Non-inverting-Amp bias current: understanding output error

Question

I have a book (Tietze, Schenk: Electronic circuits) that shows this equivalent circuit of a non-inverting amp. \$I_B\$ is the bias current, \$I_o\$ the offset current (difference of bias currents).

The output is given as $$V_o = \left(1+\frac{R_N}{R_1}\right)V_i + I_B\left(R_N - \frac{R_g(R_1 + R_N)}{R_1}\right) + \frac{I_o}{2} \left(R_N + \frac{R_g(R_1 + R_N)}{R_1}\right)$$

Edit & 2. Edit

Although I somehow understand the formula, I do not understand how to derive the parts of the equation with \$I_B\$ and \$I_o\$, see below

I tried to do the math but maybe something with my initial equations is wrong. I tried the system of these equations $$ V_p = V_i + R_g I_B$$ $$ V_n = \frac{R_1}{R_1+R_N}V_o + \frac{R_1 R_N}{R_1 + R_N} I_O + \frac{R_1 R_N}{R_1 + R_N} I_B$$ $$ V_p = V_n$$ which gives something similar, but not the same: $$V_o = \left(1+\frac{R_N}{R_1}\right)V_i + \frac{R_1 R_g - R_1 R_N + R_g R_N}{R_1}I_b - R_N I_O$$

Answer 1

Interesting, because when I use a Superposition Theorem for this circuit

The equation for \$V_o\$ if I left \$I_O\$ alone (\$V_{IN}\$ is short and \$I_B\$ are open) is:

$$V_O = I_O*R_N$$

for \$I_B\$ we have:

$$ V_O = I_BR_N - I_BR_g\left(\frac{R_N}{R_1}+1\right) = I_B\left(R_N - \frac{R_g(R_1+R_N)}{R_1}\right)$$

So finally we have:

$$ V_O = \left(\frac{R_N}{R_1}+1\right) V_{IN}+I_B\left(R_N - \frac{R_g(R_1 + R_N)}{R_1}\right)+I_OR_N$$

What do you think?

EDIT

After some thought about this I get to this conclusion:

\$I_B = \frac{I_P+I_N}{2}\$

\$I_O = |I_P+I_N| \$

Where \$I_P\$ is a Non-Inverting input bias current and \$I_N\$ - Inverting Input bias current.

From this I get:

\$ I_P = I_B-0.5I_O \$

and

\$ I_N = I_B+0.5I_O \$

And the circuit diagram will look like this:

And for this circuit this equation is correct ( from Tietze, Schenk: Electronic circuits)

$$V_o = \left(1+\frac{R_N}{R_1}\right)V_i + I_B\left(R_N - \frac{R_g(R_1 + R_N)}{R_1}\right) + \frac{I_o}{2} \left(R_N + \frac{R_g(R_1 + R_N)}{R_1}\right)$$

Try superposition and noticed that the \$V_P\$ voltage is equal to:

\$V_p=-(I_B-\frac{I_o}{2})R_g + V_i\$

EDIT 2

Let us try to find \$V_o\$ for these two cases

The voltage at \$V_p\$ node is

$$ V_p = -I_B*R_g$$

and the output voltage is \$V_O' = V_p*A_v\$ and the Non-Inverting gain is \$A_v = (1+\frac{R_N}{R_1}) = \frac{R_1+R_N}{R_1}\$ therefore

$$V_O' = -I_B*R_g*\frac{R_1+R_N}{R_1}=-I_B \frac{R_g(R_1+R_N)}{R_1} $$

And for the second case, we have

$$V_O'' = I_B*R_N$$

And finally

$$V_O = V_O'+V_O'' = -I_B \frac{R_g(R_1+R_N)}{R_1} + I_BR_N = I_BR_N -I_B \frac{R_g(R_1+R_N)}{R_1} = $$ $$=I_B\left(R_N - \frac{R_g(R_1 + R_N)}{R_1}\right) $$

If we repeat this for \$I_o\$ current we will get the plus term because now \$V_p\$ voltage is positive.

Answer 2

Why the minus before the second part of the \$I_B\$ part?

That minus sign means that if \$R_g\$ is perfectly equal to the parallel resistance formed by \$R_N\$ and \$R_1\$ then there will be no effect due to bias currents: -

\$\dfrac{R_N R_1}{R_N+R_1} = R_g\$ and rearranging, \$R_N = \dfrac{R_g(R_1+R_N)}{R_1}\$.

You could work it out a more long-winded way but I prefer, for simplicity, to try and see what that part of the equation actually means.

Where does the factor 1/2 in \$Io\$ come from?

It looks like \$I_O\$ is shared equally (but with opposite signs) on both inputs and the same procedure is used but this time, because of the negative value of \$I_O/2\$ transferred to the non-inverting input, the part of the equation inside the brackets has a positive sign and not negative.

As I said earlier, if you want to do a page of math to work this out, that's your call.