Circuit Planning with AD9643

Hi TomerWIS, 

No additional 200 Ω resistors are needed. Your schematic already includes the feedback resistors connected through the ADA4930-2 FB pins.

Best Regards,
Ian

Dear Ian, 

Thank you for your response. I did a change in the wiring, because I have noticed that the FB pins are connected to the inputs through the 200ohm resistors, instead to the outputs. Currently, the FB pins are connected to the corresponding OUTPUT pins through the 200ohm resistors, and the previous wrong connections have been lifted.

Still, instead of seeing the signal I am creating through the waveform generator (1MHz, 1Vpp, 0 offset), I see a 1MHz square-ish wave, of 1.8Vpp around 0.9v (which is also Vocm).

Do you have any idea what could cause this?

Thank you, 

Tom

Hi TomerWIS, 

The FB pins are internally connected to corresponding OUT pins, therefore the 200 Ω feedback resistors should be connected from FB+ to -IN and from -FB to +IN, as in your original schematic. 

Connecting the FB pins to the corresponding OUT pins through 200 Ω leaves the input summing nodes without negative feedback. The amplifier therefore operates open-loop and clips, producing the square-like output you observed. Also, one additional point: each output is centered around the 0.9V common mode, so measuring each OUT pins with respect to GND will not show differential output. Please use a differential probe if you haven't yet.

I have attached an LTspice simulation demonstrating this behavior:




ADA4930-2_SIM.asc
 

Best Regards,
Ian

Dear Ian. 

Thank you for your help in this issue. I think that this will solve the problem. I noticed (can be seen in the image of my design in the original post) that from the beginning the FB are connected to the wrong polarity of the input, and I assume that that was the only problem with the circuit.

Tom