Leakage Current Activating SSR

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fifty60

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USA
I have a controller that provides a 9V DC output that is connected to an SSR module which is in turn activating an SSR that is controlling a heater. The when my controller is not calling for output, I have 80V across the SSR module output, and 40V across the external SSR controlling the Heater. The SSR controlling the heater is activating, turning the heater on when there is no call for output.

I know I need to add a bleeder resister, but I do not understand what the bleeder resistor is doing. Putting a resistor in parallel will reduce the overall resistance of the load (the load in this case being the SSR directly controlling the heater, and the source will be the output of the SSR module controlling the external SSR). So when this happens, I will be drawing more current from the source, the SSR module?

It makes sense if I look at it like a voltage divider, the reduced resistance from the bleeder resister/SSR input would cause less voltage to be dropped across the SSR input, therefore preventing it to activate when there is no output.

I suppose I do not fully understand why the triacs output of the SSR is not dropping the full 120V, and is dividing in the first place with the external SSR input. Why am i seeing 80V at the SSR module output, and 40V at the external SSR input? What does adding the bleeder resister actually do to restrict the voltage available at the external SSR input?
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
140917-2338 EDT

fifty60:

You need to study circuit theory to a greater extent. and the characteristics of various commerical products.

You also need to respond to the responses to your previous question.

An AC SSR, such as an OAC5, has a Triac or back-to-back SCRs as the power switching element. This type of SSR usually includes a snnuber of some sort across the actual solid state switch. The snnuber is possibly a capacitor in series with a small resistor in parallel with the switch device. Further there is need for power to apply a trigger current to the switching device gate. All of these can provide a shunt path around the actual switch element. This is an impedance that allows leakage current when the SSR is OFF.

Why is one SSR supplying the control voltage to another SSR? In other words why two cascaded SSRs?

.
 
Location
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There are way smarter guys then me here, but I am curious. You have 120v control feeding through the output of one SSR to the input of another. With the following V measurements.
SSR 1. SSR2
L1------------] [---------------I----------L2
^---80v---^ ^--40v--^

^---------------------120V---------------^

Hmmm.

Well shucky darn. Move SSR1 and SSR2 to the right so they are above ] [ and I, respectively.
Move 80v and 40v to the right as well.
 
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fifty60

Senior Member
Location
USA
My controller has 9VDC output voltage. The mini input module accepts the 9VDC signal as an input, but the output of this mini input module is only able to handle 3A. My heater load is 20A. So I am using the mini module to energize the input of an SSR that has 110V input, and an output that has a 30A to control the 230V heaters.

The mini input module uses an opto-isolated input, but I typically still use another SSR in between the controller and anything line voltage above 110V. That is where the 2nd SSR comes in to play. I do not have a single SSR available of the 9VDC input that can handle the 30A at 230V.

The leakage current for the mini input module is 5mA. The typical input current that the SSR would see in the on state is also 5mA. Even though my voltage is well below the minimum on voltage for the SSR, the typical input current is indeed there so the SSR is energizing.

I'm wrestling with how the bleeder resistor actually works. I know how parallel resistance works, and how current behaves through parallel resistors. What I am not sure of is if the input of the SSR should be treated as a pure resistance? Also, should I consider the output of the mini input module and the input of the SSR to be a voltage divider? Will lowering the resistance by adding the parallel resistor lower the voltage because of the voltage divider effect.

Or would a better way to look at it be the 5mA leakage current will always be there. Adding the bleeder will at a maximum take 5mA. A 2KOhm bleeder resistor in parallel with the input of the SSR would then clip the voltage available to the input of the SSR at 10V...
 

GoldDigger

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Retired PV System Designer
Your last way if looking at the problem is probably the most useful one. And it is valid.
Although I would not use the word "clip" for this action, since that implies a non linear effect.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
140918-1442 EDT

fifty60:

Look at the Crydom SMR 24 90 -x with proper heat sinking for direct drive from your 9 V source. Thus, only one SSR.
http://www.crydom.com/en/Products/Catalog/s_mr-6.pdf
This is non-zero crossing I believe. I don't see any specific reference to zero-crossing SSRs. You can call them to find out.

Heat sinking is important at high currents. You can consider paralleling SSRs, if you add a little series resistance to each SSR.

With a solid state relay and a resistive load you should not have much noise fed back into your source. If a problem add a filter.

Your RS485 and RS232 problems are likely mosty common mode ground path problems. For this you need galvanic isolation between input and output. I have a product, I232, that can help in this respect with RS232.

My I232 system provides optical isolation at both ends of the 232 path. Short time peak voltage limit is 2000 V. I have applied a 1000 V RMS 60 Hz sinewave relative to earth (ground) to one of the wires between two isolaion modules with no transmission errors at 115 kbaud. The RS232 connections at the computer and CNC machine are referenced to their own EGCs at the respective machines. These EGCs may be be at a considerabe voltage difference bwtween each other.

RS485 connections may not have isolation, and then are only good for a few volts of common mode voltage. Isolation is needed.

Most Ethernet system now use transformer isolation.

I think you may be too worried about your data lines crossing perpendicular to your power wires. But they should be kept apart.

.
 

petersonra

Senior Member
Location
Northern illinois
Occupation
engineer
Or would a better way to look at it be the 5mA leakage current will always be there. Adding the bleeder will at a maximum take 5mA. A 2KOhm bleeder resistor in parallel with the input of the SSR would then clip the voltage available to the input of the SSR at 10V...

Yes, BUT a 2 kOhm resistor at 115V is almost 7 Watts so you will need a ten Watt resistor.
 

fifty60

Senior Member
Location
USA
The input of the SSR going to be a base of a transistor right? Is the input impedance of a transistor base typically high?
 

GoldDigger

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Location
Placerville, CA, USA
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Retired PV System Designer
The input of the SSR going to be a base of a transistor right? Is the input impedance of a transistor base typically high?

The base input impedance of a transistor is typically quite low (like a forward biased diode). The basic transistor driven at the base is a current amplifier rather than a voltage amplifier.
But the SSR input could be anything from the trigger input of an SCR or triac to the insulated gate of an FET just as well as the base of a transistor. And there would be protective (current and voltage limiting) circuitry in there too.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
140918-1847 EDT

fifty60:

There can be many different inputs to an SSR as there can be many different SSRs. The typical SSR used to switch an AC load controlled by a DC input consists of a Triac or a back-to-back SCR pair. The switch (Triac or whatever) is usually isolated from the input by some type of optical coupler. In its simplest form a typical optical coupler is nothing more than an LED at the input.

A simple OAC5 has an LED at the input. I tested an old OAC5 made by P&B about 1983. This was tested in a connector assembly that includes an external LED also in series. Internally there is a current limiting resistor.

The measurements were:

5.09 V 12.2 mA
4.50 V 9.59 mA
4.00 V 7.06 mA
2.98 V 2.25 mA
2.02 V 0.61 mA

Turn on was at 3.75 V 6.0 mA, and
turn off was at 3.70 V 5.4 mA.

Off AC leakage at 124 V was 1.5 mA.

.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
140918-1919 EDT

fifty60:

Some more information on the OAC5 leakage current.

Using a Powerstat and its voltage scale the leakage current varied as follows:
120 is about 124 V.

120 1.57 mA
60 0.76 mA
30 0.35 mA
15 0.17 mA

Looks fairly linear as would be expected.

From the earliest days of programmable controllers its has been necessary to provide shunt loading on some low current loads because of the SSR leakage current. This can result in a lot of wasted power. Quite often I used a 2000 ohm shunt resistor rated at a minimum of 10 W, but usually 20 W to reduce surface temperature.

Your approach is a bad design and no need for two SSRs cascaded together. You have made no mention that the first SSR is required because it is internally part of some pre-existing equipment.

If you used a first SSR with no internal snubber, then your leakage current would be much smaller.

Following is a very good general paper you should read:
http://www.crydom.com/en/Service/Crydom_TechTraining.pdf

.
 
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