How to set up equations to solve for v_os and input bias current?

[dla]

Homework Statement

A particular inverting op-amp has a gain of -100 V/V, and an input resistor 100 K. The output voltage is 9.32 V when measured with the input open and has a 9.10 V. when measuring with the input grounded. What is the input offset voltage and bias current?

The circuit is attached

Homework Equations

The Attempt at a Solution

From the gain and the input resistor I know that R_2 has to be 10 MΩ. But I just can't seem to set my equations up properly.

I get confused because bias current flows from output to V_1. But then offset voltage is at the inverting terminal. Since current is flowing opposite of V_os, I thought it'd be simply V_os - I_B1 * R_2 for when it is grounded. I don't know how to get the equation for when the input is open though since we're not given V_in? I feel like I'm complicating this way more than needed.

When grounded I got this..

$\large 9.10={v_{os}}- I_{B1}{10MΩ}$The answer for the equations are actually, for open input
$\large 9.32 = v_{os} + 10000 I_{B1}$

and for ground..
$\large 9.10 = v_{os}*101 +10000 I_{B1}$

I don't get how they got those directions, all the terms are positive. And for when the input is grounded how did they get the gain of 101 but for input there is gain of only 1 in front of v_os?

If anyone could explain I would appreciated very much, really want to get this concept down.

 

[rude man]

1. The circuit is attached

Where?

 

[dla]

Sorry I just attached it. It's just a simple inverting configuration. Where R_2 is 1 MΩ and R_1=100kΩ

 

[rude man]

OK, so write nodal equations for two cases:
1. where the input resistor is floating (no input: nothing connected to the input.)
2: ground the input as shown on your diagram.

In both cases you have a bias current flowing into the - input of the op amp. Other than that your op amp is ideal (infinite gain, no offset voltage). In paticular, this means the voltages at the op amp input terminals are equal. The offset voltage as depicted in your diagram acts as an external input voltage to the + input. It's also present in both cases.

 

[gneill]

It looks as though they've used a value for R2 that's inconsistent with the problem as described. Your value of 10MΩ is fine.

Their equations are correct in form, but they've used an incorrect value for R2 in generating them (perhaps the problem was "updated" at some point with a different gain value, but they failed to recalculate R2).

To find the equations I suggest applying KCL at the R1\R2 junction.

 

[rude man]

Yes, the problem appears to be misstated. The solution of the problem as given is for the offset voltage to be huge and for the bias current to be vanishingly small.

And if you reverse the shorted-input and the open-input output voltages the result is the same.