Optocoupler TLP-291 - Output voltage not dropping down to zero

Thread Starter

arjun sureshbabu

Joined Apr 17, 2017
7
Hello All,

I want to use the TLP-291 optocoupler (OC) as a switch and also isolate the input and output. The output of the OC will activate the Solid state relay (SSR) as shown in the attached circuit.

Problem:
1. If the input of the OC is low i.e 0 V , the collector voltage is 24 V that gets divided through RL and the internal resistance of the SSR. Hence by voltage divider rule, the voltage across the SSR input is around 13.5V.
2. But if the input of the OC is high i.e.5V, the voltage across the input of the SSR should drop down to zero but it is now. When I measure, the voltage is around 4 to 5V.

I don't understand the 2nd point mentioned above. Can any one help me out?
 

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JohnInTX

Joined Jun 26, 2012
4,601
You need more input current. You have about 8ma input. A rough minimum CTR of 100% gives you 8ma out. 8ma through 1K only drops 8 volts. Since you get around 4v, it means that the CTR is better than minimum but still not enough to get the 24ma needed to saturate the output transistor with a 1K collector resistor. Rough calculations, it is late.

Good luck!
 

Thread Starter

arjun sureshbabu

Joined Apr 17, 2017
7
Thanks for the response John..

So, if I reduce the input resistance to 230 ohms, the input current increases to 16mA. This should increase the drop across the 1K resistor.

Now regarding the 1st point,
1. If the input of the OC is low i.e 0 V , the collector voltage is 24 V that gets divided through RL and the internal resistance of the SSR. Hence by voltage divider rule, the voltage across the SSR input is around 13.5V.

If the input of the OC is low i.e. OV, I need at least 20V to the input of the SSR. To get this, I should reduce the RL to less than 400 ohms. If this is the case, how will the OC behave if its input is high?
 

JohnInTX

Joined Jun 26, 2012
4,601
Thanks for the response John..

So, if I reduce the input resistance to 230 ohms, the input current increases to 16mA. This should increase the drop across the 1K resistor.

Now regarding the 1st point,
1. If the input of the OC is low i.e 0 V , the collector voltage is 24 V that gets divided through RL and the internal resistance of the SSR. Hence by voltage divider rule, the voltage across the SSR input is around 13.5V.

If the input of the OC is low i.e. OV, I need at least 20V to the input of the SSR. To get this, I should reduce the RL to less than 400 ohms. If this is the case, how will the OC behave if its input is high?
Well, if you want 20V at the SSR with R=400 ohms then you have 10ma to drive the relay input - that works. But then if you want to pull the relay input down to zero you have to drop 24V across 400ohms = 60ma. That exceeds the absolute max collector current of the opto (as well as the LED current). There are a couple of ways to go here
1) review the SSR specs here. For a minimum drive level, you'll see that >16V turns it on and you only have to drop it to <10v to turn it off. If that is true, then you can try setting RL to provide the minimum voltage into the 3.4K (24V / 7ma) input (although it is probably not completely linear). Then see if the opto can handle it. A very rough example that does not take into account device tolerances / variances would be:
a) Relay turns on at 16V. With a calculated input R of 3.4K that yields a current of 4.7ma. At that current, RL drops the other 8V so it needs to be 8/.0047 =1700 ohms.
b) With RL = 1700 ohms, we now want to drop the SSR input voltage to 10V (its specified OFF point). That means dropping 14V across RL so 14/1700 ~=8.3ma collector current in the opto. If the CTR is 100% (and who knows what it is, they vary) you need 8.3ma in the LED input. Vf is 1.4V max so (not counting driver drops) the input resistor is (5-1.4)V / .0083A ~=434ohms.

Note that this is only an example. To make such a scheme actually work in practice you'll have to ensure that tolerances are taken into account: voltages, resistances, opto etc, and make sure that the relay actually works like I think it does from the spec. You can build a bit more drive into the opto since it has an absolute max rating of 50ma input and output. You can increase the relay drive by dropping RL but it must limit to the max collector current of the opto. At a safe 40ma, that gets RL ~= 600 ohms and 20V at the relay when the opto is off. At 100% CTR you need 40ma into the LED so that is 90 ohms input resistor.

Your real problem is a lack of gain in your switch (the opto). If this were my circuit, I would use an NPN transistor to do the switching. Connect the SSR like (+24)--SSR-- NPN collector -- NPN emitter to ground. NPN base to the collector of the opto. Now RL can be much smaller (a couple of hundred times smaller depending on the beta of the transistor). A 2n2222, 2N3904 or any garden variety NPN would do. You could also use a 2N7700 FET. Then RL can be huge. The bigger RL is the lower the LED input current is so that's your choice. The BIG payoff is that you no longer have to worry about variances in the relay, minimum turn on/off points, CTR etc. The extra gain from the transistor will bang the relay hard enough to eliminate all of the worry from running things close to their tolerances/limits with much lower control currents. What's not to like about that?

That's about what you are looking at.

As for the opto itself, presumably you are trying to isolate the 24V supply from the 5V supply.. otherwise its not needed since the SSR has an isolated input.

Good luck!
 
Last edited:

crutschow

Joined Mar 14, 2008
28,204
Why do you need so much current into the SSR input?
Can't you just increase the value of the 1k resistor to reduce the required input current?
 

Thread Starter

arjun sureshbabu

Joined Apr 17, 2017
7
Well, if you want 20V at the SSR with R=400 ohms then you have 10ma to drive the relay input - that works. But then if you want to pull the relay input down to zero you have to drop 24V across 400ohms = 60ma. That exceeds the absolute max collector current of the opto (as well as the LED current). There are a couple of ways to go here
1) review the SSR specs here. For a minimum drive level, you'll see that >16V turns it on and you only have to drop it to <10v to turn it off. If that is true, then you can try setting RL to provide the minimum voltage into the 3.4K (24V / 7ma) input (although it is probably not completely linear). Then see if the opto can handle it. A very rough example that does not take into account device tolerances / variances would be:
a) Relay turns on at 16V. With a calculated input R of 3.4K that yields a current of 4.7ma. At that current, RL drops the other 8V so it needs to be 8/.0047 =1700 ohms.
b) With RL = 1700 ohms, we now want to drop the SSR input voltage to 10V (its specified OFF point). That means dropping 14V across RL so 14/1700 ~=8.3ma collector current in the opto. If the CTR is 100% (and who knows what it is, they vary) you need 8.3ma in the LED input. Vf is 1.4V max so (not counting driver drops) the input resistor is (5-1.4)V / .0083A ~=434ohms.

Note that this is only an example. To make such a scheme actually work in practice you'll have to ensure that tolerances are taken into account: voltages, resistances, opto etc, and make sure that the relay actually works like I think it does from the spec. You can build a bit more drive into the opto since it has an absolute max rating of 50ma input and output. You can increase the relay drive by dropping RL but it must limit to the max collector current of the opto. At a safe 40ma, that gets RL ~= 600 ohms and 20V at the relay when the opto is off. At 100% CTR you need 40ma into the LED so that is 90 ohms input resistor.

Your real problem is a lack of gain in your switch (the opto). If this were my circuit, I would use an NPN transistor to do the switching. Connect the SSR like (+24)--SSR-- NPN collector -- NPN emitter to ground. NPN base to the collector of the opto. Now RL can be much smaller (a couple of hundred times smaller depending on the beta of the transistor). A 2n2222, 2N3904 or any garden variety NPN would do. You could also use a 2N7700 FET. Then RL can be huge. The bigger RL is the lower the LED input current is so that's your choice. The BIG payoff is that you no longer have to worry about variances in the relay, minimum turn on/off points, CTR etc. The extra gain from the transistor will bang the relay hard enough to eliminate all of the worry from running things close to their tolerances/limits with much lower control currents. What's not to like about that?

That's about what you are looking at.

As for the opto itself, presumably you are trying to isolate the 24V supply from the 5V supply.. otherwise its not needed since the SSR has an isolated input.

Good luck!

Thanks a lot for the detailed explanation John.. It was really helpful.. I will try out as you said..
 

crutschow

Joined Mar 14, 2008
28,204
No I cannot.. because the SSR input has 3.4K resistance, the voltage gets divided across the input of SSR.
How much input voltage/current does the SSR need?

You would be better off inverting the signal and connecting the opto output transistor in series with the SSR input so that when the opto output is on, then the SSR will also be on.
That way the opto output current would be the same as the SSR input current.
 

Thread Starter

arjun sureshbabu

Joined Apr 17, 2017
7
How much input voltage/current does the SSR need?

You would be better off inverting the signal and connecting the opto output transistor in series with the SSR input so that when the opto output is on, then the SSR will also be on.
That way the opto output current would be the same as the SSR input current.
Yeah.. I understand.. thanks for the response. But I need inverted output signal.
 
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