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I am analyzing ways to make a current limiting circuit, and here I'm having an unexpected problem. The schematic is as follows:

enter image description here

The problem I am encountering is twofold:

  1. Even when node Vin is connected to ground, diode D1 still glows, albeit faintly.
  2. Increasing the V5 generator up to 8V works, and the diode glows brighter linearly. BUT, the moment I go over 8V of input, the diode suddenly drops luminescence by about 80% (almost off) and stays there. The moment I go back below 8V, it turns back on bright, and falls off linearly as I decrease further below 8V.

Another remark: Putting the diode right below the transistor (anode at emitter, cathode negative opamp input) gets rid of problem 2). Problem 1) still remains.

As I understand, increasing voltage too much should make Q2 quickly turn on and off at high speed, regulating and keeping voltage constant, not drop it and make the diode go almost off. The circuit is made to limit current to a value a bit over 20mA. When I go over 8V of input, the current suddenly drops to about 7mA.

VC = 12V

VE= Ground

Opamps are LM358.

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  • \$\begingroup\$ What are VC and VE and how are they connected in this circuit??? Obviously if the LED lights, there is current flow, and if the schematic is correct, it must flow through that transistor. I'd be double-checking my connections, I'd double-check the transistor pinout is what I think it is, and I'd verify it's function out-of-circuit. If the opamp is disconnected, and the transistor base tied to ground, does the LED still light??? \$\endgroup\$
    – Kyle B
    Commented Oct 6, 2020 at 21:31
  • \$\begingroup\$ I believe this is a take on the two-transistor constant current source. merghart.com/p/27/Two-Transistor-laser-current-source Negative feedback would (should) limit the opamp output to just enough current to make it all balance w/o blowing up the transistor (If I'm right...) \$\endgroup\$
    – Kyle B
    Commented Oct 6, 2020 at 21:36
  • \$\begingroup\$ OK - So VE = GND --- Is this a rail-to-rail output opamp??? It may not be able to drive all the way to GND, i.e. it may not be able to turn the transistor off completely. \$\endgroup\$
    – Kyle B
    Commented Oct 6, 2020 at 21:42
  • \$\begingroup\$ Sorry I added that info. It's a LM358, it should be rail to rail. I connected the bjt base to ground, and the diode lights up even more than before! (It even ever so slightly flickers in the rythm of me changing the input votage, as if it was still connected). The circuit is connected properly I checked it several times. \$\endgroup\$
    – math101
    Commented Oct 6, 2020 at 21:47
  • 1
    \$\begingroup\$ LED = light emitting diode. LED diode = light emitting diode diode. \$\endgroup\$
    – JRE
    Commented Oct 6, 2020 at 21:57

1 Answer 1

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There are a couple of things wrong with the circuit.

  1. If you need the input of the circuit to go up to (or beyond 8V) it would be better to divide the input voltage down to say a range of 0-5V.

Currently, when you get to 8 or 9V the emitter of Q1 is forced to the same voltage and there is no voltage to drive the LED. This is probably why the LED is getting dimmer.

Put a resistor divider between op-amp 1 and the input of opamp 2. Say two 4.7k resistors. You will need to reduce the value of R2 by the same amount to say 100 ohms.

  1. As shown you will be overloading the output of the first op-amp by shorting it's output to ground with the transistor Q2. The resistive divider from part 1 will fix that as well as Q2 will only have to pass the current through the resistive divider, 1-2mA.

What is the design range of the input voltage?

To ensure the LED goes off when the input is at zero the op-amps need to be ones that ground is a legitimate value.

Q2 will just limit the current linearly when it starts conducting, it will not oscillate on and off.

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  • \$\begingroup\$ Regarding the voltage at emitter: To make the negative opamp input same as positive (i.e. 8 or 9V) won't the opamp drive the transistor so it conducts about 32mA and so that 32mA*(220+31) equals said 8V? If so won't such a current keep the LED on, even burn it, regardless of emitter voltage? That is, isn't the voltage at negative terminal the product of bjt letting through a high enough current? \$\endgroup\$
    – math101
    Commented Oct 6, 2020 at 22:09

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