Thermal Pad

To continue, let’s assume that the relay is a 100A GN SSR with a Rjb rating of 0.155ºC/W and a 1.1Vf. As before, it is mounted to a 1.0ºC/W heat sink and powering a 50A load in a 40ºC ambient.

Tdie = 40ºC + (1.0ºC/W x (50A x 1.1Vpk)) + (0.155ºC/W x (50A x 1.1Vpk))Tdie = 40ºC + (1.0ºC/W x (50A x 1.1Vpk)) + (0.155ºC/W x (50A x 1.1Vpk))

Tdie = 40ºC + (55ºC)+ (8.525ºC)Tdie = 40ºC + (55ºC)+ (8.525ºC)

Tdie = 103.5ºCTdie = 103.5ºC

To ensure accuracy, we should also add the thermal pad, which typically adds 0.1ºC/W to the assembly.Tdie = 103.5ºC + (55W x 0.1ºC/W)Tdie = 103.5ºC + (55W x 0.1ºC/W)

Tdie = 109ºCTdie = 109ºC

Now that we have determined that the SCR die should operate at less then their maximum rated temperature, a quick thermal analysis of the assembly must be performed to verify the accuracy of the calculation. This is important since any deviation in one or more of the variables will lead to significant differences between the calculated and actual die temperature.

Since it is not always feasible to attach a thermocouple to the die of an SSR, temperatures can be measured at the base plate of the SSR to verify the accuracy of the estimate. Unfortunately, this method still leaves a level of uncertainty in the analysis since we must calculate the differential between the die and the base plate. However, as this calculation has the least impact in total temperature rise, and given the accuracy of measured power over estimated power dissipation, the end result will be fairly accurate.

Thermocouple inserted into a groove milled in the top of the heat sink. The groove should be slightly larger then the diameter of the TC to allow the SSR to mount flush with the heat sink.

Tdie = actual base plate + (specified Rjb x actual power)

To guarantee reliability, never let the base plate exceed 100ºC and allow the SSR to stabilize for a few hours before taking the final measurement!

www.crouzet-usa.com

Forced Air vs. Convection CoolingThermal Properties

Calculating the thermal impedance of a heat sink with forced air is a little more difficult since there are a few more variables and intangibles involved. There is, however, a simple formula that can give an estimate of the thermal impedance, which can then be verified through evaluations.

For simplicity, let’s assume that there is minimal obstruction to the airflow and that the “open” area in the heat sink is roughly the same size as the area of the fan.

First, we must convert the CFM rating to linear feet per minute (LFM);LFM = (CFM / (area / 144)) x 70% (70% derate for back pressure)LFM = (40 / ((3” x 3”) / 144)) x .7LFM = (40 / .0625) x .7LFM = 448 (Derate down to 400LFM)

Forced Vertical Airflow

3” x 3” 40 CFM Fan

1.0ºC/W Heat Sink(Convection)

www.crouzet-usa.com


Forced Vertical Airflow

3” x 3” 40 CFM Fan

1.0ºC/W Heat Sink(Convection)

Once the approximate LFM rating is known (400LFM), a correction factor can be applied to the heat sink to determine the thermal impedance with airflow.Velocity (LFM) Correction Factor100 .757200 .536300 .439400 .378500 .338600 .309700 .286800 .268900 .2521,000 .239

So our heat sink would have a .378ºC/W thermal impedance with 400 LFM of airflow.

(1.0ºC/W x .378)

www.crouzet-usa.com


To demonstrate the increase in the efficiency of the heat sink provided by the 40CFM fan, we can calculate how much more current (I > 50 Amps) the SSR would have to carry in order to obtain the same 109.0ºC die temperature as before. To ensure adequate derating, we will round the impedance of the heat sink up to 0.4ºC/W.

109.0ºC = Tamb + (Rs-a x (Vf x I)) + (Rjb x (Vf x I))109.0ºC = 40ºC + (.4ºC/W x 1.1X) + (.255ºC/W x 1.1X)109.0ºC = 40 + .44X + .281X69.0 = .721X X = 95.7 Amps (Increase of 45.7 Amps)

Always evaluate an assembly that is to be cooled by forced air before the customer begins using the assembly in their production. Forced air cooling systems are tricky at best and SSR failure may result from an inadequate understanding of the systems parameters. Installing assemblies in the customers equipment for thermal testing is the best way to ensure overall reliability.

www.crouzet-usa.com

A standard heat sink profile listed in an extruder’s catalog will typically have the thermal impedance specified when cut to a length of three inches. A rough determination of the impedance of an extrusion profile cut in different lengths can be obtained with a correction factor. Multiplying the Rs-a/3” by the correction factor for the desired extrusion length will give the thermal impedance of that profile when cut to that length. This is a valuable tool when designing prototype assemblies, but the correction factor will vary slightly for each profile due to various spacing and lengths of the fins.

Thermal Properties

Cut-to-Length Extrusions

Extrusion Length Correction Factor1” 1.802” 1.253” 1.004” .875” .786” .737” .678” .649” .6010” .5811” .5612” .54

A 1.0ºC/W 3” profile cut to a length of 6” would have a thermal impedance of 0.73ºC/W.

www.crouzet-usa.com

Thermal Properties

Useful FormulasTo Calculate SCR Temperature: Tdie = Tamb + (Rs-a x Power) + (Rjb x Power)Tdie = Tbp + (Rjb x Power)

To Calculate Heat Sink Thermal Impedance: Rs-a = (Tbp - Tamb) / PowerRs-a = ((Tdie - (Rjb x Power)) - Tamb) / Power

To Calculate Minimum Required Heat Sink Thermal Impedance: Rs-a-min = (Tdie-max -Tamb - (Rjb x Power)) / Power

To Calculate Maximum Allowable Current Given Heat Sink Impedance: Imax = (Tdie-max - Tamb) / ((Rs-a x Vf) + (Rjb x Vf))

To Calculate Base Plate Temperature: Tbp = Tdie - (Rjb x Power)Tbp = Tamb + (Rs-a x Power)

To Calculate Rjb: Rjb = (Tdie - Tbp) / Power

www.crouzet-usa.com

Always verify any calculation. Catastrophic field failures may occur if the actual value of a variable shifts even a slight amount from the estimate. Evaluate every new assembly in an environment as close as possible to the actual application or in the actual equipment for which it is intended.

Thermal Properties

General Guidelines

Round up every number in your calculations and use maximum value specifications whenever possible. If the test data yields results that are better than originally estimated, and target pricing is maintained, then everyone wins.

The ambient temperature given by the customer may be misleading. The room temperature outside of the panel, or panel temperature when the system is not running, will not help much when calculating minimum heat sink requirements. Ensure that the temperature of the air moving through the fins of the heat sink is the value that is used in the estimates.

If something bad can possibly happen, assume that it will. Loss of airflow, 100% duty cycle operation, heavy surge currents, and excessive ambient temperatures, are just examples of anomalies that may occur in any application. If the customer has experienced them before, then he will most certainly experience them again.

www.crouzet-usa.com

Please send requests for additional SSR related training material directly to;Doug Sherman - [email protected] Ronnie Haiduk - [email protected]

introduction introduction basic structure of a solid-state relay inputoptical isolation trigger...

Documents