Peak voltage detector schematic

Such a divider wouldn't be much of an advantage since the typical input resistance of a modern DVM is about 10 megohms. The DVM would be 10 times slower at discharging the cap than your divider.

You could put 9 resistors of 10 megohms each in series to make a 90 megohm resistor and then put the meter at the bottom of the string for a total of 100 megohms resistance across the cap. Then multiply the reading by 10. If you have a couple of DVMs, use a second one on the ohms range to measure the input resistance of the other (set to volts range) to make sure it's really 10 megohms. If it's not exactly (to 1% or so) 10 megohms, use the measured value to figure your divider ratio. You might also measure the 10 megohm resistors in the string to see what their total is, and use that value in figuring your divider ratio.

The 100 meghohm load on a 1 uF cap should be a slow enough discharge to get a reasonable reading.

Reply to
The Phantom
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I explained what to do in another post:

" Such a divider wouldn't be much of an advantage since the typical input resistance of a modern DVM is about 10 megohms. The DVM would be 10 times slower at discharging the cap than your divider.

You could put 9 resistors of 10 megohms each in series to make a 90 megohm resistor and then put the meter at the bottom of the string for a total of 100 megohms resistance across the cap. Then multiply the reading by 10. If you have a couple of DVMs, use a second one on the ohms range to measure the input resistance of the other (set to volts range) to make sure it's really 10 megohms. If it's not exactly (to 1% or so) 10 megohms, use the measured value to figure your divider ratio. You might also measure the 10 megohm resistors in the string to see what their total is, and use that value in figuring your divider ratio.

The 100 meghohm load on a 1 uF cap should be a slow enough discharge to get a reasonable reading."

If the DVM you would be using to measure the voltage at the bottom of the divider isn't 10 megohms, then find out what it is by measuring it with a second DVM, and use the actual input resistance of the DVM to figure out what your divider ratio is. If the actual input resistance of the DVM is Rdvm and if you have used 9 resistors of 10 megohms in series, then you would multiply your reading by (Rdvm + 90,000,000)/Rdvm.

Reply to
The Phantom

It absolutely would. That's how voltmeters work; you're just increasing the multiplier resistance (as it's called).

Think about it. If your DVM has 10 megohm input resistance and you are measuring 1 volt, the same current is seen by the meter as if you add 90 megohms additional and then measure 10 volts. If the meter can do the first case, it can do the second case.

Reply to
The Phantom

An easier way to calibrate would be this: Take one of your power supplies and set it to its highest voltage, maybe 20 or 30 or even 50 volts. Measure it with a DVM, and write down the voltage. Put the 90 megohms in series with the DVM and measure the voltage. Divide the reading without the 90 megohms by the reading with the 90 megohms. That is your multiplier factor and it takes into account the input resistance of the DVM and the actual values of the 90 megohm string.

Reply to
The Phantom

I would very much like to make some sort of a "peak voltage detector". A device that, when plugged into a circuit, after a while, builds up an adequate representation of peak voltage. That would be again for my IGBT inverter.

That's for peak voltages under, say, 1 kV.

I tried making one from just a cap and diode, but multimeters would discharge it too quickly and getting reliable readings is difficult.

After reading the art of electronics, something like this comes to mind:

Make a small capacitor charged by diode (and a series resistor to limit current). After a while, voltage would build up there.

Connect a voltage divider (say, 1 megohm and 1 kOhm in series). To the cap. That would be a 1:1001 divider.

Connect that as input to some sort of FET based amplifier (that takes voltage and does not require input current) that produces same voltage, but at higher current. That would be the input ot a regular multimeter.

The multimeter reading would need to be multiplied by 1000 to get the actual voltage.

Alternatively to all this, does anyone know if higher end multimeters like Fluke 8050A (which I have) can measure voltage without discharging the cap so much? I have one lying around somewhere.

i
Reply to
Ignoramus20878

Maybe I'm wrong, but I thought most multimeters today came with 10 megs input impedance. That's 10 times more than you propose.

You have an oscilloscope, as I recall. Can't you set up the triggering to give a trace when the voltage exceeds some value you set with the trigger level control?

Good luck.

John

Reply to
John - KD5YI

AIRC, Art of Electronics shows a peak detector with op amps. Why not use what they show?

Ed

Reply to
ehsjr

I have two handheld MMs, one is a nice but old Extech and another is a $3.99 harbor freight unit.

Both discharge the caps rather quickly, I can barely get the reading.

I also have a Fluke 8050A, which seems to have 100K resistance. (I downloaded its manual today).

I suppose I can, yes. I am open to suggestions. Tek 2245A.

i
Reply to
Ignoramus20878

Sure, I guess I missed it or did not get to it. Do you know what chapter or page it was on?

i
Reply to
Ignoramus20878

Is an oscilloscope out of the question? It would be easier to record peaks. Most ohmeters only measure up to 10Meg, but there are some that go to 100 Meg. Perhaps their impedance is also higher. In any case, adding series resistance will increase input impedance.

greg

Reply to
GregS

Thanks Phantom. I am not sure if a divider with such high resistance (10 MOhm) would produce currents that are high enough to allow the DMM to reliably measure voltage. I will try your suggestion however.

i
Reply to
Ignoramus20878

There is a peak detector included in the later pages of....

formatting link

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

For just peak detector purposes, delete the rectifier sections and input your (attenuated) signal into the plus input of U3.

Increase R4 to suit your required droop rate.

Use another OpAmp section, connected for unity gain, to buffer "PEAK"... then your multimeter will do no loading at all.

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

My favorite OpAmp (when I last did discrete designs) was the TL084 (TI) or MC34084 (Motorola).

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

Should be. I've followed this thread loosely, so I don't know your exact requirements, but the ...084 will run on up to ~30V total supply, so you should be able to get a easily readable voltage.

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

Page 217

Reply to
ehsjr

thanks, will check tonight.

i
Reply to
Ignoramus5455

Jim, sorry if I am missing the obvious, but this peak detector on page

4 would quickly droop (discharge cap) if connected to a multimeter, right? I am talking about the part starting with U3. i
Reply to
Ignoramus5455

Right.

That's exactly what I am wondering about, what opamp would be most suitable for this purpose (infinite input impedance, voltage driven or whatever it is called). To avoid any droop altogether. Something in DIP package or just a piece with contacts would be best.

i
Reply to
Ignoramus5455

Yes, looks just what I need, thanks a lot!

i
Reply to
Ignoramus5455

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