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Old 23rd February 2004, 09:49 PM   (permalink)
Default Division of Voltages

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.
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Old 23rd February 2004, 09:53 PM   (permalink)
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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
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Old 23rd February 2004, 10:01 PM   (permalink)
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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
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Old 23rd February 2004, 11:44 PM   (permalink)
Default 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
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Old 24th February 2004, 12:28 AM   (permalink)
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How do you get impedance by dividing two voltages? Is the object of the exercise to find the impedance, or is that just part of the problem? What is the problem?
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Old 24th February 2004, 12:42 AM   (permalink)
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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.
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Old 24th February 2004, 01:28 AM   (permalink)
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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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Old 24th February 2004, 02:48 AM   (permalink)
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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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