Project: A circuit to allow use of ammeters with different internal resistances

Thread Starter

richard3194

Joined Oct 18, 2011
173
Hi. I believe that for a microammeter to read correctly in a circuit that has been designed with a microammeter with a particular internal resistance, a replacement microammeter of the same FSD must have the same value of internal resistance to show the correct current. Where a replacement microammeter has an internal resistance lower than the original, you can of course simply add a sereis resistor to bring up the resistance to match the original. However, if you have a meter with a larger internal resistance it's not possible to use resistors that will enable the replacement meter to read correctly as the original.

What I think I need then is a circuit which will allow any microammeter to read the same as the original, no matter what it's internal resistance.

My original 100 uA meter has an internal resistance of 260R. I can buy a replacement 100 uA meter that has an internal resitance of 800R. But, I need a circuit to enable me to use it so it reads the same as the original meter in the circuit designed for the original meter. Meter is DC meter.

There may be a circuit out there on the WWW. But, I'm not sure what to put as a search string.
 
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dl324

Joined Mar 30, 2015
16,845
My original 100 uA meter has an internal resistance of 260R. I can buy a replacement 100 uA meter that has an internal resitance of 800R. But, I need a circuit to enable me to use it so it reads the same as the original meter in the circuit designed for the original meter.
Use a current divider.
 

Thread Starter

richard3194

Joined Oct 18, 2011
173
I'm assuming no shunt or series resistor in the original meter. Perhaps I should check that out first.

P.S. Oh, I don't think it matters what is in the original meter, a shunt or series resistor.
 
Find the resistance of the naked meter and re-do it.

it's kinda easy. Suppose you have an unknown meter.
You want a potentiometer set at the highest value and a power supply to get what you think si a good deflection. 50 uA and 1 uA were good meter currents.

Then find a nice potentiometer and PS in series with the meter and adjust the pot for mid-scale. You need to measure the voltage across the potentiometer and the resistance of the potentiometer to get the current.

The value if the potentiometer is the internal R of the meter. the current is the 1/2 scale deflection current. Go from there.

I might be missing a lot of details.
 

Thread Starter

richard3194

Joined Oct 18, 2011
173
The meter is actually from Test Set Type 193A a quartz crystal activity meter.

The meter face plate says "Res 260 Ohms"

But, I'd like to use a much smaller meter. And it seems I need a circuit with a transistor if a replacement meter has say an IR of 800R.

However: It just occurred to me. that the key meter would be the replacement meter. If that 800R was in fact largely made up of a series resistance, I could use that by altering the series resistor. But, I've no way of knowing unless I buy that 800R meter . :)
 
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MrChips

Joined Oct 2, 2009
30,711
It may not be as simple as replacing one meter with another.
You need to know the impedance of the circuit that is driving the meter.
Can you tell us the instrument or show a photo of the instrument?
 

Thread Starter

richard3194

Joined Oct 18, 2011
173
I seem to have determined (with an ohmeter) that the original meter has 100R in series with a coil resistance if 160R. Thus a total of 260R as indicated on the meter plate.

A replacement meter cannot have a coil resistance above 260R. Anything less than that I could add a series resistor. This would be true if both the oriinal meter unit and the replacement one (i.e. the coil assembly only) were equally sensitive.
 
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AlbertHall

Joined Jun 4, 2014
12,345
A circuit using an op-amp is possible which would have an input resistance of 260Ω and an output which will drive a current equal to the current in the input.
 

Thread Starter

richard3194

Joined Oct 18, 2011
173
Here is the essential circuit. One sets grid current to 50uA by adjusting screen voltage. R7 is 100K. Internal resistance of M1 is 260R. V2 is an EF50. The grid must I presume go positive and draw current thru M1 and R7. M1 has it's positive terminal connected to ground. Current will be probably be varying DC I imagine. Possibly smoothed by C9, which is 0.1 uF. Frequency of operation 3-10 Mhz.. Somone may need to explian to me how we get current flowing thru M1.

TestSetAM193  Xtal Osc Circuit 20pf position.jpg
 
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AlbertHall

Joined Jun 4, 2014
12,345
As the meter has a 100k resistor in series with it, the resistance of the meter really doesn't matter as it is very low compared with that 100k.
It does need to be the same full-scale current to get the same readings, but that is all.
 

Thread Starter

richard3194

Joined Oct 18, 2011
173
As the meter has a 100k resistor in series with it, the resistance of the meter really doesn't matter as it is very low compared with that 100k.
It does need to be the same full-scale current to get the same readings, but that is all.
Oh, I now see the great importance of knowing the entire circuit. :)

That's about it then. I can use a variety of meters having different internal resistances (within reason) and they will all make acceptable replacements. No need for an electronic meter circuit in this instance.
 
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