# Analog Devices 'Circuits from the Lab' Question

Discussion in 'General Electronics Chat' started by dannybeckett, Apr 18, 2013.

1. ### dannybeckett Thread Starter Senior Member

Dec 9, 2009
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0
Hello again.

I have a question on a new analog devices reference design. This circuit employs transimpedance gain switching, which is what I am currently researching. They use the ADG633 CMOS switch to swap out the feedback network of the amplifier. I understand that the switch has an on resistance of about 50ohms, which in the feedback loop will add a 50ohm error to the gain. What I don't understand is how connecting another switch at the output of the device, nulls this error? This is demonstrated about 1/3 the way down the page:

http://www.analog.com/en/circuits-from-the-lab/CN0312/vc.html

Any clarity on this matter would be greatly appreciated!

2. ### joeyd999 AAC Fanatic!

Jun 6, 2011
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5,931
Yes, this is strange. The output would have an impedance of Ron, and the Vo would be dependent on the subsequent load. Seems the only way the Rons would cancel would to force Iout = Iphotodiode.

Am I wrong?

3. ### crutschow Expert

Mar 14, 2008
21,336
6,116
Since the left side of the switches are inside the feedback loop, their resistance will not affect the output at the junction of the switch pair. The op amp will force its output voltage to whatever is needed to overcome the switch resistance to maintain the voltage at the junction of the switch pair equal to Id x Rf.

But the output load at Vo must be a very high impedance (such as a non-inverting CMOS op amp circuit) to minimize any error due to load current through the output side of the switches.

4. ### Ron H AAC Fanatic!

Apr 14, 2005
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The left hand switch puts Ron inside the feedback loop. The second switch selects which feedback loop you are monitoring. If you need lower output impedance, add a voltage follower.

EDIT: Carl beat me to it. He also explained it better than I.

5. ### joeyd999 AAC Fanatic!

Jun 6, 2011
4,092
5,931
Dang...you're right, Crutshow...I made the mistake of concentrating on what the text was saying, and not on the picture.

6. ### dannybeckett Thread Starter Senior Member

Dec 9, 2009
170
0
OK that makes sense - but, would the following circuit not achieve exactly the same outcome but use less components, and have a lower output impedance?

7. ### dannybeckett Thread Starter Senior Member

Dec 9, 2009
170
0
nope it wouldn't, the 52 ohm offset would exist in that configuration! cheers for clearing this up =]