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I'm modifying a current-limiting circuit to work with a transistor instead of a switch. This circuit amplifies a voltage across a sensing resistor via a differential amplifier op-amp configuration. This signal is then compared to a reference created by a voltage divider. If the signal is higher than the reference the output of the second op-amp goes positive. This voltage is felt by a transistor that then allows current to be drawn away from the main circuit (thus turning it off). However, once the circuit is turned off, the second op-amp is no longer positive and current is permitted to flow, resulting in a flickering.

Here's the part I'm not sure about

To prevent this, I'm inserting a second transistor which feels a current at the base if the circuit is over-current. when this happens, current will flow through this transistor into the base of itself and the other transistor. Since Vcc is on both the normally open and normally closed terminals of the relay, current will always flow through this transistor and the switch will remain off until the main power supply is turned off.

Will this work or will the delay between the switch of base voltage cause the strategy to fail? Are there any other issues I'm missing?

Here's a close-up of the area in question followed by the full circuit. (Vcc=24V)

T2 is the transistor I'm questioning and T3 is the transistor that needs a base voltage to pull power away from the relay coil. enter image description here enter image description here

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  • \$\begingroup\$ Can you just replace T3 with a thyristor? Once those start conduct, they do not stop conducting even if the base signal is removed. They only stop conducting if something else brings the current to zero while the base signal is removed. A 2N5060, for example. Your scheme will not work because no base-emitter current will flow in T2. \$\endgroup\$ – DKNguyen Apr 23 '19 at 20:52
  • \$\begingroup\$ Yea I guess so! All I need is the thyristor to trigger at the same time as T3 and a base current will always be applied to T3. This is the simplest solution so I'll probably go with it. \$\endgroup\$ – noneqyou Apr 23 '19 at 21:06
  • \$\begingroup\$ How's that? T1 is normally conducting so I would need it off once I have the current sensing circuit go positive. The only option would then be T3. So, when T3 is activated it keeps current away from the base of T1 \$\endgroup\$ – noneqyou Apr 23 '19 at 21:30
  • \$\begingroup\$ Ah it seems we were writing that at the same time. T3 is the one that pulls current away. \$\endgroup\$ – noneqyou Apr 23 '19 at 21:31
  • \$\begingroup\$ Ah, I see what's happening now. T3 redirects the base current away from T1 to turn it off, and since T1 drives the coil that turns the relay off. When I looked back I misinterpreted that you were directly rerouting the current away from the load. T3 is the one you want to be a thyristor. I was right the first time. \$\endgroup\$ – DKNguyen Apr 23 '19 at 21:44
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No, you can't do that. A BJT can't feed current from its own emitter back to its base.

If you want a latching effect, you need to create a bistable circuit with two transistors.

schematic

simulate this circuit – Schematic created using CircuitLab

Q2 and Q3 form a bistable multivibrator. C1 makes sure that the circuit powers up with Q2 off and Q3 and Q1 on. If the output of the current sense circuit ever goes high, it turns on Q2, which then turns off Q1 and Q3. Q3 being off makes sure that Q2 stays on until the power is interrupted.

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  • \$\begingroup\$ Whoa this is a really neat circuit! Thanks for posting it! I'm going to just use a thyristor as per the comment by Toor about, however, a thyristor is essentially a bistable circuit with two transistors so I'll just go ahead and accept your answer. \$\endgroup\$ – noneqyou Apr 23 '19 at 21:10
  • \$\begingroup\$ Actually I may hedge on that thyristor. Can you elaborate on C1? Is this to account for any spikes when turning on? \$\endgroup\$ – noneqyou Apr 23 '19 at 21:19
  • \$\begingroup\$ C1 holds the base of Q2 low long enough for Q3 to turn on during power-up. Otherwise, it's a crap shoot as to which one "wins". \$\endgroup\$ – Dave Tweed Apr 23 '19 at 21:36
  • \$\begingroup\$ Assuming there's no overcurrent on startup this would not be an issue though right? \$\endgroup\$ – noneqyou Apr 23 '19 at 21:38
  • \$\begingroup\$ No, it definitely IS an issue, regardless of overcurrent. \$\endgroup\$ – Dave Tweed Apr 23 '19 at 21:38

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