Current sensor

I'm struggling to see how the output of this circuit is $$V_{0} = {I_{s} R_{s} R_{L}/{5k}} $$

My work thus far:

I understand the voltage into the base of the BJT is $$I_{s}R_{s}$$ I understand the voltage into the collector is Vin

Now what I am struggling to see is where the 5k term and the R_l term come from.

I see that the IsRs is the voltage across the the two leads of the opamp, this detects the current. Where the 5k and how Rl interacts I don't understand.

My guess would be the 5k is a result of voltage division.

Can someone guide me to understanding this circuit?


1 Answer 1


The 5k term comes from the internal resistor on the positive input to the opamp, through which current to the BJT's collector flows.

The voltage into the base of the BJT is, in fact, not IsRs because of the negative feedback via the BJT. The opamp increases the voltage to the BJT's base until the collector draws enough current from the positive input via the 5k resistor to make the two inputs equal. Since the current into the collector is equal to the current out of the emitter (plus the base's negligible contribution), the same current will flow out of the OUT pin. Rl converts that current to a voltage.

  • \$\begingroup\$ So what is the voltage at the output of the op amp? \$\endgroup\$
    – Adam
    Commented Jun 26, 2015 at 15:36
  • \$\begingroup\$ @Adam It's not terribly important what it is, but it will be one Vbe above the emitter voltage. The feedback loop ensures it's whatever is needed to make that particular transistor conduct the right amount of current, which makes the transistor's exact gain and Vbe irrelevant. \$\endgroup\$ Commented Jun 26, 2015 at 15:51
  • \$\begingroup\$ I guess I'm trying to use my knowledge from my intro circuits class to figure this out math mathematically. The voltage across the inputs of the op amp is Is*Rs correct? \$\endgroup\$
    – Adam
    Commented Jun 26, 2015 at 15:55
  • \$\begingroup\$ The voltage across the inputs is 0, because the Opamp acts to make it so via the negative feedback - in fact, when analyzing Opamps in feedback circuits you can take that as a given to start with. The voltage across the resistor is IsRs. \$\endgroup\$ Commented Jun 26, 2015 at 15:56
  • \$\begingroup\$ okay, so the bjt is turned on when there is a voltage difference across IsRs. So then the current from Vin flows through the 5k resistor to the Out port? \$\endgroup\$
    – Adam
    Commented Jun 26, 2015 at 15:59

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