circuit senses high-side current

The original current sensor circuit is here

Was that circuit an ED Ideas for Design, or similar? Ref, please?

The high-side circuit is a two-stage current mirror, fed from a low- side current source. Any voltage drop across the current-sensing shunt resistor unbalances the mirror by a current Ix = Is Rs/R1.

The low-side circuit is a servo that senses the difference between the current source and the mirror's output, and adds a current Ix through the MOSFET to rebalance the mirror. This current is the aforementioned Ix = Is Rs/R1, and is sensed by an output resistor Rg, so the output voltage Vo = Is Rs Rg/R1 shows the current Is.

The MOSFET (which could have been a BJT instead) is operated at very low currents, and should therefore be a small-die part (such as the suggested 2n7000, not a large-die part like an IRF150), to keep the capacitance down.

Contrary to the drawing, the opamp does not need to be a rail-rail opamp. Its inputs should work to the Vee rail, e.g., like an LM324.

Hi Mr. Hill, thanks for u r reply. Yes, exactly it is from ED design ideas. In that the operation is not explained in detail. If u could expalin more in detail it would be appreciated. RAF

I'm sorry, RAF, I thought I had done so, e.g., Vo = Is Rs Rg/R1, etc. Could you tell us exactly what issue of ED's Design Ideas it came from? Thanks!

Hi , I am searching for a current sensor for motor application (high side) which can withstand high common mode voltages( 100V). The Circuit in the ED's design ideas is meant for avoiding high common mode voltages which the current mirrors do. So i need detailed operation of each component so that i can proceed further. Thanks and Regards RAF

In the ED'd design on March1,2001, EDN123 was published.