TS9222 i

[KR.5]

Hi, Below is the design

1) input voltage at inverting side is 8V with the 4.7k series connecting to inv terminal

2) voltage at non inverting terminal is fixed at 3.5V.

3) the difference between inputs is greater than 1V(Vid=1V).

4) My question is whether OPAMP have clamp diode at both inputs?. If it is there, kindly share that schematic

5) As per datasheet 1mA is specified as the max input current. pls share one example calculation for calculating input current

[Peter BENSCH]

Welcome, @KR.5​, to the community!

Well, the differential voltage in your specific case is 4.5V and is thus considerably greater than the permissible 1V. There are no protective diodes between the inputs that can be used for this purpose, but you can assume that there are parasitic pn junctions that can only tolerate a maximum current of 1mA.

With the existing 4.7k resistor, you limit the current to ~957µA, which is below the 1mA limit, but it is still recommended to connect resistors with half the value to each the input lines. This has the advantage of balancing the voltage drops caused by the bias current.

Does it answer your question?

Regards

/Peter

[KR.5]

Hi Peter,

Thanks for the support. I provided same value of resistor to non inverting terminal as well. 957uA close to the 1mA max limit. I need to increase the value a bit.

Regards

Kesavan R

[Peter BENSCH]

OK, good.

If the problem is solved, please mark this thread as answered by selecting Select as best, as also explained here. This will help other users find that answer faster.

Good luck!

/Peter

[KR.5]

do i need to consider the parasitic diode forward drop in the current calculation?

[Peter BENSCH]

You could take the forward voltage of the parasitic diodes into account (approx. 0.65V), but you have more reserve if you take it as zero. In your example with differential voltage 4.5V you get the (total) resistances:

• forward voltage = 0.65V
  • R = (4.5V-0.65V)/1mA = 3.85k
• forward voltage = 0
  • R = (4.5V-0V)/1mA = 4.5k

In the second case, the presence of the forward voltage results in a smaller current of approx. (4.5V-0.65V)/4.5kohms = 0.855mA, which is well below the limit.

Good luck!

/Peter