Offset voltage comes to saturation in coupling amplifier circuit using OpAmp with Gain=100

Question

I did get in troubles with my student project circuit below!

The purpose is to amplify difference between two inputs. Specifically, 20mV and 10mV are connected with input to get 1V at the last OpAmp's output.

The problem is when two input are grounded,the output offset voltage of OA1 and OA2 comes to saturation in reality. I attempted to find out what problem is but it doesn't make sense.

As I figured, I used a potentiometer to adjust the offset voltage because I thought the offset voltage is just 20-30 mV, but in fact, it wasn't.

What's the cause of this problem?

Here is my circuit on proteus.

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that is exactly what you asked me to do..right?

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EDIT

Let me explain how I calculated the component values in the circuit. First, with \$U_1\$ we have that $$ \begin{split} V_- &=\frac{(V_1-V_{01})R_2}{R_1+R_2}+V_{01}\\ V_+ &=\frac{(V_{02}-V_{01})(R_3+R_4)}{R_3+R_4+R_5}+V_{01}\\ \\ \implies & \frac{(V_1-V_{01}) R_2}{R_1+R_2}=\frac{(V_{02}-V_{01}).(R_3+R_4)}{R_3+R_4+R_5}\quad (\text{because }V_+=V_-) \end{split} $$
$$ \implies V_{01}\left[\frac{R_2}{R_1+R_2} - \frac{R_3+R_4}{R_3+R_4+R_5}\right]=V_1\frac{R_2}{R_1+R_2}-V_{02}\frac{R_3+R_4}{R_3+R_4+R_5}\label{1}\tag{1} $$ As for U1, the same holds for U2: we have $$ \begin{split} V_- &=\frac{(V_2-V_{02})R_7}{R_7+R_6}+V_{01}\\ V_+ &=\frac{(V_{01}-V_{02})(R_4+R_5)}{R_3+R_4+R_5}+V_{02}\\ \end{split} $$
And again, since \$V_+\$ is also equal to \$V_-\$, $$ V_{02}\left[\frac{R_7}{R_7+R_6} - \frac{R_4+R_5}{R_3+R_4+R_5}\right]=V_2\frac{R_2}{R_1+R_2}-V_{01}\frac{R_4+R_5}{R_3+R_4+R_5}\label{2}\tag{2} $$ Assuming that \$R_3=R_5, R_1=R_6, R_2=R_7\$, from \eqref{1} and \eqref{2} we have $$ V_{02}-V_{01}=\frac{R_2(2R_3+R_4)}{R_4(R_2+2R_1)+2R_3R_1}(V_1-V_2)\label{g}\tag{G} $$ (I tested this equation on Proteus and it seems to be right) So now... how can I chose resistor’s value suitably to make this circuit work well?

Answer 1

Edit: by using the circuit formulas for the input voltages \$V_+\$ and \$V_-\$ of the OpAmps \$U_1\$ and \$U_2\$ and taking into account that their differential gain \$A_d\$, however being very high, is finite i.e. we have $$ V_{01}=A_{d}(V_{+}-V_{-})\big|_1\quad\qquad\quad V_{02}=A_{d}(V_{+}-V_{-})\big|_2 \label{ad}\tag{AD} $$ I found that the circuit is unstable, no matter how the values of resistors \$R_3, R_4, R_5\$ are chosen and I show below how to prove it. Of course, the topology of the positive feedback loop could be modified in order to stabilize the circuit, but then you should answer to this very basic question: "Why do a positive/negative feedback mix is needed for this application?"

The problem with your circuit is that the resistors \$R_3,R_4, R_5\$ add a positive feedback loop to the couple of operational amplifiers \$U_1\$ and \$U_2\$: when the two inputs are grounded, the input stage of your amplifier looks as below

^{simulate this circuit – Schematic created using CircuitLab}

This implies that every circuit asymmetry (not only the offset voltage, but perhaps also noises or leakage currents) make the outputs of the two amplifiers rising up to \$\pm V_{o_\mathrm{sat}}\$ (the sign of the output depends on the sign of the maximum circuit asymmetry your circuit experiences). And the problem cannot be solved simply by changing the values of the feedback resistors, as the analysis below shows.
The starting point are the expressions of the voltages at the input of the OpAmps: $$ U_1\:\left\{ \begin{split} V_- &=\frac{(V_1-V_{01})R_2}{R_1+R_2}+V_{01}\\ V_+ &=\frac{(V_{02}-V_{01})(R_3+R_4)}{R_3+R_4+R_5}+V_{01}\\ \end{split}\right. \\ U_2\: \left\{ \begin{split} V_- &=\frac{(V_2-V_{02})R_7}{R_7+R_6}+V_{02}\\ V_+ &=\frac{(V_{01}-V_{02})(R_4+R_5)}{R_3+R_4+R_5}+V_{02}\\ \end{split}\right. $$ Then, by using \eqref{ad} we get $$ \begin{split} V_{01} & = A_d (V_+ - V_-)\big|_1\\ & = \left[\frac{R_2}{R_1+R_2}(V_1 - V_{01}) - \frac{R_4+R_5}{R_3+R_4+R_5}(V_{02} - V_{01})\right] A_d\big|_1\\ \\ V_{02} & = A_d (V_+ - V_-)\big|_2\\ & = \left[\frac{R_7}{R_7+R_6}(V_1 - V_{01}) - \frac{R_4+R_5}{R_3+R_4+R_5}(V_{01} - V_{02})\right]A_d\big|_2. \end{split} $$ Now, assuming \$A_d\big|_1=A_d\big|_2\$ and \$R_1=R_7\$, \$R_2=R_6\$ by the above formulas we get $$ \begin{split} (V_{01} - V_{02}) &\left[1 + \left( \frac{R_2}{R_1+R_2} - 2 \frac{R_4+R_5}{R_3+R_4+R_5}\right)A_d \right]\\ & = \frac{A_d R_2}{R_1+R_2}(V_1 - V_2) \end{split}\label{dg}\tag{DG} $$ From equation \eqref{dg} we see that in order for the feedback to bee negative, i.e. for the term multiplying the output voltage \$(V_{01} - V_{02})\$ to be \$>1\$, the following condition needs to be fulfilled $$ \frac{R_2}{R_1+R_2} \gg 2 \frac{R_4+R_5}{R_3+R_4+R_5}\label{fbc}\tag{FBC} $$ However, for every physically admissible value of the resistances involved, condition \eqref{fbc} is always false: to see this, note that $$ \frac{R_2}{R_1+R_2}=\kappa <1 $$ and if we assume \$R_5 = R_3\$, \eqref{fbc} implies that $$ \begin{split} 2\frac{R_4+R_3}{2R_3+R_4} & \ll \kappa\\ 2 (R_4+R_3) & \ll \kappa (2R_3+R_4)\\ &\Updownarrow\\ 2(1-\kappa)R_3 &\ll (\kappa -2)R_4 <0 \end{split} $$ which is clearly false.

Possible solutions

• Change the topology of your circuit in order to reproduce the standard instrumentation amplifier topology. This will assure you have a high gain and an adequate CMRR i.e. you can
• Assuming that you really need a circuit with positive feedback as the one you designed, you have to change the structure of your feedback networks: for example you can connect two resistors between the reference ground and each positive input of \$U_1\$ and \$U_2\$. Then \eqref{fbc} is no more requried and, using the same method of analysis shown above you should find a new on and see if is physically realizable.

EDIT: In order to keep the overall gain as high as required, an easy way to proceed is to change the gain of the second stage: for example, putting \$R_8=R_{10}=1.5\mathrm{k}\Omega\$ guarantees that the second stage of the circuit, i.e the differential amplifier made of \$U_3\$ has a voltage gain \$A_v\$ of \$100\$, making the overall voltage gain strictly greater. Note however that nothing is for free in circuit design: rising the gain of a stage lowers its bandwidth since the \$GBW\$ is kept (customarily) constant by circuit design, thus you can amplify weaker but slower signals. After all, you are using the almighty LM741.

Answer 2

Try this circuit for U1 & U2 (same U3 differential amplifier as in your circuit, but increase that 100 ohm load to at least a few K).

Looks like you've got positive feedback in your circuit that is causing the op-amps to rail.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 3

Your Op Amps have more input voltage offset than the DC input applied. So null adjust would be required.

Both circuits work but the non-inverting input has better COmmon Mode rejection even with this Ideal Op Amp and ideal resistors.

Here I applied the same 10 mVdc differentials but added 1Vp sine common mode.

here with AC common mode removed. The output DC is almost correct and errors due to insufficient gain used in my model. (100k) just FYI.

Whenever signals out do not make sense examine all the voltages at every node and see why it makes sense. Then correct your offset with null pot on pins 1,4 to Vee.

Read up on INA IC's which use laser matched R's