simplest rectifier-amplifier for a current inductive sensor

Just correcting a minor typo: the sensor outputs 10 mV every 3 Amps drawn by the load.

Build a precision rectifier with gain. The basic idea is this:

Use a cheap opamp, LM324 or LM321 if you don't need the other sections elsewhere. Mind the input common mode range, a couple hundred mV negative can be done but no more. Also, this one isn't RR output but with a pull down might work, depending on what kind of spikes your ADC input pin presents.

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Regards, Joerg

http://www.analogconsultants.com

Don't rectify it; just amplify then sample with a moderately fast ADC and crunch the samples.

John

That would be overkill for my purposes, and I also need something very simple in order to make in work on a small very low power PIC. I'm thinking about something around a single small mosfet, plus few other parts, which rectifies the signal and amplifies it enough to drive the ADC. I believe I've seen something like that in the past but can't remember where. If I can't find it I'll probably try the opamp solution Joerg suggested. I'm not sure if it can be adapted to a single 5V powered circuit, but it looks like the most simple and widely used out there.

Tnx to all of you for the responses.

5V single supply will not be a problem. You can buy opamps in SOT23-5L packages. Very tiny, cheap. But you have to look at your ADC. Some are unbuffered and must be driven hard because of capacitor charging going on at the input. Got to go through that section of the uC manual with a fine comb.

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Regards, Joerg

http://www.analogconsultants.com

Are you sure the PIC adc isn't good enough? The cheapest, and most accurate, rectifier is software.

John

This one's nearly simple and there's a lot of gubbins but it isn't plagued by opamp offset voltages. A 30mV~ (ie=10A~)input will give a 300mV DC output (at point "Pos+"). You would still need a rail-to-rail-input-output opamp, (eg MCP 6021 )to get this voltage up to a more ADC friendly level, such as 4V. [Sod it, just realised the first transistor would have been better as a plain vanilla AC amp, wouldn't then need the bootstrapping. Too late now!] +5V | .-. | | 1k| | '-' | o-----. | |100uF .-. --- | | --- +5V 1k| | | | Diodes=IN4148 etc 10uF '-' | | || | | |/ ,---o----||----------, o---------| BC546 | | || | | | |>

.-. .-. | | | | 47k| | | |47k +5V | | | | | | | | | '--------o-----o '-' '-' .-. | | +Pos| |Neg- | | | .-. - V | |33k | | | ^ - '-' | 1k| | ___ | | || | |/ '-' ~ACin o--|___|----o---o----||-----o-----| BC546 | 1k || | |> --- o | | 0V | 10uF | |---, --- .-. .-. | 0V | | | | --- | | 470| | ---100u 10k'-' '-' | | | | --- --- --- 0V 0V 0V

(created by AACircuit v1.28 beta 10/06/04

You need to filter as well, or does the microcontroller have its own sample/hold? Let me suggest a multiplier that gives current-squared output, made with a transconductance amplifier (see the LM13700 data sheet), followed by either a lowpass filter (averaging) made with the second section of the dual chip. Your current measurement would then be RMS, but the microcontroller will have to do the square root function.

It sounds complicated, but it's just a few resistors, an amplifier chip, and a capacitor.

Not quite, the so-called sensor is a simple current transformer, and, in the case of the ASM-010, it requires a 50R 'burden' resistor, supplied by the user, to develop that particular Volts/Amp. Since the amplitude is so small, peaking at 33mV, you can use a simple single supply op-amp with gain, but the ac-input impedance must be 50R.