Circuit senses high-side current

Bob Bell and Jim Hill, On Semiconductor, Phoenix, AZ -- EDN, 3/1/2001

The accurate, high-side, current-sense circuit in Figure 1 does not use a dedicated, isolated supply voltage, as some schemes do. Only the selected transistors limit the common-mode range. The circuit measures the voltage across a small current-sense resistor, R_S. The operation of the circuit revolves around the high-side current mirror comprising Q₁ and Q₂. All the circuit components have one overall function: to make the collector currents equal in Q₁and Q₂. The additional current mirror using Q₃ sets the values of the collector currents. The collector current is (V_CC-0.7)/(R₅+R₆)

100 µA. You can best calculate the gain of the circuit by analyzing the loop formed by R₁, R_S, R₂, Q_1B (emitter base), and Q_1A (base emitter). In Figure 1, the currents are I_S, the high-side measurement current; I₁ and I₂, the mirror currents of Q_1A and Q_1B; and I₃, a branch current from the emitter of Q_1A.

When you sum the currents around the loop, (I_S·R_S) + (I₂·R₂)+VQ_1B(e-b)-((I₁+I₃)·R₁)=0. Because I₁=I₂, R₁=R₂, and the emitter-base voltages are equal, I₃=I_S·R_S/R₁. Looking at the remaining circuitry, the op amp keeps the transistors' collector currents equal by controlling I₃ through Q₄. Therefore, the overall transfer function is V_OUT=I_S·R_S·R_G/R₁. For R_G=1 k?, the transfer function is V_OUT=0.5·I_S. The circuit can operate over a common-mode input range of approximately 10V to several hundred volts, limited by the selected transistors.