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Division of Voltages

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Oluvision

New Member
I have a project due in 4 weeks time. I need to divide two voltages to get a measure of impedance of a DUT. Some one suggested a microproccessor, but I do not know how to use a microproccessor.
Is there any other way of doing this division.
 

Styx

Active Member
You could get two log amps and pass the two voltages through the log amp. Use a diff anp to get the difference and then inv-log amp to get

V1/V2 since Log(V1) - Log(V2) = Log(V1/V2)

that is an analogue way other wise microprocessor
 

bmcculla

New Member
If you want to do analog division you could look into a attenuator with programmable attenuation. The numerator would be the input and the denominator would set the attenuation. Gain of .5 = devide by 2. You might be able to find an opamp circuit for this on the internet. You can use a transistor and resistor as a voltage devider where transistor works as a programmable resistor. If you can find a way to use an opamp to linearize this effect you would have a functional circuit.

All that said it still might be easier to learn about microcontrollers. The solution is trivial in a microcontroller.

Brent
 

Oluvision

New Member
How do I use a Transistor

To: BmCulla

How do I use a transistor in conjuction with a resistor to divide. I thought but I did not get a clue

Oluvision
 

Oluvision

New Member
To: Russlk
I am trying to measure the magnitude of the impedance of any Device. I am using a transresistance amplifier to mesaure the current. I need scheme to divide the voltage by the current.
 

bmcculla

New Member
This is what I was thinking of. This schematic doesn't have any biasing will only work for a small range of inputs and only with small signals. With some creative use of opamps you should be able to get better performance. Ive seen this same idea used in the feedback path of an opamp to provide adjustable gain.

The transistor is working like a programmable resistor.

Hope this helps
Brent
 

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Roff

Well-Known Member
In your transimpedance amplifier, use the unknown Z as the feedback element. Call the known stimulus Vin, and call the resistor from Vin to the summing node Rs.

Then Vout=-Vin*Z/Rs.

If Vin/Rs=1 (or .001, etc.), then Vout=-Z (times a scale factor).
 
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