AD620 INPUT PROTECTION

[lordmag]

in the ad620 datasheet the max differential input is 25 V so i want t o make protection for my input amplifier what do you think about this circuit is that the right one ?

my application is that i want to measure the VOLTAGE of a contact RESISTANCE like what we see in the attach 2

 

[(*steve*)]

That will limit either input to prevent them from exceeding either supply rail by more than 0.7V.

If you need better than this, you can replace the diodes with zener diodes, and as long as their voltage is significantly greater than 1/2 of your supply voltage (greater then say, 12V in this case), you can limit the excursion to some number of volts away from either supply rail.

Another approach is to place a pair of diodes across the op amps inputs. This will prevent them from differing by more than 0.6V, which is plenty in your application. You still need a way of ensuring that both inputs are not taken further than the supply rails though. You could add these diodes to your existing circuit.

Beware leakage currents through the diodes though. In our case the input resistance should be low enough that this won't have any substantial impact.

 

[lordmag]

Can you explain for me more
this type of amplifier protection !!
and why we use here resistance 10 k

 

[lordmag]

hi ,
any one can explain for me this type of ad620 amplifier protection !!!

 

[lordmag]

i' m using 1N4148 diode

 

[lordmag]

this is my application

 

[duke37]

As *steve* says, a diode will conduct if the input exeeds the voltage of the power supply by 0.7V. This limits the input voltage to the amplifier.

That circuit looks very complicated, the voltage measured will be 0V when the contact is closed and 13.5V when it is open. Depending on the detail of what is going on, you could do the same with a resistor and a LED.

 

[lordmag]

duke , you want to say some thing like this

 

[(*steve*)]

Why did you remove three of the four protection diodes in that last circuit?

Also why a gain of 495? Each volt measured on the voltmeter will be 0.00202020202 volts across the switch, hardly a neat number.

I would go a gain of 100, or 1000, or some similar number giving a more easily interpreted number.

With the current setup, the voltmeter will read about 12V with the switch open simply because that's likely your supply voltage. It would try to swing the output to about 6000 volts.

 

[lordmag]

oki

look at this circuit :
if R contact is ON
R_contact = 0.2 ohm
current I = 10 mA
==> U_ contact = 2 mV.

for the AD620 RG = 100 ohm ===> G = 495

so that Vout = G * U_contact = 0.99 V
if R_ contact is OFF

?? what is happening )

 

[lordmag]

so what do you think steve

 

[(*steve*)]

I explained what I thought was wrong in my previous post. Nothing has changed