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I have a device which can be modeled with a resistor in parallel with a capacitor. Now I want to measure its resistance and capacitance as a function of voltage bias (from -2V to 2V) and frequency (from DC to 10kHz). Originally I was thinking of using a LCR meter such as hand held Agilent U1732C which can measure in 100Hz, 120Hz, 1kHz and 10kHz. But the problem is I am not sure how to apply a bias and measure with this equipment at the same time. On the other hand, I don't need very high precision. I think 5% precision should be okay. But more frequency points from DC to 10kHz will be helpful. So my question is:

  1. How can I apply bias on the device and measure its C and R using Agilent U1732C? Can I do that?

  2. If 1 doesn't work, can I do it with a not-so-complicated DIY kit rather than resorting to a high-end LCR meter?

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If it were me , I would not only measure C & tan delta but ESR as well and test to 10MHz.

You can use a scope rear sweep signal to drive an FM generator, but 50 Ohm is not the best source.

If you cant scrape a simple AC current source design to generate cap impedance as the output voltage, use a voltage source (ie. darlington emitter follower to drive the cap with an adjustment for DC offset +/-2 V and monitor current with a 1 Ohm shunt. The scope can show the DC bias on one channel and ac coupled envelope synced to scope sweep. This is a quick & dirty method. XY mode of V vs I is another method.

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Why not use a Digital Volt Meter to measure its resistance?

Once you have that value you can calculate capacitance from the impedance reduction by applying an oscillator thru a known resistor or use a signal generator with a known output impedance. At low frequencies the impedance will look like just the resistance. At high frequencies the capacitor will dominate and the impedance will be much lower.

Simple formula can take you to the answer.

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  • \$\begingroup\$ I'm thinking that his device must have a voltage dependent resistance and capacitance? \$\endgroup\$ – Alfred Centauri Oct 2 '13 at 21:54
  • \$\begingroup\$ @AlfredCentauri or.... He's just a noob. I'll modify my answer if it is. \$\endgroup\$ – Andy aka Oct 2 '13 at 22:13
  • \$\begingroup\$ @Andyaka Thanks for your answer. Maybe I am missing something, but your method may only work well for some resistance and compacitance. If I have a "leaky" device, for example, which has 1kOhm resistance and 1nF capacitance under 1kHz and some bias, the capacitance impedance is 160kOhm. The reduction in impedance from DC to 1kHz is not that obvious (assuming the capacitance doesn't change significantly), and the result can be influenced by the instrument noise. And yes, the resistance and capacitance is voltage dependent. \$\endgroup\$ – shva Oct 2 '13 at 23:57
  • \$\begingroup\$ @shva - I think you need to edit your question to make it clear that the values are dc level dependent. Who mentioned 1kHz? Signal generators are quite capable of generating tens of MHz and applying a variable dc voltage to get the dc characteristics is so easy. Then you apply the same dc voltage range but also apply a signal generator in parallel to understand the capacitance effects with dc changes. Quite easy really. \$\endgroup\$ – Andy aka Oct 3 '13 at 7:14

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