Below circuit is the negative rail of a dual tracking power supply.

How do I detect voltage drop on Node1 and pull down Node2 to limit the current without using an op-amp?

Edit: I just want to define a limit that U1 couldn't output more than 200 mA for example, when shorted to ground.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ You should probably have drawn the circuit inverted so that positive voltages are at the top as is the convention. "How do I ... pull down Node2 ...?" Wouldn't you want to pull it high - more positive? Would powering the Node2 op-amp (can you see why we use component designators OA1, R1, etc.) negative supply from NODE1 do the trick? \$\endgroup\$
    – Transistor
    Commented Nov 3, 2020 at 9:35
  • \$\begingroup\$ @Transistor The voltage at Node2 is positive (0-2.5V), I want to pull that to 0V. isn't that pulling down? \$\endgroup\$ Commented Nov 3, 2020 at 9:51
  • \$\begingroup\$ Yes it is, but won't pulling it down make the LM337 (U1 ...) output go even more negative, the opposite of what you want. \$\endgroup\$
    – Transistor
    Commented Nov 3, 2020 at 9:53
  • \$\begingroup\$ @Transistor I know I'm wrong but to my understanding and also on the breadboard, the second op amp inverts the positive voltage to negative(so the output of op-amp will go more positive at 0V), so with 0V at Node2 op-amp will adjust the output of LM337 to 0V. and that's what I want, drop the output voltage to limit the current. \$\endgroup\$ Commented Nov 3, 2020 at 9:57
  • \$\begingroup\$ @Transistor It has to go more positive at 0V because of the LM337 -1.25V reference voltage. \$\endgroup\$ Commented Nov 3, 2020 at 10:07

1 Answer 1


enter image description here

Figure 1. Original circuit.


simulate this circuit – Schematic created using CircuitLab

Figure 2. Simplified circuit. OA2 has been removed because it is only a unity gain buffer so it has no effect in this analysis.

  • OA1 is configured for negative feedback. If all goes well it will settle down with NODE2 very close to 0 V (which has been applied to the non-inverting input. With the R5/R4 ratio applied the U1 output voltage should be \$ V_- = -\frac {R_4}{R_5} V_+ = -\frac {10k}{1k} V_+ = -10V_+ \$.

See if that helps your understanding before going back to the original problem.

  • \$\begingroup\$ I thought when it's just negative rail there's no need to switch it upside down, thanks for that. about the buffer; because the voltage at positive rail can go up to ~30V and adding big resistors as divider was making the circuit susceptible to noise, I thought adding the buffer stage will reduce the current draw when pulling it down to ground and also I will not have the noise problem anymore. \$\endgroup\$ Commented Nov 3, 2020 at 10:49
  • \$\begingroup\$ O and I made another mistake, the positive side of divider is connected to positive regulator voltage reference, I edited the original question as well. \$\endgroup\$ Commented Nov 3, 2020 at 10:56
  • \$\begingroup\$ Please see the edit to the schematic, I made several mistake when drawing the schematic and corrected them now, sorry for that... \$\endgroup\$ Commented Nov 3, 2020 at 11:07
  • \$\begingroup\$ That doesn't really change anything. You are feeding 0 to 2.5 V to the OA2 buffer so that's the same as feeding 0 to 2.5 V to the top of my Figure 2, R5. My answer still stands. \$\endgroup\$
    – Transistor
    Commented Nov 3, 2020 at 12:16
  • \$\begingroup\$ Why in your equation the LM337 -1.25V reference voltage doesn't matter? I mean that's right that the input voltage should be 0V which is 0V, but the output of OA1 depends on the R5 input voltage. which at 0V OA1 output has to go positive in order to keep two inputs at 0V. is that correct? \$\endgroup\$ Commented Nov 3, 2020 at 12:23

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