Introduction

1. Introduction Introduction to Solid-State Relays

2. Introduction Basic Structure of a Solid-State Relay 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. Load AC Mains Input Optical Isolation Trigger Circuit Switching Circuit Protection 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.

3. Introduction Basic Structure of a Solid-State Relay 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. 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.

4. Introduction 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). The control system may control a single SSR, or a bank of multiple SSR’s. Control System; PLC, Temp. Controller, Pressure Switch, etc. + SSR AC Line (24-575Vac) - Load 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.

5. 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. Control System; PLC, Temp. Controller, Pressure Switch, etc. 0 volts + OFF SSR AC Line (24-575Vac) - Load 0 volts Control System; PLC, Temp. Controller, Pressure Switch, etc. AC Current Flow 5 volts + SSR ON AC Line (24-575Vac) - Load Line Vac

6. Solid-State Relays Introduction Basic Structure Explained Regulated AC Input Circuit Example: Current regulator maintains constant current level SSR LED Full-Wave Bridge and input capacitors convert the AC input into a DC signal. Input current flows through the input LED embedded in the coupler, which gates on the utput device in the coupler. Series Resistors

7. Solid-State Relays Introduction Basic Structure Explained Unregulated AC Input Circuit Example: No Current Regulator No LED Full-Wave Bridge (some AC input SSR’s might utilize a half-wave bridge, which is essentially a diode, resistor, and capacitor. Series Resistors

8. Solid-State Relays Introduction Basic Structure Explained Regulated DC Input Circuit Example: ( + ) LED Current Regulator Circuit ( - )

9. Solid-State Relays Introduction Basic Structure Explained Unregulated DC Input Circuit Example: No LED No Current Regulator

10. Solid-State Relays Introduction Basic 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.

11. Solid-State Relays Introduction Basic Structure Explained 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 Coupler • 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 • 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 Pros; Pros; Cons; Cons;

12. Transistor coupler output circuit Triac coupler output circuit Solid-State Relays Introduction Basic Structure Explained Optical Isolator and Trigger Circuit:

13. Solid-State Relays Introduction Basic Structure Explained Output & Trigger Circuit: GN SSR Output Circuit Inductor Transient Suppressor Optical Isolator Output Back-to-Back SCR Assembly (mounted directly to the base plate) Current Limiting Resistor

14. A K G Basic Structure Explained 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. + Direction of Current Flow with voltage applied to the gate. - LOAD Trigger Circuit SCR 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.

15. A K G Solid-State Relays Basic Structure Explained Output & Trigger Circuit: 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. 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. A 1) + 2) - 1 2 1 2 Current Flow AC line B 1) - 2) + LOAD The SCR prevents current from flowing when A is negative with respect to B AC Line Since current only flows in one direction through an SCR, the power to the load is effectively reduced by 50%. No gate voltage Gate Signal Voltage applied to gate Load Voltage / Current

16. A G K SCR 1 SCR 2 G K A LOAD Solid-State Relays Basic Structure Explained 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. SCR #1 Terminal #1 Line or Neutral 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 #2 Terminal #2 Line or Neutral SCR #1

17. Solid-State Relays Basic Structure Explained One Operation Cycle: GN SSR Output Circuit 2) Terminal 1 swings ‘positive’ with reference to terminal 2 (gate current) 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. SCR1 SCR2 4) As the AC line crosses zero, SCR1 turns off and the process repeats with SCR2. 1) Input power turns on coupler (load current)

18. Solid-State Relays Basic Structure Explained One Operation Cycle: Voltage across the SSR output terminals during normal conduction of a zero-cross solid-state relay. Line Voltage Trigger Voltage Forward Voltage Drop Voltage across the terminals Switching ‘Window’

19. Solid-State Relays Basic Structure Explained Protection: 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.)

20. SCR 1 SCR 2 A G K G K A Solid-State Relays Basic Structure Explained GN SCR Assembly: Output Terminal Output Terminal Lead-Frames Base plate Anode (A) (trace) Output Terminal Gate Lead (G) Ceramic Insulator SCR Die Cathode Lead (K)

21. Solid-State Relays Basic Structure Explained GN PCB (480Vac) Assembly: Input Terminals Input Current Limiting Transistors LED Input Status Indicator Optical Isolators (2 are used to achieve the 1200V rating) Internal Transient Protection (across each coupler) Snubber Capacitors (no longer used in GN) Gate Connection to B-B Assembly

22. Thermal Properties Thermal Properties of Solid-State Relays

23. Thermal Properties

24. Thermal Properties P=IE 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. To calculate the amount of power (P) being dissipated, multiply the load current (I) by the voltage drop (E). Forward Voltage Drop

25. A K G Thermal Properties P=IE E=IR 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. 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. +24Vdc Vf = 1.1V Current Flow 24 Ohms -VCC 1) Load current = 22.9V / 24 Ohms = 0.95 amps 22.9V Drop (24V - 1.1V) 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

26. A G K SCR 1 SCR 2 G K A 10kW Thermal Properties 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. 240Vac 1) Load current = 10kW / 240Vac = 41.7 amps 2) SCR power dissipation = 41.7 amps x 1.1Vpk = 45.87 Watts

27. Solid-State Relays 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: Two SCR Die (Soldered to copper traces) Copper tracing (fused to substrate) Ceramic substrate Copper tracing (fused to substrate and soldered to base plate) Base plate

28. Solid-State Relays 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. Solder Joints Substrate Base Plate SCR Junction (core) Heat Sink

29. Solid-State Relays 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. 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 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 Temperature Tamb = Ambient Temperature Rs-a = Heat Sink Rating Rjb = Specified SSR Thermal Impedance I x Vf= Power Dissipated by the SSR (as discussed previously in this presentation)

30. Solid-State Relays Thermal Properties Tdie = Tamb + (Rs-a x (I x Vf)) + (Rjb x (I x Vf)) 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))

31. Base Plate Temperature Tamb = 40ºC Solid-State Relays 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. 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 Tbp = Tamb + (Rs-a x (I x Vf)) Tbp= 40ºC + (1.0ºC/W x (50A x 1.1Vf)) Tbp= 95.0ºC Convection Airflow

32. Thermal Heat Sink Compound or Thermal Pad 1.0ºC/W Heat Sink Tamb = 40ºC Solid-State Relays Thermal Properties 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 = Tamb + (Rs-a x (I x Vf)) + (Rjb x (I x Vf)) Tdie = 40ºC + (1.0ºC/W x (50A x 1.1Vpk)) + (0.155ºC/W x (50A x 1.1Vpk)) Tdie = 40ºC + (55ºC)+ (8.525ºC) Tdie = 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 = 109ºC Convection Airflow

33. 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. 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! 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)

34. 1.0ºC/W Heat Sink (Convection) 3” x 3” 40 CFM Fan Thermal Properties Forced Air vs. Convection Cooling 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. Forced Vertical Airflow 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 .7 LFM = (40 / .0625) x .7 LFM = 448 (Derate down to 400LFM)

35. 1.0ºC/W Heat Sink (Convection) 3” x 3” 40 CFM Fan Thermal Properties Forced Air vs. Convection Cooling 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. Forced Vertical Airflow Velocity (LFM)Correction Factor 100 .757 200 .536 300 .439 400 .378 500 .338 600 .309 700 .286 800 .268 900 .252 1,000 .239 (1.0ºC/W x .378) So our heat sink would have a .378ºC/W thermal impedance with 400 LFM of airflow.

36. Thermal Properties Forced Air vs. Convection Cooling 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 + .281X X = 95.7 Amps (Increase of 45.7 Amps) 69.0 = .721X 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.

37. A 1.0ºC/W 3” profile cut to a length of 6” would have a thermal impedance of 0.73ºC/W. Thermal Properties Cut-to-Length Extrusions 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. Extrusion LengthCorrection Factor 1” 1.80 2” 1.25 3” 1.00 4” .87 5” .78 6” .73 7” .67 8” .64 9” .60 10” .58 11” .56 12” .54

38. qp pq ¹ l = h / (m x v) Thermal Properties Useful Formulas To 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) / Power Rs-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) E=h x f 2X + 3Y = Z To Calculate Rjb: Rjb = (Tdie - Tbp) / Power

39. Thermal Properties General Guidelines 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. 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.

40. Additional Questions? Please send requests for additional SSR related training material directly to; Doug Sherman - shermand@us.crouzet.com or Ronnie Haiduk - haidukr@us.crouzet.com