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In our company, we have developed our own ampere-meter that transfers the measured data to a PC via USB. An employee who no longer works for us created this. The circuit board of the measuring device also has a circuit for voltage measurement, but the firmware was never developed for it. Now it's up to me to write the firmware for it.

My question is if and how voltage can be measured with this circuit. Vinput and Voutput are the two inputs for voltage measurement (like on a multimeter the "COM" and "Voltage Input" jacks).

(The current measuring circuit works independently, so you don't have to consider it)

This is the voltage measurement circuit:

Circuit

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    \$\begingroup\$ This is just averaging Vinput and Voutput. \$\endgroup\$
    – SteveSh
    Feb 20, 2023 at 13:54
  • \$\begingroup\$ Strange naming: "Vinput and Voutput are the two inputs for voltage measurement.". Where are these nodes really connected to? Please provide the schematic. \$\endgroup\$
    – devnull
    Feb 20, 2023 at 14:01
  • \$\begingroup\$ @devnull They are connected to a screw terminal. You connect these to the load you want to measure. Initially, the device was thought for Current measuring, that's where the names come from \$\endgroup\$ Feb 20, 2023 at 14:19
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    \$\begingroup\$ @Dario.Casciato what does "these" mean? Are you saying that you connect both inputs across the terminals of a (floating) load? \$\endgroup\$
    – Designalog
    Feb 20, 2023 at 14:27
  • \$\begingroup\$ @Dario.Casciato. The voltage source and the load (assuming you need to measure the voltage at this load) must be included in the schematic for this question to be answerable. \$\endgroup\$
    – devnull
    Feb 20, 2023 at 14:39

2 Answers 2

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The basic function is a non-inverting voltage summer.

The output voltage is simply given by (for DC, after the R37 & R6 voltage divider):

$$ V_{out} = \frac{R_{37}}{R_{37}+R_6}\left(V_{input}\left(\frac{R_5}{R_5 + R_{33}}\right) + V_{output}\left(\frac{R_{33}}{R_{33}+R_5}\right)\right) $$

The 220ohm resistor has been added, most likely, to isolate the input and output capacitances of the op-amp (@ V(-) or node 4 and node 1).

R37 and R6 form a 0.5V/V voltage divider.

If we fill in the numbers, the output voltage is, then:

$$ V_{out} = 0.25(V_{input} + V_{output}) $$

So, can it measure voltage? Yes it can. Basically, scales both input signals by 1/4x.

Why would you want to do this? I'll need more details from your system to tell you.

EDIT:

Since you say that both inputs are akin to a V and COM DMM inputs, then, this circuit will simply sense the average of both inputs \$ = 0.5(V_{input} + V_{output})\$, and it will further scale this expression by 0.5, assuming both inputs are connected to a well-defined voltage (even if it's 0V, but not floating).

That being said, its measurement will not be the same as a multimeter, as this circuit cannot measure the voltage difference between 2 points as a typical voltage measurement mode of a DMM. The best description I could give of it is that of a common-mode voltage measurer/sensing point; measures the average voltage between 2 node voltages*.

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  • \$\begingroup\$ OP said this is for an ampere (amp) meter. Maybe Vinput and Voutput are just the two voltages across a current sense resistor? \$\endgroup\$
    – SteveSh
    Feb 20, 2023 at 14:21
  • \$\begingroup\$ @SteveSh Waiting for the full picture from the OP. Meanwhile, I think the transfer function I derived is correct. \$\endgroup\$
    – Designalog
    Feb 20, 2023 at 14:26
  • \$\begingroup\$ Yes, I wasn't taking issue with what your presented. \$\endgroup\$
    – SteveSh
    Feb 20, 2023 at 14:28
  • \$\begingroup\$ @SteveSh actually, I'm more confused now. The OP question is if the circuit can measure voltage, but he also added that it was thought to be part of an anmeter system. I guess we need more details to know what was the original designer's intention... \$\endgroup\$
    – Designalog
    Feb 20, 2023 at 14:31
  • \$\begingroup\$ @SteveSh perhaps he meant that it's not an ammeter anymore and it's only meant to measure voltage... somehow \$\endgroup\$
    – Designalog
    Feb 20, 2023 at 14:33
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It matters a great deal what Vinput and Voutput are connected to. If the source (or whatever is connected) has much internal impedance then the behavior will be different.

For example, if "Voutput" is left open, the ADC will see Vinput/2. If it is grounded the ADC will see Vinput/4.

There's also a bit of low pass filtering, cutoff frequency is ~6.8 kHz

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