simulate this circuit – Schematic created using CircuitLab

Can somebody please explain me how a transistor mirror circuit works and how can it be used as over current detector?

I badly need a crystal clear explanation for the above circuit. Anybody can please help with this.


1 Answer 1


Note: this answer was to a previous version of the question, and is not applicable to the schematic above.

A basic BJT current mirror has two transistors, which are designed to be as close to identical as possible.

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The bases and emitters are tied together, so the base-emitter voltages are equal. Since we assume the transistors themselves are identical, equal Vbe means equal current into the base. And equal current into the base means equal current from collector to emitter. Thus by setting the current through one, by the selection of voltage and resistance, you also set the current through the other.

One way of doing an overcurrent detector would be to put a second resistor in the mirrored leg, and measure the voltage across that resistor. If it rises above some fixed level, the current must have exceeded your setpoint. I'm not immediately sure why that would be better than measuring the voltage on the primary leg, though. At first glance, this seems like a pretty round-about way of doing an overcurrent limit.

  • \$\begingroup\$ Thanks Remiel. I have a supply and the load between which this circuit is connected to monitor the safe current that is sourced by the supply. So I need to use this circuit to detect the over current drawn from the supply. So is there anyway that i can do with this? \$\endgroup\$
    – IamDp
    Commented Mar 20, 2013 at 17:02
  • \$\begingroup\$ Yes, you could put a resistor between the secondary source and the secondary transistor, and measure the voltage across that resistor. But again, a shunt resistor in the primary would be simpler and should have nearly identical effect. \$\endgroup\$ Commented Mar 20, 2013 at 18:21
  • 2
    \$\begingroup\$ Current mirrors are often used in overcurrent protection because it allows a shift of the voltage reference from Vin to Gnd. Use a high-side current mirror (PNP transistors instead of NPN) and put the "secondary load" between the collector of the secondary transistor and ground. Now you have a ground-referenced voltage that's proportional to the high-side load current. Note that the value of the resistor can be varied to determine the "gain" of the circuit. \$\endgroup\$
    – Dave Tweed
    Commented Mar 20, 2013 at 19:33

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