Monitor high-side current without an external supply

Vijay Damle, Digitronics, Pune, India -March 01, 2001

Typical high-side current-sensing circuits require a dc source that is 2.5 to 13V greater than the V+ high-bus voltage (Figure 1). Generating this supply is painful in many situations. For example, in power supplies for TV transmitters, the main SMPS (switch-mode power supply) output supplies the power amplifier, and a series switching regulator steps down the main SMPS output to drive the exciter. The system must remotely display the currents of both of these supply outputs, with 0 to 50A corresponding to 0 to 5V referred to sense V-. Because of the presence of a series switch, the V- lines of both outputs are common. Thus, you cannot use shunts in the V- line and amplify. Shunts are necessary on the positive bus of both the outputs. The main output supplies 30 to 45V at 30A, and the exciter supply outputs 22 to 26V at 10A. You need costly Hall-effect sensors to achieve the proportional output, though isolation is not required.

Figure 1 Typical high-side current-sensing circuits require a dc source that is 2.5 to 13V greater than the V+ high-bus voltage; generating this supply can be difficult.

An alternative approach for this application takes advantage of low-offset op-amp characteristics to design a circuit that works with a wide voltage range and needs no other supply. The V+ and inverting and noninverting terminals of the OP07 op amp need a minimum of approximately 2 to 2.5V to function properly. Thus, you can pull the op amp's inputs by more than 2.5V below the positive-supply connection and tie the op amp's V+ pin to shunt V+ (Figure 2).

Figure 2 A modified current-monitoring circuit pulls the op-amp inputs below the positive supply voltage.

In the circuit, IC2 with R10 and R11 generate a 15V output. The R3 and R6 pair and R5 and R8 pair form dividers such that the op amp's inverting and noninverting inputs are approximately 3V less than the V+ supply of the op amp. You can use R7 and R9 to trim the offset to avoid the need for potentiometers. Op amp IC1 and Q1 generate a current that is proportional to the shunt voltage. R12 generates a voltage that is proportional to the drop across shunt R4. R1 trims the gain.

If you use this circuit at less than 25V, then you can delete IC2, R10, and R11. You should also ground IC1's V- pin by shorting R2, and you can replace R2 with a constant-current source to reduce the power due to bus-voltage variation (Figure 3).

Figure 3 At voltages less than 25V, you can replace R2 with a constant-current source.

This circuit was tested for 0 to +55°C, and it maintained proportional output within ±1% for a bus-voltage variation of 25 to 45V over this temperature range.

This approach has many advantages. An external supply is unnecessary. The circuit is suitable for bus voltages of 5 to 60V with component changes. Other circuits have limitations due to op-amp absolute-maximum voltage ratings. The circuit acts as the minimum load that SMPS outputs normally require, which eliminates or reduces high-wattage resistance across the output. You can easily scale the circuit for different proportional outputs. You can add a buffer amplifier to reduce the output impedance, and the buffer can derive its supply across R2, which increases its operating supply range by 15V or more. One limitation is that, in the case of a short circuit, the current-proportional output drops to zero.

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