Transcript No Slide Title

Introduction
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Introduction
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.
Basic Structure of a Solid-State Relay
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.
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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.
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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.
+
AC Line (24-575Vac)
SSR
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.
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Motor / Pump, Heating
Element, Solenoid Valve,
Transformer, Ballast, etc.
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.
OFF
Control System;
PLC, Temp.
Controller,
Pressure Switch,
etc.
0 volts +
AC Line (24-575Vac)
SSR
Load
0 volts
ON
Control System;
PLC, Temp.
Controller,
Pressure Switch,
etc.
AC Current Flow
5 volts +
AC Line (24-575Vac)
SSR
Load
Line Vac
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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
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
Introduction
Basic
Structure Explained
Regulated DC Input Circuit Example:
(+)
LED
Current Regulator Circuit
(-)
Introduction
Basic
Structure Explained
Unregulated DC Input Circuit Example:
No LED
No Current Regulator
Basic Structure Explained
Introduction
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.
Basic Structure Explained
Introduction
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;
Pros;
Cons;
Triac Driver
•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
Pros;
Cons;
Transistor Coupler
•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
Basic Structure Explained
Introduction
Optical Isolator and Trigger Circuit:
Triac coupler output circuit
Transistor coupler output circuit
Basic Structure Explained
Introduction
Output & Trigger Circuit:
GN SSR Output Circuit
Inductor
Optical Isolator
Output
Transient
Suppressor
Back-to-Back SCR
Assembly (mounted directly
to the base plate)
Current Limiting
Resistor
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.
+
A
Direction of Current Flow with
voltage applied to the gate.
Trigger
Circuit
G
K
LOAD
-
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.
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Basic Structure Explained
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.
Output & Trigger Circuit:
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) -
A
Current Flow
G
K
LOAD
1 2
1 2
AC line
B 1) - 2) +
The SCR prevents current from flowing when A is negative
with respect to B
AC Line
No gate
voltage
Voltage applied to gate
Gate Signal
Since current only flows in one direction through an
SCR, the power to the load is effectively reduced by
50%.
Load Voltage / Current
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
SCR #2
A
G
K
SCR 1
G
SCR 2
K
A
Terminal #1
Line or Neutral
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
LOAD
SCR #1
Terminal #2
Line or Neutral
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
1) Input power
turns on coupler
4) As the AC line crosses zero, SCR1 turns off and
the process repeats with SCR2.
(load current)
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’
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.)
Basic Structure Explained
GN SCR Assembly:
Output Terminal
Output Terminal Lead-Frames
A
G
K
SCR 1
G
SCR 2
K
A
Base plate
Anode (A)
(trace)
Output Terminal
Gate Lead (G)
Ceramic Insulator
SCR Die
Cathode Lead (K)
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)
Gate Connection to
B-B Assembly
Snubber Capacitors
(no longer used in GN)
Thermal Properties
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Thermal Properties
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 forwardvoltage 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
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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.
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
A
Vf = 1.1V
G
Current Flow
K
24 Ohms
1) Load current = 22.9V / 24 Ohms = 0.95 amps
2) SCR power dissipation = 1.1V x 0.95 amps = 1.05 Watts
22.9V Drop
(24V - 1.1V)
2) Load power dissipation = 22.9V x 0.95 amps = 21.76 Watts
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-VCC
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
A
SCR 1
G
1) Load current = 10kW / 240Vac = 41.7 amps
G
K
SCR 2
K
A
2) SCR power dissipation
= 41.7 amps x 1.1Vpk = 45.87 Watts
10kW
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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
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
SCR Junction
(core)
Base Plate
Heat Sink
Thermal Properties
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
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))
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)
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.
(Rjb x (I x Vf))
Tamb + (Rs-a 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.
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
Base Plate Temperature
Tbp = Tamb + (Rs-a x (I x Vf))
Tbp= 40ºC + (1.0ºC/W x (50A x 1.1Vf))
Tbp= 95.0ºC
Tamb = 40ºC
Convection Airflow
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
Thermal Heat Sink
Compound or Thermal Pad
1.0ºC/W Heat Sink
Tamb = 40ºC
Convection Airflow
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)
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.
1.0ºC/W Heat Sink
(Convection)
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)
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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.
Velocity (LFM)
100
200
300
400
500
600
700
800
900
1,000
Correction Factor
.757
.536
.439
.378
.338
.309
.286
.268
.252
.239
(1.0ºC/W x .378)
So our heat sink
would have a
.378ºC/W thermal
impedance with 400
LFM of airflow.
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Forced Vertical Airflow
1.0ºC/W Heat Sink
(Convection)
3” x 3” 40 CFM Fan
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.
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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.
A 1.0ºC/W 3” profile cut to a length of 6”
would have a thermal impedance of Extrusion Length
0.73ºC/W.
1”
2”
3”
4”
5”
6”
7”
8”
9”
10”
11”
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Correction Factor
1.80
1.25
1.00
.87
.78
.73
.67
.64
.60
.58
.56
.54
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)
To Calculate Rjb:
Rjb = (Tdie - Tbp) / Power
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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.
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Please send requests for additional SSR related training material directly to;
Doug Sherman - [email protected]
or Ronnie Haiduk - [email protected]
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