Hi,

I want to implement a similar circuit as seen in AN-306 and I'm trying to figure out the transfer function given in the note. To me the circuit looks like a simple voltage divider, but the writers of AN-306 have come up with a different equation whos derivatoin seems to elude me. Any help would be appreciate.

Thanks,

Ben

Hi Ben,

I'm sorry the application note was not very clear about what was going on with the AD630. The AD630 data sheet has a good application section that explains how the part works.

Here is an@@ analysis leading to the equation:

Peak current into RTEST is Vp/10,000

Voltage across RTEST is (Vp/10,000) x RTEST

The AD524 is connected for G = 1000

Gain of IC2 is 100

Total gain is 10^5

Voltage out of IC2 is (Vp/10,000) x RTEST x 10^5 = Vp x RTEST x 10

The voltage out of IC2 is a sinewave with a peak value of Vp x RTEST x 10

The AD630 performs a synchronous demodulation (same as full-wave rectification) on the sinewave:

So the mean value of the signal out of the AD630 is (2/pi) x Vp x 10 x RTEST

The output lowpass filter, IC3, has a gain of 1, and so the final DC (mean) output is

Vout = (2/pi) x Vp x 10 x RTEST

Vp = 10V, so the final output is

Vout = (2/pi) x 10 x 10 x RTEST = (200/pi) x RTEST

RTEST = (pi/200) x Vout = 0.0157 Vout, which is the final equation in the AN.

I hope this answers your question, and let us know if we can help you in the future.

Best Regards,

Walt Kester

Analog Devices