I have a measuring bridge and want to connect it to an ADC with an AD8227 instrumentation amplifier (Datasheet). For the reference voltage, I have connected a voltage divider to an LP358 as an impedance converter (Datasheet).

How can I now calculate the current or power consumption of the op amp and instrumentation amplifier?

Below is the schematic of my circuit.


simulate this circuit – Schematic created using CircuitLab

I have found a short calculation for an op amp here, but I don't know if I have to use the output and supply current mentioned in the Datasheet or does it only show the maximum possible current provided by the op amp? Or does it depend what is connected to the LP358 output, in my case the reference pin of the AD8227? Does the gain resistor of the AD8227 (not in the schematic) have an impact on the current consumption?

  • \$\begingroup\$ If you are that concerned about power in Op Amp, why use a BJT OA? When your bridge is drawing 5mW which could be reduced to 1mW by lowering the differential voltage. \$\endgroup\$ Aug 25, 2020 at 12:53

2 Answers 2


To determine the power consumption of your LP358 based voltage buffer (I would not call that an impedance converter) you need to know two things:

  • the current consumption of the opamp itself, when there is no current flowing into or out of the output. You can find this number in table 6.5: \$I_{CC}\$ Supply current which is typically 54 uA. This is for a supply voltage of \$V_{CC}\$ = 5 V.

  • the current that flows out of the opamp's output. If there is a current flowing out of the output, then the opamp is supplying current to another circuit and that current needs to come from somewhere, it comes from the supply so this current adds to the current consumption of the opamp itself (the 54 uA).

In your circuit, the output current of the LP358 flows into the reference voltage input of the AD8227 so we'll have to look in the datasheet of the AD8227 what is connected there and how much current it takes. Table 2 shows what we need: REFERENCE INPUT, \$I_{IN}\$ has a typical value of 12 uA.

Now simply add both: 54 uA + 12 uA = 66 uA. Note that these are typical values so the actual value can be somewhat lower or higher. Depending on the circumstances the LP358 can consume up to 125 uA (max value in the datasheet).

Does the gain resistor of the AD8227 have an impact on the current consumption?

The AD8227's datasheet only mentions figures for a Gain = 5. My guess is that the AD8227's current consumption might be affected by the Gain you choose. I highly doubt that the current into the reference voltage input depends on the Gain though. To be sure you need to measure it.


The reference current depends on whether the reference voltage is lower or higher than the voltage U (RG). If both the input voltage and the reference are 2.5V, the current direction is from the AD8227 to the opamp. (U (RG) is the input voltage + one pnp B-E voltage: Uin + 0.64V.) ad1

When the reference voltage is greater than U (RG), the current direction is reversed and goes from the opamp to the AD8227. Its value depends on the reference voltage. This consumption is definitely negligible. ad2

Consumption is independent of Gain resistance. The largest consumer may be the load on the output of the AD8227. ad3

The deviation of the reference voltage from 2.5V is the same as the offset voltage of the LP358 (2mV). The voltage of the unconnected reference pin is the same as the voltage U (RG) /3.14V/. The resistance up to U (RG) is 60k.


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