I think the answer is on the page linked below, but I can't 'translate' the math, to change my 100mA meter to read volts. The example uses a 1mA meter. Me bad at math 
http://ourworld.compuserve.com/homepages/Bill_Bowden/page11.htm
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http://ourworld.compuserve.com/homepages/Bill_Bowden/page11.htm
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A voltmeter is suppose to have infinite input resistance. AKA no current draw. Getting an ammeter to read volts will be loading the circuit you are trying to measure. Just go and buy the proper voltmeter.
hi redrabbit
To answer you we need to know two things.
The internal resistance of the meter (just measure it with an ohm meter) which will probably be very small and the number of volts you want for a full scale reading.
Simple example (ignoring meter resistance)
Say you want to measure up to 100 volts which is handy since you wouldn't have to change the meter scale.
R=E/I
R=100/0.1
R= 1000 ohms
Power rating of resistor, P = I*I*R
P=0.1*0.1*1000
P=10 Watts
That's a very insensitive meter so it will only be really useful for reading the power supply output voltage. Too much loading for in circuit measuring.
Cheers
Rob
To answer you we need to know two things.
The internal resistance of the meter (just measure it with an ohm meter) which will probably be very small and the number of volts you want for a full scale reading.
Simple example (ignoring meter resistance)
Say you want to measure up to 100 volts which is handy since you wouldn't have to change the meter scale.
R=E/I
R=100/0.1
R= 1000 ohms
Power rating of resistor, P = I*I*R
P=0.1*0.1*1000
P=10 Watts
That's a very insensitive meter so it will only be really useful for reading the power supply output voltage. Too much loading for in circuit measuring.
Cheers
Rob
Why not just an opamp set up for voltage controlled current output? Much more elegant solution, probably cheaper than the 10watt resistor and best of all, will hardly load the source.
Sam.
Sam.
Interesting concept.
I would like to see the circuit analysis. Given that the op-amp has to source 100 mA, what device did you have in mind?
Cheers
Rob
I would like to see the circuit analysis. Given that the op-amp has to source 100 mA, what device did you have in mind?
Cheers
Rob
In simple terms you have the meter connected across Vout to -Vin and then resistor Rf from -Vin to ground. The voltage sensing is from the +Vin of the opamp.
The input voltage is directly proportionate to the voltage across Rf developed by the current through the meter and Rf (since the opamp ideally draws no current from -Vin).
This solution is also independent of the resistance of the meter.
In realistic terms you'd be sampling voltages much higher than the opamp so just add a voltage divider (two resistors) to +Vin.
Here's an example:
Let's assume the peak voltage is 100V. A voltage divider of 10k and 90k ohms feeds 0 to 10V into +Vin. Rf must match the peak voltage on +Vin when the peak current flows through it, giving:
Rf = 10V / 100mA = 100 ohms
Now you just have to find a general purpose opamp that will output 100mA. I haven't looked into this myself but I'm sure you can find one.
I've attached a pdf from my study guide. Look for the VCIS circuit. There's also an example the of using just a BJT with a few passive components. Doing a discrete solution like the BJT means you can easily get 100mA.
Sam.
The input voltage is directly proportionate to the voltage across Rf developed by the current through the meter and Rf (since the opamp ideally draws no current from -Vin).
This solution is also independent of the resistance of the meter.
In realistic terms you'd be sampling voltages much higher than the opamp so just add a voltage divider (two resistors) to +Vin.
Here's an example:
Let's assume the peak voltage is 100V. A voltage divider of 10k and 90k ohms feeds 0 to 10V into +Vin. Rf must match the peak voltage on +Vin when the peak current flows through it, giving:
Rf = 10V / 100mA = 100 ohms
Now you just have to find a general purpose opamp that will output 100mA. I haven't looked into this myself but I'm sure you can find one.
I've attached a pdf from my study guide. Look for the VCIS circuit. There's also an example the of using just a BJT with a few passive components. Doing a discrete solution like the BJT means you can easily get 100mA.
Sam.
Forget it. A 100mA meter is completely unsuitable for your purpose. You can pick up a 1mA for peanuts and add a series resistor as rpapps describes to convert it to a 100V meter. The only use for a 100mA meter is... ...measuring current up to 100mA.
Edit: Since your 100mA meter is virtually useless, you won't be worried about possibly destroying it. If you open it up from the back I wouldn't be in the least surprised to find that it's actually a 1mA meter with a low resistance shunt resistor across the terminals. If so, it may be possible to convert it into a voltmeter.
Edit: Since your 100mA meter is virtually useless, you won't be worried about possibly destroying it. If you open it up from the back I wouldn't be in the least surprised to find that it's actually a 1mA meter with a low resistance shunt resistor across the terminals. If so, it may be possible to convert it into a voltmeter.
Two other possibilities:
If this meter is salvaged from some bit of equipment, it may just have a 100mA scale, but really be something else. A quick check with an ohmmeter will tell all- sometimes, there's a tiny legend on the scale itself that can tip you off.
And on the surplus and salvage market in the US, you're about equally likely to find a 1mA FS and a 50uA FS. If the latter, that's perfect for voltmeter application.
If this meter is salvaged from some bit of equipment, it may just have a 100mA scale, but really be something else. A quick check with an ohmmeter will tell all- sometimes, there's a tiny legend on the scale itself that can tip you off.
And on the surplus and salvage market in the US, you're about equally likely to find a 1mA FS and a 50uA FS. If the latter, that's perfect for voltmeter application.
SY said:
100mA meters will almost always have a shunt resistance -- and as SY suggests are 1 ma, 100uA or 50uA full scale. The shunt may be internal in which case you will have to pry the thing apart --
a favorite beginner project is using an MPF102 JFET to drive the meter movement -- this provides a high input impedance -- Ray Marston had one such design on his series in Nuts n Volts -- and thankfully N&V has a free public archive -- the article can be found here: http://www.nutsvolts.com/~downloads/fetjun.pdf
fwiw -- there are some very high quality VOM's (Simpson) and vacuum tube voltmeters (like HP 412's) on the bay --
a word of caution about hooking a DVM to a current meter... if it is a low-current meter damage may result from overdriving the motion with the signal from the test equipment (I killed a 10mA meter this way, using the resistance function on a handheld multimeter)
WoW...thanks for the great suggestions. I have yet to research all of them.
It is a vintage Simpson, 0-100mA, no other writing on the face. It's has a bit of weight to it, if that adds to any speculation about it.
Yes, I will save this meter for a more elegant situation ; my need for a meter is to quickly measure volts and load for:
*4 variable PSUs (in a single box),
*up to 12v / 200mA.......to power guitar fx,
*using a three-position momentary switch on/off/on... (volts/off/milliamps)
*another rotary switch points to which supply output I want to adjust/check.
...the reason for the momentary switch is to keep the meter out of the circuit, except when needed.
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BTW: Lackinnj....funny, I was just searching for Jfets today, at my local surplus store....2n5457......they didn't have any.
It is a vintage Simpson, 0-100mA, no other writing on the face. It's has a bit of weight to it, if that adds to any speculation about it.
Yes, I will save this meter for a more elegant situation ; my need for a meter is to quickly measure volts and load for:
*4 variable PSUs (in a single box),
*up to 12v / 200mA.......to power guitar fx,
*using a three-position momentary switch on/off/on... (volts/off/milliamps)
*another rotary switch points to which supply output I want to adjust/check.
...the reason for the momentary switch is to keep the meter out of the circuit, except when needed.
=RR=
BTW: Lackinnj....funny, I was just searching for Jfets today, at my local surplus store....2n5457......they didn't have any.
World politics left aside for the moment, there are several companies producing low-cost panel meters that are getting down into the $20-$30 range IIRC. And for cheaper than that you can get a DMM from an auto parts store to mount and leave, dedicated, in-circuit.
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