Hello all,

I have an inverter stage that I need to monitor its output voltage, a friend suggested the circuit below but I didn't understand how it works and why he chooses these values in particular.

What circuit do the opamps form? What would the output look like?
What is the purpose of the resistors and capacitors (red circle)? is it just for the filtering! or for opamp input protection since the output voltage is supposed to be high?


Thanks,

 

They form voltage followers (non-inverting unity gain amplifiers).

They have a high input impedance and a low output impedance.

 

If they were unity gain the 10Ω resistors would not be in the feedback path. They are attenuators. The attenuation factor for the last amplifier in the chain looks like:

\( \cfrac{10\Omega}{1k\Omega||1k\Omega}\;=\;0.02 \)

 

Start at the input. The input signal is divided by about 300.
First amp has a gain of 1. (buffer) So the signal will not be loaded down by the next stage.
The two 1K resistors add 1.25V and divide by 2.
Then the signal is buffered again.
It looks like this circuit drives an ADC that can measure 0 to 2.5V. (just a guess)

 

Start at the input. The input signal is divided by about 300.
First amp has a gain of 1. (buffer) So the signal will not be loaded down by the next stage.

I didn't think about buffers because of the resistor on the feedback loop! it through me away! how is it a unity buffer when it has resistors?

It looks like this circuit drives an ADC that can measure 0 to 2.5V. (just a guess)

The signal will be fed to an isolated amplifier and then to another opamp stage (similar to the one above in the green circle) and then to an ADC through a cable.

 

I didn't think about buffers because of the resistor on the feedback loop!

The resistor has no function. (as to gain) There is no current in the resistor. (well maybe nA which is almost zero)
You need two resistors. Do the gain formula with one resistor = open. Gain=1.

fed to an isolated amplifier

The iso-amplifier probably also works at 0 to 2.5V.

 

The iso-amplifier probably also works at 0 to 2.5V.

Do you think the iso-amplifier is needed? Is there a possibility for the high voltage to get to the ADC if i decided to not use the iso-amplifier ?

 

When the signal is a high voltage, shown as up to 1200 volts, and the measuring device is an A/D input, possibly on a processor, adding opamps is a cheap protection scheme in case there is a fault in the input attenuator, which does seem to be designed to pass spikes as well as DC. And certainly those amplifiers provide some gain.

 

Do you think the iso-amplifier is needed? Is there a possibility for the high voltage to get to the ADC if i decided to not use the iso-amplifier ?

It's possible that the high voltage common is isolated (at a potential difference) from the ADC signal common. If this is the case, the iso-amplifier or some other means of galvanic isolation is needed.

 

When dealing with the power line: You will never know how some fool connected the wires.

In your isolated frontend, measurements are made from "power line" to "ground". (which might be from ground to power) It does not matter what voltage the microcomputer is sitting on.

If you do not have isolation, your main microcomputer is sitting on some voltage. (maybe ground or neutral) If the wiring is wrong then the micro might be at power line. Or maybe you want to measure Power to Neutral but you are measuring Power to Ground. (small error)

Isolation is safe. If a human can touch the computer safe is better. If the computer talks to another computer or the internet you need to be safe. If the computer is totally in a plastic box with no outside connection, then isolation is not important.

 

Sorry for reviving this thread but I have some updates and some concerns..
I simulated the circuit with LTspice and you can see the graphs and circuit in the attached images.
I gave an 800VAC, 50Hz input (which is a line to neutral voltage), and the first resistor stage (R1..R4) is bringing the voltage way down to the mV range (+/-12mV)!! this signal will be the input to the opamp which can support higher voltages! is it necessary to attenuate the voltage up to the mV range?
And yes according to the opamp1_in and opamp1_out signals the opamp is just a buffer. in this case, can it be removed completely !?
the second resistor stage (R6, R7) is just adding a dc offset of 1.25V and then the second opamp is just a buffer after that a third stage of resistors that brings the voltage back to 288mV!! this is probably because the isolation amplifier has differential inputs that have a range of +/-200mV!? what makes me wonder is why this attenuation and amplification of voltage (from mV to V to mV again)?
and also the isolation amplifier has a +/-200mV input range, does it mean that each input has to be 400mV peak to peak or the range between the 2 inputs has to be +/-200mV?
PS: there are two circuits like the one attached (one for the phase and one for the neutral) and the output of these circuits will be a differential input for the isolation amplifier.

 

Reduction of the maximum input to a millivoly range is probably not required, UNLESS the following input is only able to accept millivolt signals, such as from strain-gages and thermocouples. The use of an opamp does provide a convenient means of gain adjustment, and, possibly more valuable, a less expensive "fuse" in case of an over-voltage incident.

 

I didn't think about buffers because of the resistor on the feedback loop! it through me away! how is it a unity buffer when it has resistors?

The only reason for putting an impedance in the feedback stage of a buffer is to impedance match the other input, for high precision circuits. Not so critical for '082 type opamps which have a JFET input stage. Often the impedance match is needed in high precision instrumentation amplifiers with bipolar input stages with low input offset. A JFET input has a rather high input offset of 3 to 10 mV and this offset varies because of thermal effects. I agree, in this application it makes no sense. And besides the 10 Ohms is no where close to a "match".

 

My guess now is that the circuit the TS was given is of an amplifier for another application and happened to be handy.
One serious caution is that "common" in higer voltage circuits and system is seldom quite the same as "common" in analog inputs to logic packages. A buffer amplifier arrangement will not help, an isolation scheme of some variety is required. If that 800 volts (nominal) is part of an electric vehicle charging system then certainly an actual isolation scheme is required. Those are available commercially, the last ones that I used were less expensive than building and calibrating 100 of the assemblies.

 

A buffer amplifier arrangement will not help, an isolation scheme of some variety is required.

Hence the Si8920, it is used as an isolation barrier between the analog circuits and the digital circuit.

Also, is it required to have separate supply rails/grounds for each channel (meaning for each phase to neutral measurement)?

 

They form voltage followers (non-inverting unity gain amplifiers).

They have a high input impedance and a low output impedance.

Just a reminder to those who are new.... impedance means 'resists current flow'. high means resists lots of current (because it's only wanting to sense the value of the voltage), and low means accepts lots of current (usually as a driving force, like audio inputs, for example).