Is this op amp circuit a promising idea to measure 300VAC?

Question

I need to measure 300VAC. I am using the following circuit and it is working good. In the circuit, I am using 2mA/2mA current transformer. More details about this transformer are available here.

However, I want to use only one power source of 3.3V to reduce my PCB size and cost. Is the following circuit good? If it is not a workable solution, pelase suggest me what would be the better solution.

EDIT:

The transformer is a current transformer with 1:1 relation of 2mA/2mA. Also, there is a limit of secondary resistance. The secondary resistance should be less than 100 ohms. It is not a problem to short circuit the secondary. More details are available at the following link.

https://www.micro-transformer.com/2ma-2ma-voltage-transformer-ZMPT101C.html

If I use a voltage divider (solution suggested by Kuba hasn't forgotten Monica) as shown in the following circuit, the voltage swing at the output is little and hence, I cannot use the complete range of a microcontroller. So, I suppose the use of an operational amplifier is required.

Edit 2: Thanks to Kuba hasn't forgotten Monica, I understood several other mistakes in my previous circuits. Upon including Kuba's suggestions, I have the following circuit suggestion. Is it good to use a instrumentation amplifier to take full advantage of the microprocessor's ADC?

Edit 3: I came up with another circuit which seems to work fine. Please give a look at it.

Edit 4: Solution:

Finally I am able to solve it. The following circuit is working good.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

Re. Circuit #1

The op-amp is incorrectly biased.

Re. Circuit #2

1. Add a fuse! Assume that the rest of the circuit can fail shorted.

2. Accommodate the +10% high line condition by increasing the V-I resistor R1 to 165k ohm.

3. The gain is too high and the op-amp's output attempts to swing close to the rails. Recall that the full transformer output current flows into the op-amp output. After all, that's the only destination it can flow into! The op-amp won't be able to drive close to the rails at such a current. According to the datasheet, with a 2mA load the output can swing to about +/-0.4V away from the rails, i.e. between 0.35V and 2.95V:

   

   The gain has to be decreased to lower the output voltage swing, i.e. the feedback resistor value has to be increased.

4. Add transient protection to the transformer secondary. Disconnecting the input will create an inductive kick that will overstress any load, be it an op-amp or an MCU.

Re. Circuit 3

The transformer is used as a current transformer, so the output current is exactly the input current. With a 150k V-I resistor at the input, the output current will be exactly 2mA across the recommended 50 ohm load. That's only 100mV RMS. The "+/-141mA peak" annotation is incorrect: the secondary current is 2mA RMS, maximum.

The voltage divider as shown further divides this small signal by half. At the very least, you'd want to move the divider to bias the winding, so that no amplitude would be lost:

^{simulate this circuit – Schematic created using CircuitLab}

The output looks as follows:

Discrete Current Source I-V Converter

If you can accommodate a slightly smaller output voltage range than the full 0 to 3.3V, then a current mirror would be another cheap solution for an I-V converter. The circuit below presents <100 Ohm load impedance to the transformer secondary. The output voltage at nominal line is 1.8Vpp.

^{simulate this circuit}

The output is shown below.

At volume, this probably costs <$0.25.

Cheap Op-Amp I-V Converter

As for an op-amp: The good old LM358 and LM324 work with input and output down to ground, and will do the job straight from 24V supply, and is cheap. LM324 and LM358 have stock levels in millions.

^{simulate this circuit}

For filtering the supply to the op-amp, it’s OK to use a simple RC filter.

The op-amp drives about 10Vpp through the 3k feedback resistor: that's 2mA RMS into the transformer secondary.

The output is then scaled down to fit into the 0.2-3.1V output range at nominal line voltage.

The op-amp can be any jellybean part not worse than LM358, ie. TL071, TL081, LM324, etc. With LM358 as shown, you have a second op-amp you could do something with.

The output impedance of this circuit is about 1.35k ohm. The fault condition output current into the MCU pin would be 3mA peak if the output of the op-amp suddenly saturates at 20V. This fault current has a transient character, since the coupling is capacitive. So the op-amp failing with output shorted to either rail will not damage the MCU.

Answer 2

Absent the question of proper circuit protection, that should work except that you seem to be designing for 300V peak. 300VAC generally means 300V RMS; that, in turn, means about 425V peak -- your circuit, as drawn, will saturate.

When it does saturate, the op-amp will no longer be actively controlling the voltage at it's \$V_-\$ input -- you'll be forking that job onto the op-amp's protection diodes, or otherwise whatever circuitry happens to be there.

Moreover, while the thing is advertised to have a 20mA output, it doesn't specify how well it works at that much output current.

If I were to use the same front-end (\$150 \mathrm k \Omega\$ resistor and that current transformer), I would terminate the current transformer with a passive resistor to ground. Then I would use the op-amp as a plain old voltage buffer, probably with higher resistances so the output is only loaded with \$10\mathrm k \Omega\$.

I haven't actually checked this circuit, and the coffee hasn't kicked in yet this morning -- feel free to analyze, simulate, and otherwise check it carefully before use. It should give you a sine wave riding on 1.65V, scaled so that it stays within the rails when you've got a 300V RMS sinewave on the input.

^{simulate this circuit – Schematic created using CircuitLab}