Transfer function circuit

Irving

Joined Jan 30, 2016
783
Minor point, your Vee is confusing. You should draw it with + to ground and set it to 12v to reflect its a -ve voltage (as if it were a battery)

+/- input in pic 1 is your Vi in pic 2
 

Thread Starter

Zeeus

Joined Apr 17, 2019
597
Minor point, your Vee is confusing. You should draw it with + to ground and set it to 12v to reflect its a -ve voltage (as if it were a battery)

+/- input in pic 1 is your Vi in pic 2
in pic 2, the end of R3 and R1 are connected to Vi

So the +/- should both be connected to this Vi? (will try)
 

Irving

Joined Jan 30, 2016
783
Ah, that link wasn't on your original.

The test circuit in your first pic is generating a differential signal for a differential input. Your 'ideal bridge' is single ended w.r.t ground, and is expecting an input varying +/- wrt ground.

You need to connect R3 to R4 in the test circuit, and remove R1, R2. The junction of R3 and R4 is your Vin. You need to choose R3 & R4 to give a voltage divider that keeps the signal from A7 in the expected +/- range of your Vin. The output of A7 is determined by the input signal from the sweep generator, reduced by the R11/R12 voltage divider x the gain of A7.
 
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