help with AD620 - Electronic Circuits Projects Diagrams Free

raviram87 · 13th February 2008, 07:09 PM

hello people...

recently i was working with an AD620 instrumentation amplifier... it has a gain range of 1 to 1000.... tis gain can be varied by an external resistor connected between pins 1 and 8 of AD620....

gain = 49.4k/(Rg) + 1... where Rg is the resistor between pins 1 and 8...

i used two AD620s... the first one was set for a gain of 1000 followed by the next which was set for a gain of 100 by selecting appropriate resistors such that the above relation is satisfied.... the output of the first AD620 was fed into the input (pin3) of the second AD620 and its other input (pin2) was grounded... the AD620 was supplied with +/-12V

the input to the first AD620 was a 20mV 71 Hz square pulse... with the setup above the output must have saturated to 12V rite (20mV*1000*100) ??? But i observed just a 2.4V output....

Why is this happening? Is the procedure correct? Pls help...

Thanks in advance... :-)

ronsimpson · 14th February 2008, 12:19 AM

Trouble shooting.
>Pull the second AD620 to simplify things. Does that make a difference?
>Check pins 2,3,5 of first amp. Pin2=gnd, Pin3=20mV signal, Pin 5=gnd? Don’t just check for gnd with a scope probe use ohm meter.
>Check gain resistor again. 1000? Or 100?
>You will never get +-12 volts swing. +- 10.5 volts.
>Is your 2.4 volts centered around (+-1.2) or is it 8to10.5 volts?

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13th February 2008, 07:09 PM	(permalink)
raviram87	help with AD620 hello people... recently i was working with an AD620 instrumentation amplifier... it has a gain range of 1 to 1000.... tis gain can be varied by an external resistor connected between pins 1 and 8 of AD620.... gain = 49.4k/(Rg) + 1... where Rg is the resistor between pins 1 and 8... i used two AD620s... the first one was set for a gain of 1000 followed by the next which was set for a gain of 100 by selecting appropriate resistors such that the above relation is satisfied.... the output of the first AD620 was fed into the input (pin3) of the second AD620 and its other input (pin2) was grounded... the AD620 was supplied with +/-12V the input to the first AD620 was a 20mV 71 Hz square pulse... with the setup above the output must have saturated to 12V rite (20mV1000100) ??? But i observed just a 2.4V output.... Why is this happening? Is the procedure correct? Pls help... Thanks in advance... :-)

14th February 2008, 12:19 AM	(permalink)
ronsimpson	Trouble shooting. >Pull the second AD620 to simplify things. Does that make a difference? >Check pins 2,3,5 of first amp. Pin2=gnd, Pin3=20mV signal, Pin 5=gnd? Don’t just check for gnd with a scope probe use ohm meter. >Check gain resistor again. 1000? Or 100? >You will never get +-12 volts swing. +- 10.5 volts. >Is your 2.4 volts centered around (+-1.2) or is it 8to10.5 volts?