Resistor Divider Output changes when connected with AD829

Hi O. Santhosh Kanna,

Voltage divider theorem assumes that the current between the two resistors in the divider network is the same. In the case when the switch is off, there is a high impedance path between S and D, which means negligible current is flowing to the switch, therefore currents flowing through, say 82k and 22k are equal. Once the switch is on, there is now low impedance path between S and D. Because of this, whatever is the load of the switch loads the voltage divider network. The loading draws current from the voltage divider network, thus the current flowing through 22k will not anymore be equal to the current flowing through 82k. Once this happens, the voltage divider equation no longer applies.

Also, there's a mini tutorial on switches and multiplexers. You can read it here:

http://www.analog.com/static/imported-files/tutorials/MT-088.pdf

I hope this gives some enlightenment to your problem.

Best regards,

Neil

In my circuit, at "D" Side of ADG1206, I have 100KE with respect to GND and Opamp Input Terminal. Since These two are at higher impedance values than the 22KE in my voltage divider, they should have little effect on my voltage divider circuit when the channel is selected at ADG1206. But it creates a huge change in the divided amplitude value. i do not understand, why does this happen.

Hi,

May I know your expected values and measured values for the voltage divider networks? Relative to the resistors in the voltage divider, the 100k is not big enough to have negligible effect. Could you try changing it to 10M perhaps? Also, the input impedance of AD829 is quite low for your application. FastFET op amps such as the ADA4817 and AD8065/6 offers higher input impedance and might be better choice for your application.

I hope this helps.

Neil

Hi,

          Thanks for you. I try with the any of the mentioned op amps.