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- Thread starter William loh
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- #2

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Following proceedure 5 in the multimeter handbook minimum closed circuit voltage at thermocouple is 8 mV

The point is that many transducers in control engineering produce voltage in the 0 - 200mV range.

- #3

vk6kro

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You need to know the resistance of the meter, but suppose it was 100 K ohms. This would mean that when the meter is reading 200 mV, the current in the meter will be 2 microamps.

So, if you wanted the meter to measure 20 volts, the current would still need to be 2 microamps so, you would need a series resistor of 10 megohms minus the 100 K of the meter, so 9.9 megohms.

ie 20 volts / 0.000002 amps = 10 000 000 ohms.

If you wanted to measure 200 volts maximum, you could use a 100 Megohm resistor, but it would be better to just use the 9.9 Meg ohms resistor as above, but put a 11.11 k resistor across the meter.

This would take 18 μA when there was 200 mV across the meter, (200 mV / 11111 ohms = 18 μA), so the total current would be 20 μA. Then if you put 200 volts across the 9.9 Meg plus meter in series, the meter would read close to full scale which you could interpret as 200 volts.

Some meters let you move the decimal point so the reading makes more sense.

- #4

FOIWATER

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I agree with voltage generated by thermocouples, seebeck voltages I would say that's the cause

- #5

vk6kro

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If you put a resistor in parallel with the meter, the meter will read full scale if the voltage across the resistor is 200 mV

So, if you put a 1 ohm resistor across the meter, the meter would read 200 mV if the current was 200 mA.

With a 0.1 ohm resistor, the meter would read 200 mV if the current through the resistor was 2 amps.

Meters that measure the voltage out of a thermocouple would normally be calibrated in units of temperature like °C or °F.

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I get it now, Thank you so much, you guys are really helpful.

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jim hardy

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