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let's consider this two Op-Amp amplifier:

enter image description here

From what I can understand, it consists of two amplifiers:

  • A current to voltage low pass filter (at left)
  • A non-inverting amplifier (at right)

Well, there are some things I can't understand:

  1. Why is there a feedback resistor R1? From my analysis (see next), the circuit behaviour would be the similar without it (it would be always a current to voltage low pass filter followed by a non-inverting amplifier)?

  2. What is the role of the DC voltage Vs? From my analysis, it seems it's unuseful since it doesn't affect the neither first amplifier (the current I1 depends only on Vout) nor the second one.

MY ANALYSIS:

Let's evaluate the transfer function (transimpedance Vout/Iin) in the frequency domain.

Such an amplifier consists of two amplifying blocks which cannot be analyzed as totally indipendent circuits since there is a feedback resistor R1 that connects them.

For the first block, one can say that:

• The negative feedback provides 0V at - input pin (virtual short).

• By KCL, current on the parallel impedance Ra//Ca is equal to $$I_{a}=I_{IN}-I_{1}=I_{IN}+\frac{V_{OUT}}{R_{1}}$$

• By KVL and Ohm's Law, the output voltage of the first op-amp is equal to $$V_{a}=-I_{a}Z_{a}=-\frac{R_{a}}{1+j\omega C_{a}R_{a}}\left(I_{IN}+\frac{V_{OUT}}{R_{1}}\right)$$

For the second block:

• The negative feedback provides Va at - input pin (virtual short).

• By KVL and Ohm's Law, the final output voltage is equal to $$V_{OUT}=V_{a}\left(1+\frac{R_{3}}{R_{2}}\right)$$

Therefore:

\begin{cases} V_{a} & =-\frac{R_{a}}{1+j\omega C_{a}R_{a}}\left(I_{IN}+\frac{V_{OUT}}{R_{1}}\right)\\ V_{OUT} & =V_{a}\left(1+\frac{R_{3}}{R_{2}}\right) \end{cases}

So, by putting the first equation inside the second one: $$V_{OUT}=-\frac{R_{a}}{1+j\omega C_{a}R_{a}}\left(I_{IN}+\frac{V_{OUT}}{R_{1}}\right)\left(1+\frac{R_{3}}{R_{2}}\right)$$ that means $$V_{OUT}+\frac{R_{a}}{1+j\omega C_{a}R_{a}}\frac{V_{OUT}}{R_{1}}\left(1+\frac{R_{3}}{R_{2}}\right)=-\frac{R_{a}}{1+j\omega C_{a}R_{a}}I_{IN}\left(1+\frac{R_{3}}{R_{2}}\right)$$ that means$$ \frac{V_{OUT}}{I_{IN}}(j\omega)=\frac{-\frac{R_{a}}{1+j\omega C_{a}R_{a}}\left(1+\frac{R_{3}}{R_{2}}\right)}{1+\frac{R_{a}}{R_{1}\left(1+j\omega C_{a}R_{a}\right)}\left(1+\frac{R_{3}}{R_{2}}\right)}=-\frac{\frac{R_{a}\left(1+\frac{R_{3}}{R_{2}}\right)}{1+\frac{R_{a}}{R_{1}}\left(1+\frac{R_{3}}{R_{2}}\right)}}{1+j\omega\frac{C_{a}R_{a}}{1+\frac{R_{a}}{R_{1}}\left(1+\frac{R_{3}}{R_{2}}\right)}}$$

that is the final transimpedance transfer function of this amplifier, which behaves as a first order low-pass filter. If, as written at the beginning, R1=10k, R2=3k, R3=5k, R4=1k, R1=10k, the resulting transfer function will be:$$\frac{V_{OUT}}{I_{IN}}(j\omega)=-\frac{\frac{R_{a}\frac{8}{3}}{1+\frac{R_{a}}{10k\Omega}\frac{8}{3}}}{1+j\omega\frac{C_{a}R_{a}}{1+\frac{R_{a}}{10k\Omega}\frac{8}{3}}}$$

You may see that, if I'm not wrong:

  1. R1 doesn't change so much the transfer function. If it were infinite (i.e. no R1), the circuit would again represent a low pass filter/amplifier. What is its purpose?

  2. I don't see any role in Vs. It's absent in the transfer function and I can't understand why it is included in the circuit (together with R4).

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  • \$\begingroup\$ It might be more difficult to explain than it is to simulate. Be sure to use uic, or startup: it's an oscillator with a triangle and a square wave. The output of one is connected to the input of the other. \$\endgroup\$ – a concerned citizen Apr 25 at 13:55
  • \$\begingroup\$ Is this a DAC output stage and 4V is the reference voltage ? \$\endgroup\$ – tobalt Apr 25 at 13:57
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It's too late for math.

enter image description here

The second opamp is non-inverting with gain 1+R3/R2 = 2.66666 ; I labeled "op1" the output of the first opamp, so V(out) = 2.666*V(op1). Since the input of the first amp is a virtual ground, current through R1 (the resistor on top) is v(out)/R1 = v(op1)*2.6666/R1, so I remove the second opamp, connect R1 to the output of the first, and divide its value by 2.666 to keep the same current. The simulator agrees.

Since we have two resistors in parallel, I remove one and replace the other with their parallel resistance, giving the same result.

The result is a standard integrator. It will behave a bit differently at HF when the opamp poles stick out, though.

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    \$\begingroup\$ I am impressed by the other answerers' ability to derive transfer function analytically. However I have never seen the point of that exercise in my career. This answer shows how much faster one can get results with a sim + intuition ;) \$\endgroup\$ – tobalt Apr 30 at 14:16
  • \$\begingroup\$ Yeah I am impressed by the math too lol. \$\endgroup\$ – bobflux Apr 30 at 16:31
  • \$\begingroup\$ Math is the foundation of the simulation. The simulation is just doing the math for you. Understanding the math gives you insight on how and to what degree the component values and operational parameters influence the behavior of the circuit. \$\endgroup\$ – Pau Coma Ramirez Apr 30 at 17:42
  • \$\begingroup\$ Equations are the same whether you write them in math language or in schematic language. All I did was a change of variable. \$\endgroup\$ – bobflux Apr 30 at 18:00
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I've not seen this particular topology before, nor have I considered it before. Here's a modified diagram that I made before analyzing it:

enter image description here

Perfunctory calculations are as follows. (I used iin as your input current, ia as the output current from the left side amplifier, and ib as the output current from the right side amplifier.)

eq1 = Eq( va/(1/s/ca) + va/ra, vi/(1/s/ca) + vi/ra + ia )
eq2 = Eq( vb/r3 + vb/r2, vo/r3 )
eq3 = Eq( vo/r4 + vo/r3 + vo/r1, vs/r4 + vb/r3 + vi/r1 + ib)
eq4 = Eq( vi/(1/s/ca) + vi/ra + vi/r1, va/(1/s/ca) + va/ra + vo/r1 + iin)
eq5 = Eq( va, vb )
ans = solve( [eq1, eq2, eq3, eq4, eq5], [ia, ib, va, vb, vo] )
ans[vo].subs({vi:0})/iin

    -r1*ra*(r2 + r3)/(ca*r1*r2*ra*s + r1*r2 + ra*(r2 + r3))

t0 = ra*ca
w0 = (r1*r2+ra*(r2+r3))/r1/r2/t0
K = (-r1*ra*(r2+r3)/r1/r2/t0)/w0
simplify( K )

    -r1*ra*(r2 + r3)/(r1*r2 + ra*(r2 + r3))

That's not all that helpful. But at least the algebra is right and I don't have to worry about manual mistakes.

So, by inspection, I now find:

$$\begin{align*} \tau_{_0}&=R_a C_a\\\\ A_v&=1+\frac{R_3}{R_2}\\\\ R_v&=A_v\,R_a\\\\ \omega_{_0}&=\bigg[1+\frac{R_v}{R_1}\bigg]\cdot \frac1{\tau_{_0}}\\\\ R&=R_1 \mid\mid R_v\\\\ \frac{V_o}{I_i}&=-R\frac{\omega_{_0}}{s+\omega_{_0}}\end{align*}$$

The full system transimpedance \$R\$ has units of \$\Omega\$.

There's a lot to be gained from the above.

  1. The voltage gain opamp on the right provides a unitless voltage gain of \$A_v\$.
  2. \$A_v\$ amplifies the left opamp's transimpedance, \$R_a\$, in order to provide a new transimpedance, \$R_v\$.
  3. The system transimpedance is dominated by the smaller of \$R_1\$ or \$R_v\$.
  4. The 1st order corner frequency, \$\omega_{_0}\$, is shifted away from \$\frac1{\tau_{_0}}\$ by the factor, \$1+\frac{R_v}{R_1}\$.
  5. If \$R_v\gg R_1\$ then \$R_1\$ sets the system transimpedance, while allowing \$\omega_{_0}\$ to be independently adjusted by the factor \$1+\frac{R_v}{R_1}=1+A_v\frac{R_a}{R_1}\$.

Note that those opamps have their negative reference grounded and their outputs cannot go negative with respect to ground. So the direction of your \$I_{in}\$ current must be directed oppositely as shown on your diagram.

The purpose of \$R_1\$ is to set the transimpedance, assuming \$A_v\cdot R_a\gg R_1\$.

The transfer function is indeed independent of \$V_s\$ and \$R_4\$ and I honestly don't have a suggestion except the idea that the opamps are open-collector types. Is that possible?

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First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course).

Well, we are trying to analyze the following opamp-circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$ \begin{cases} \text{I}_4=\text{I}_1+\text{I}_7\\ \\ \text{I}_7=\text{I}_2+\text{I}_3\\ \\ 0=\text{I}_1+\text{I}_2+\text{I}_8\\ \\ \text{I}_9=\text{I}_5+\text{I}_{10}\\ \\ 0=\text{I}_3+\text{I}_6+\text{I}_{10} \end{cases}\tag1 $$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_1-\text{V}_2}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_1-\text{V}_4}{\text{R}_3}\\ \\ \text{I}_5=\frac{\text{V}_3}{\text{R}_4}\\ \\ \text{I}_5=\frac{\text{V}_4-\text{V}_3}{\text{R}_5}\\ \\ \text{I}_6=\frac{\text{V}_\text{n}-\text{V}_4}{\text{R}_6} \end{cases}\tag2 $$

Substitute \$(2)\$ into \$(1)\$, in order to get:

$$ \begin{cases} \text{I}_4=\frac{\text{V}_1-\text{V}_2}{\text{R}_1}+\text{I}_7\\ \\ \text{I}_7=\frac{\text{V}_1-\text{V}_2}{\text{R}_2}+\frac{\text{V}_1-\text{V}_4}{\text{R}_3}\\ \\ 0=\frac{\text{V}_1-\text{V}_2}{\text{R}_1}+\frac{\text{V}_1-\text{V}_2}{\text{R}_2}+\text{I}_8\\ \\ \text{I}_9=\frac{\text{V}_3}{\text{R}_4}+\text{I}_{10}\\ \\ \text{I}_9=\frac{\text{V}_4-\text{V}_3}{\text{R}_5}+\text{I}_{10}\\ \\ 0=\frac{\text{V}_1-\text{V}_4}{\text{R}_3}+\frac{\text{V}_\text{n}-\text{V}_4}{\text{R}_6}+\text{I}_{10} \end{cases}\tag3 $$

Now, using an ideal opamp, we know that:

  • $$\text{V}_1=0\space\text{V}\tag4$$
  • $$\text{V}_x:=\text{V}_+=\text{V}_-=\text{V}_2=\text{V}_3\tag5$$

So we can rewrite equation \$(3)\$ as follows:

$$ \begin{cases} \text{I}_4=\frac{0-\text{V}_x}{\text{R}_1}+\text{I}_7\\ \\ \text{I}_7=\frac{0-\text{V}_x}{\text{R}_2}+\frac{0-\text{V}_4}{\text{R}_3}\\ \\ 0=\frac{0-\text{V}_x}{\text{R}_1}+\frac{0-\text{V}_x}{\text{R}_2}+\text{I}_8\\ \\ \text{I}_9=\frac{\text{V}_x}{\text{R}_4}+\text{I}_{10}\\ \\ \text{I}_9=\frac{\text{V}_4-\text{V}_x}{\text{R}_5}+\text{I}_{10}\\ \\ 0=\frac{0-\text{V}_4}{\text{R}_3}+\frac{\text{V}_\text{n}-\text{V}_4}{\text{R}_6}+\text{I}_{10} \end{cases}\tag6 $$

Now, we can solve for the transfer function:

$$\mathcal{H}:=\frac{\text{V}_4}{\text{I}_4}=-\frac{\text{R}_1\text{R}_2\text{R}_3\left(\text{R}_4+\text{R}_5\right)}{\text{R}_1\text{R}_2\text{R}_5+\text{R}_4\left(\text{R}_2\text{R}_3+\text{R}_1\left(\text{R}_2+\text{R}_3\right)\right)}\tag7$$

Where I used the following Mathematica-code:

In[1]:=Clear["Global`*"];
V1 = 0;
V2 = Vx;
V3 = Vx;
FullSimplify[
 Solve[{I4 == I1 + I7, I7 == I2 + I3, I5 == I1 + I2 + I8, 
   I9 == I5 + I10, 0 == I3 + I6 + I10, I1 == (V1 - V2)/R1, 
   I2 == (V1 - V2)/R2, I3 == (V1 - V4)/R3, I5 == V3/R4, 
   I5 == (V4 - V3)/R5, I6 == (Vn - V4)/R6}, {I1, I2, I3, I5, I6, I7, 
   I8, I9, I10, Vx, V4}]]

Out[1]={{I1 -> (I4 R2 R3 R4)/(R2 R3 R4 + R1 (R2 + R3) R4 + R1 R2 R5), 
  I2 -> (I4 R1 R3 R4)/(R2 R3 R4 + R1 (R2 + R3) R4 + R1 R2 R5), 
  I3 -> (I4 R1 R2 (R4 + R5))/(R2 R3 R4 + R1 (R2 + R3) R4 + R1 R2 R5), 
  I5 -> -((I4 R1 R2 R3)/(R2 R3 R4 + R1 (R2 + R3) R4 + R1 R2 R5)), 
  I6 -> ((I4 R1 R2 R3 (R4 + R5))/(
    R2 R3 R4 + R1 (R2 + R3) R4 + R1 R2 R5) + Vn)/R6, 
  I7 -> I4 - (I4 R2 R3 R4)/(R2 R3 R4 + R1 (R2 + R3) R4 + R1 R2 R5), 
  I8 -> -((I4 R3 (R2 R4 + R1 (R2 + R4)))/(
    R2 R3 R4 + R1 (R2 + R3) R4 + R1 R2 R5)), 
  I9 -> -(((I4 R1 R2 ((R4 + R5) R6 + R3 (R4 + R5 + R6)))/(
     R2 R3 R4 + R1 (R2 + R3) R4 + R1 R2 R5) + Vn)/R6), 
  I10 -> -(((I4 R1 R2 (R4 + R5) (R3 + R6))/(
     R2 R3 R4 + R1 (R2 + R3) R4 + R1 R2 R5) + Vn)/R6), 
  Vx -> -((I4 R1 R2 R3 R4)/(R2 R3 R4 + R1 (R2 + R3) R4 + R1 R2 R5)), 
  V4 -> -((I4 R1 R2 R3 (R4 + R5))/(
    R2 R3 R4 + R1 (R2 + R3) R4 + R1 R2 R5))}}

My equation was also confirmed using LTspice.


When we want to apply the derivation from above to your circuit we need to use Laplace transform (I will use lower case function names for the functions that are in the (complex) s-domain, so \$\text{y}\left(\text{s}\right)\$ is the Laplace transform of the function \$\text{Y}\left(t\right)\$):

$$\text{R}_2=\frac{1}{\text{sC}}\tag8$$

So, we can rewrite the transfer function as:

$$\mathscr{H}\left(\text{s}\right)=-\frac{\text{R}_1\cdot\frac{1}{\text{sC}}\cdot\text{R}_3\left(\text{R}_4+\text{R}_5\right)}{\text{R}_1\cdot\frac{1}{\text{sC}}\cdot\text{R}_5+\text{R}_4\left(\frac{1}{\text{sC}}\cdot\text{R}_3+\text{R}_1\left(\frac{1}{\text{sC}}+\text{R}_3\right)\right)}=$$ $$-\frac{\text{R}_1\text{R}_3\left(\text{R}_4+\text{R}_5\right)}{\text{R}_1\text{R}_5+\text{R}_4\left(\text{R}_3+\text{R}_1\left(1+\text{CR}_3\text{s}\right)\right)}\tag9$$

Now, when working with sinusoidal signals we can use \$\text{s}:=\text{j}\omega\$ (where \$\text{j}^2=-1\$ and \$\omega=2\pi\text{f}\$ with \$\text{f}\$ is the frequency of the input signal in Hertz). So, we get:

$$\underline{\mathscr{H}}\left(\text{j}\omega\right)=-\frac{\text{R}_1\text{R}_3\left(\text{R}_4+\text{R}_5\right)}{\text{R}_1\text{R}_5+\text{R}_4\left(\text{R}_3+\text{R}_1\left(1+\text{CR}_3\text{j}\omega\right)\right)}=$$ $$-\frac{\text{R}_1\text{R}_3\left(\text{R}_4+\text{R}_5\right)}{\text{R}_1\text{R}_5+\text{R}_3\text{R}_4+\text{R}_1\text{R}_4+\text{C}\text{R}_1\text{R}_3\text{R}_4\omega\text{j}}\tag{10}$$

So, the absolute value if given by:

$$\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\left|-\frac{\text{R}_1\text{R}_3\left(\text{R}_4+\text{R}_5\right)}{\text{R}_1\text{R}_5+\text{R}_3\text{R}_4+\text{R}_1\text{R}_4+\text{C}\text{R}_1\text{R}_3\text{R}_4\omega\text{j}}\right|=$$ $$\left|-1\right|\cdot\frac{\left|\text{R}_1\text{R}_3\left(\text{R}_4+\text{R}_5\right)\right|}{\left|\text{R}_1\text{R}_5+\text{R}_3\text{R}_4+\text{R}_1\text{R}_4+\text{C}\text{R}_1\text{R}_3\text{R}_4\omega\text{j}\right|}=$$ $$\frac{\text{R}_1\text{R}_3\left(\text{R}_4+\text{R}_5\right)}{\sqrt{\left(\text{R}_1\left(\text{R}_5+\text{R}_4\right)+\text{R}_3\text{R}_4\right)^2+\left(\text{C}\text{R}_1\text{R}_3\text{R}_4\omega\right)^2}}\tag{11}$$

And the argument:

$$\arg\left(\underline{\mathscr{H}}\left(\text{j}\omega\right)\right)=\arg\left(-\frac{\text{R}_1\text{R}_3\left(\text{R}_4+\text{R}_5\right)}{\text{R}_1\text{R}_5+\text{R}_3\text{R}_4+\text{R}_1\text{R}_4+\text{C}\text{R}_1\text{R}_3\text{R}_4\omega\text{j}}\right)=$$ $$\arg\left(-1\right)+\arg\left(\text{R}_1\text{R}_3\left(\text{R}_4+\text{R}_5\right)\right)-\arg\left(\text{R}_1\text{R}_5+\text{R}_3\text{R}_4+\text{R}_1\text{R}_4+\text{C}\text{R}_1\text{R}_3\text{R}_4\omega\text{j}\right)=$$ $$0+0-\arctan\left(\frac{\text{C}\text{R}_1\text{R}_3\text{R}_4\omega}{\text{R}_1\text{R}_5+\text{R}_3\text{R}_4+\text{R}_1\text{R}_4}\right)=$$ $$-\arctan\left(\frac{\text{C}\text{R}_1\text{R}_3\text{R}_4\omega}{\text{R}_1\left(\text{R}_5+\text{R}_4\right)+\text{R}_3\text{R}_4}\right)\tag{12}$$


Using your values we get:

  • Absolute value: $$\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{10000\text{R}_1}{\sqrt{\left(3750+\text{R}_1\right)^2+\left(3750\text{CR}_1\omega\right)^2}}\tag{13}$$
  • Argument: $$\arg\left(\underline{\mathscr{H}}\left(\text{j}\omega\right)\right)=-\arctan\left(\frac{3750\text{CR}_1\omega}{3750+\text{R}_1}\right)\tag{14}$$
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    \$\begingroup\$ I think you forgot to answer the OP's actual questions. They already had the transfer function. \$\endgroup\$ – Elliot Alderson Apr 30 at 17:39

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