simulate this circuit – Schematic created using CircuitLab

Assuming I'm giving a step input( from 0 to Vs) what could be the response?

I analyzed it thus, but I'm not sure if my approach is correct:

I reduced the above circuit to :


simulate this circuit

Now this is a first order RC circuit with tau as RC, Vout will be :

\$\ V_\text{s} + \text{Vs}\left(1-e^{-\frac{t}{\text{RC}}}\right)\$

(Vin = Vs after t= 0)

Is this analysis correct?

  • \$\begingroup\$ your Vout formula doesn't make sense, Vin/R is a current, but you need a voltage. \$\endgroup\$ Aug 26 '16 at 9:48
  • \$\begingroup\$ IN your second schematic, R2 doesn't make a difference – it's parallel to an ideal voltage source. \$\endgroup\$ Aug 26 '16 at 9:49
  • \$\begingroup\$ Changes made, I was going for Vin + Vc = Vout \$\endgroup\$ Aug 26 '16 at 9:59
  • \$\begingroup\$ @MarcusMüller So is my second schematic wrong, to be derived from the first? \$\endgroup\$ Aug 26 '16 at 10:00
  • \$\begingroup\$ what is Vc? You don't specify that. \$\endgroup\$ Aug 26 '16 at 10:02

This circuit is actually a common configuration and the only factor that cannot be determined due to the ideal op amp is the eventual high-frequency roll-off.

It's perhaps easier to recognize when redrawn like this:

enter image description here

I've used a TL071 as the op amp just because it's easier for me than getting an ideal op amp in play and used \$R_f\$ (for feedback resistor) instead of \$R_1\$ and \$R_g\$ (for gain resistor) instead of \$R_2\$ because I find this makes it easier to make sense of the gain expressions when they get complex.

This circuit produces the following frequency characteristic:

enter image description here

The right-most gain roll-off after 1 MHz is due to the real frequency response of the TL071, so would not be present using an ideal op amp.

However, the single pole, single zero characteristic would appear just as it does here at roughly 1.6 kHz and 3.2 kHz respectively.

The pole appears at the frequency where \$X_c = R_f\$ which is:

$$ \frac{1}{2 \pi R_f C} = \frac{1}{2 \pi 100 \cdot 1e-6} \approx 1592 Hz $$

The zero appears at twice that frequency.

For a 1V step, this produces a response that looks to me like the result you proposed:

enter image description here

I haven't verified that you got precisely the right time constant and I'm not sure I would have chosen the equivalent circuit you did (it seems to neglect the effects of feedback). But you seem to have arrived at the correct qualitative behavior, if not the exact behavior.

The additional gain and bandwidth of an ideal op amp would not change the character of this step response.

The transfer function for this circuit, in "low-entropy" form is:

$$ \frac{V_{out}}{V_{in}} = \frac{R_f+R_g}{R_g} \frac{s(R_f\parallel R_g)C + 1}{sR_fC+1} $$

Substituting values for this particular circuit:

$$ \frac{V_{out}}{V_{in}} = \frac{100+100}{100} \frac{s(100\parallel 100)(.000001) + 1}{s(100)(.000001) +1} = 2\cdot \frac{.00005s + 1}{.0001s +1} $$

shows us the DC gain (2), the single pole at \$\omega_p = 1/.0001 = 10,000\$ and the single zero at twice that frequency \$\omega_z = 1/.00005 = 20,000\$. In Hertz these are 1591 Hz and 3183 Hz respectively.

  • \$\begingroup\$ Regarding the transfer function, in low entropy form: I got the same expression, except for the s(Rf||Rg)C + 1 in the numerator. I started off by writing KCL at common node of Rf,Rg,C. Current through the capacitor would be -> C d(Vout-Vin)/dt and when I took the Laplace transform, I presumed that since Vin is a constant step, it (capacitor current) will reduce to sCVout, giving me the final equation as : (Vout-Vin)/Rf + sCVout = Vin/Rg Is this approach correct? \$\endgroup\$ Aug 27 '16 at 5:38
  • \$\begingroup\$ @AmbareeshSrJa: Yes, the transfer function is perhaps most suited for frequency response, apologies, I got a little carried away in the analysis and strayed beyond your direct question a fair ways :) The transfer function can, however, give us the \$t=0+\$ and \$t=\infty\$ values readily, substituting \$s=\infty\$ and \$s=0\$ respectively, to yield 1 and 2. \$\endgroup\$
    – scanny
    Aug 27 '16 at 5:40
  • \$\begingroup\$ In that case wouldn't my analysis be wrong? Since mathematically I can only see one pole in my equation and no zeroes ? :/ \$\endgroup\$ Aug 27 '16 at 5:44
  • \$\begingroup\$ I started with the standard non-inverting amp formula $$ \frac{V_{out}}{V_{in}} = 1 + \frac{Z_f}{Z_g} = 1 + \frac{R_f \parallel \frac{1}{sC}}{R_g} $$ That provides a quick start into the algebra allowing you to skip the differential equations. \$\endgroup\$
    – scanny
    Aug 27 '16 at 5:46

Yes, using ideal components, your time domain analysis looks correct for a step input. The RC time constant is \$R_1 C_1\$, without regard to \$R_2\$.

The general frequency domain transfer function for the circuit is...

$$ V_{out}/V_{in} = 1 + \frac{R_1}{R_2 + R_1 R_2 C_1 \cdot j 2 \pi f} $$

  • \$\begingroup\$ Don't you think there is something missing in the formula? (Hint: check dimensions of both sides :-) ; or: transfer function is dimensionless). \$\endgroup\$
    – Eric Best
    Aug 27 '16 at 10:39
  • \$\begingroup\$ @Eric Best It seems fine to me. ohms * ohms * Farads * Hz = ohms so the equation is dimensionless. \$\endgroup\$
    – user4574
    Aug 27 '16 at 12:03
  • \$\begingroup\$ Really? :-) You are speaking about the denominator, I about the whole equation (or formula). My hint didn't help? (both sides?) I can see Vout on the left and it is customary that voltage used to be in volts. What is missing is either the fractional expression on the left side with Vin in the denominator or the multiplier Vin on the right, so that it gives a sense. The transfer function is usually designated as G(jω) or G(s), H(jω), H(s), etc. I have never seen Vout. \$\endgroup\$
    – Eric Best
    Aug 27 '16 at 21:35

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