AC to DC voltage mirror for reference?

[pacmann]

Whats the best way to convert an AC voltage to its equivalent DC for measurement? The output can be low current in the 10-50 mA range.

As an example, if there is 200 vpp AC I want to create a 200 vdc reference voltage. Does this kind of circuit have a name?

[pacmann]

Can the circuit be as easy as a bridge rectifier with parallel capacitor?

[Zero999]

It depends on whether the voltage drop across the rectifier diodes is a problem or not. Under light loads, a silicon diode will typically drop around 0.6V and they're two in series in a bridge rectifier giving a total drop of 1.2V.

Another thing to note is AC voltages are normally specified by the RMS (Root Mean Square) voltage. A rectifier and filtering capacitor will give you the peak voltage, which is higher by a factor of √2, so 200VAC RMS in, will give 200√2 = 283VDC out, minus the diode voltage drops.

[pacmann]

It depends on whether the voltage drop across the rectifier diodes is a problem or not. Under light loads, a silicon diode will typically drop around 0.6V and they're two in series in a bridge rectifier giving a total drop of 1.2V.

Another thing to note is AC voltages are normally specified by the RMS (Root Mean Square) voltage. A rectifier and filtering capacitor will give you the peak voltage, which is higher by a factor of √2, so 200VAC RMS in, will give 200√2 = 283VDC out, minus the diode voltage drops.

Thanks for explaining. I think the drop of 1.2v will be ok.

Can the RMS be compensated with someting like a resistor divider at the DC side? Is there a better way than a divider?

Is there a way to isolate the input without using a transformer because of the low output current requirement?

[Zero999]

It depends on whether the voltage drop across the rectifier diodes is a problem or not. Under light loads, a silicon diode will typically drop around 0.6V and they're two in series in a bridge rectifier giving a total drop of 1.2V.

Another thing to note is AC voltages are normally specified by the RMS (Root Mean Square) voltage. A rectifier and filtering capacitor will give you the peak voltage, which is higher by a factor of √2, so 200VAC RMS in, will give 200√2 = 283VDC out, minus the diode voltage drops.

Thanks for explaining. I think the drop of 1.2v will be ok.

Can the RMS be compensated with someting like a resistor divider at the DC side? Is there a better way than a divider?

Is there a way to isolate the input without using a transformer because of the low output current requirement?

If the AC signal is always a sine wave, then a potential divider can be used to divide by √2.

An analogue isolator can be used to isolate the signal, but the usual way is to digitise the signal first and pass it through a digital isolator. It's also possible to use PWM, or PFM the signal before passing it through a digital isolator.

[fcb]

What level of accuracy? Peak AC or RMS?

[wraper]

For measurement you don't normally outright rectify high voltage and smooth it. It should be a voltage divider followed by precision rectifier. You will need to add true RMS to DC converter if you want to measure RMS voltage for different waveforms.
https://circuitdigest.com/electronic-circuits/half-wave-and-full-wave-precision-rectifier-circuit-using-op-amp