0
\$\begingroup\$

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).

Non-Inverting Amp with bias current

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: $$V_o = \left(1+\frac{R_N}{R_1}\right)V_i + \frac{R_1 R_g - R_1 R_N + R_g R_N}{R_1}I_b - R_N I_O$$

\$\endgroup\$
1
\$\begingroup\$

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

Non-Inverting Amp with bias current

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:

enter image description here

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 enter image description here

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.

\$\endgroup\$
4
  • \$\begingroup\$ Nice answer, thanks! Still I am struggling with the IB part. How to get the second part after the minus? I somehow do not see the loop you are going. \$\endgroup\$
    – JLo
    Feb 17 '17 at 21:46
  • \$\begingroup\$ Notice that \$\frac{R1+R_N}{R1}\$ is equal to \$(\frac{R_N}{R1} +1)\$ and \$IB*Rg\$ is a voltage at VP node. \$\endgroup\$
    – G36
    Feb 18 '17 at 14:30
  • \$\begingroup\$ I edit my answer \$\endgroup\$
    – G36
    Feb 18 '17 at 14:54
  • \$\begingroup\$ And thanks again! I figured my problem was the voltage direction over the current sources. I assumed them in the same direction as the current but this obviously wrong in this convention. So I always got wrong sings. \$\endgroup\$
    – JLo
    Feb 18 '17 at 17:56
0
\$\begingroup\$

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.

\$\endgroup\$
3
  • \$\begingroup\$ Actually I am interested in the math :). I understand what is going on but unfortunately can not derive it in maths. Edited my question to show what I did. \$\endgroup\$
    – JLo
    Feb 17 '17 at 13:51
  • \$\begingroup\$ Unfortunately this boils down to a math question that happens to be related to electronics. Helping you derive the math isn't really on-topic for this site so, maybe ask on SE maths. You have my sympathy but unfortunately not another 30 minutes of my time. Make life easier by just examining the bias current effect to see if that part of the equation falls out. Eat smaller chunks of food etc.. \$\endgroup\$
    – Andy aka
    Feb 17 '17 at 14:36
  • \$\begingroup\$ Thanks for your help anyway! I delegate the actual derivation to Matlabs symbolic computing, so I do not think there is an error in the equation solving. Rather the error is in the initial equations describing the circuit/Opamp behavior where the math people can not help, I fear. \$\endgroup\$
    – JLo
    Feb 17 '17 at 14:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.