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I can't find it anywhere. I know it's not a darlington pair. I tried looking at the 'common emitter' configuration but the explanations of that all talk about a single transistor.

a circuit

The switch on the left is actually the collector of an optocoupler and I'm trying to switch the load (the 50 ohm resistor and the LED) so that the load is connected when the optocoupler LED is on.

I should have used a PNP and switched the load from the high side but I didn't have a suitable transistor at hand so I came up with this. I did build this circuit and it works, but I don't know whether I just got lucky.

Do you think there is anything wrong with this circuit? I don't know much about the internal workings of transistors, so I might be missing something very important.

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  • \$\begingroup\$ There is much more to say here, but since you've already accepted a answer, that would be pointless. \$\endgroup\$ Sep 5, 2015 at 11:58
  • \$\begingroup\$ Well the first answerer did answer the exact question, the name of the circuit, which enabled me to look it up on the internet. Others provided very helpful information. I'm just learning about transistors, so any input you'd care to give is much appreciated. I don't think it's pointless, 50 people have looked at this page since I asked the question, so there is some point to discussion. \$\endgroup\$
    – dodo
    Sep 5, 2015 at 13:08
  • \$\begingroup\$ I can't tell if you're being deliberately obtuse or just missing the point, which is you should wait a while to select the answer to accept. Accepting a answer says I got what I wanted, I'm done. I and others often skip over question with accepted answers, so you don't know what other people might have said once you accept. For example, I could have said a lot more than the single answer that was here and that you accepted when I first saw this question. \$\endgroup\$ Sep 5, 2015 at 13:53

3 Answers 3

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This is a cascade amplifier circuit.

enter image description here

https://en.wikipedia.org/wiki/Cascade_amplifier

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  • \$\begingroup\$ Thank you very much. My circuit doesn't have resistors between the emitters and the ground. What's their purpose? Will I get away fine with the current circuit for switching a simple digital signal? \$\endgroup\$
    – dodo
    Sep 5, 2015 at 9:51
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    \$\begingroup\$ Their purpose is to control the DC voltages and gain. As you're only interested in "on" and "off" choosing a value of 0 ohms for them (as you did!) is perfectly appropriate. \$\endgroup\$
    – user16324
    Sep 5, 2015 at 9:59
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Here is no name for this transistor configuration. It simply boosts your current driving capabilities for the 50Ohm load and inverts the logic level from your optocoupler. While this configuration might work, there is a more efficient way of controlling the load. In your circuit, when the optocoupler is not conducting, the second transistor always dissipates energy.

Optocoupler Amplifier

In this picture, you use the current driving capabilities of the optocoupler directly. Depending on the CTR (Current Transfer Ratio) of the optocoupler and your driving current of the optocoupler LED, you need a resistor (R2) to limit the current through Q1. The resistor has to be small enough to drive Q1 into saturation.

Short example:

A transistor goes into saturation when both the base-emitter and base-collector junctions are forward biased e.g when the collector voltage drops below the threshold of the base-collector voltage (about 0.4V-0.6V). The saturation voltage (collector-emitter-voltage) of a common NPN transistor like BC546 is about 0.2V. The 50Ohm load will in this case be driven with a current of (5V-0.2V)/50Ohm =96mA. The BC546 has a DC-Gain of minimal 100. From here we can determine the needed base-current to (96mA / 100) = 960uA. So you need at least 1mA base-current to drive Q1 into saturation. The normal Current Transfer Ratio of most optocoupler lies in the range of 30% to 200%. If we assume a CTR of 50% then you need to drive the LED with about 2mA to get Q1 into saturation. R1 is only for protection, so you can increase it's value depending on the collector-emitter-voltage of the optocoupler to limit the maximum current.

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  • \$\begingroup\$ I was concerned about the same thing, with the first transistor always conducting. So, this configuration is actually a darlington pair, yes? I will try to make this in the second board I'm making, thanks. I can't connect a too small resistor to the base of Q1 because the opto can take 50 ma max in the output side. 100 ohm seems to be the minimum in my case. \$\endgroup\$
    – dodo
    Sep 5, 2015 at 10:47
  • \$\begingroup\$ Yes, this configuration is actually a darlington pair.The base current depends not only on the resistor value but also on the Current Transfer Ratio (CTR) of the optocoupler. I updated my answer with some additional explanations. \$\endgroup\$
    – Martin
    Sep 5, 2015 at 11:56
  • \$\begingroup\$ Now I understand the purpose of R2 much better, thank you. In my circuit, the CTR is about 60% and I'm driving the opto LED with 5 ma. This gives me a 3 ma sinking capability on the opto output side. So, as I understand it, the internal resistance of the optocoupler already limits the current to 3ma but R2 is there to slightly increase the resistance to allow for variations due to temperature, manufacturing etc. is this correct? \$\endgroup\$
    – dodo
    Sep 5, 2015 at 13:31
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    \$\begingroup\$ Overall good answer, +1. However, your configuration is not a darlington since the collectors aren't tied together. The OP is getting too hung up on names. Also R2 is rather low. It would be better to not count on the CTR of the optocoupler to limit the base current. 1 kOhm would be better. That limits the base current to about 4 mA even when the opto is saturated, without detracting from the case where the current is limited by the CTR and whatever the LED is being driven with. \$\endgroup\$ Sep 5, 2015 at 13:49
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    \$\begingroup\$ Yes, exactly. R2 is used as a safety measure for variations temperature, tolerances and also in the event of short-circuits. As @OlinLathrop suggests, a value of 1k should be sufficient even in the case of full saturation of the optocoupler. \$\endgroup\$
    – Martin
    Sep 5, 2015 at 14:04
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It's one inverter (common emitter amplifier, but not biased into linear operation) connected to another.

The first transistor provides no amplification, only logic inversion.

The purpose of the circuit appears to be to switch the 5V to whatever that circle represents (with 50 ohms in series)- it should turn on when the switch is closed. It will probably do that well with common transistor types. You could also just put the switch in series with the load and save the 1K resistors and transistors.

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  • \$\begingroup\$ Using the switch in series was what I wanted to do at first, but my optocoupler can't take that much current, it's a 4n25. Sorry for the lack of detail. \$\endgroup\$
    – dodo
    Sep 5, 2015 at 10:08
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    \$\begingroup\$ In that case it's important to note that the opto must be well saturated (<< 0.6V) at 5mA to reliably turn the transistor off. You might want to increase the left 1K. \$\endgroup\$ Sep 5, 2015 at 10:11
  • \$\begingroup\$ Thank you, I had only used optos for simpler things and never realized the CTR would be important for digital signals. I am driving the opto LED with 5ma, so 1.7K would be the correct resistance for the first base? (CTR=0.6 for 5ma LED current according to the datasheet) \$\endgroup\$
    – dodo
    Sep 5, 2015 at 10:30
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    \$\begingroup\$ 1.7K might be the minimum that will work, probably a bit more that 1.7K since CTR is not specified in deep saturation (usually with several volts across the transistor). Maybe use 20K. \$\endgroup\$ Sep 5, 2015 at 10:34

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