enter image description here

While performing simple practical with LM741 IC in voltage follower mode. Connections as shown in the figure below. I looked online for the input and output impedance calculation and all I got is the calculations for inverting and non-inverting or emitter follower configuration methods. I am bit stuck here while calculating input and output impedance in buffer or voltage follower configuration.

  • \$\begingroup\$ What exactly have you tried so far in calculating this output and input impedance? \$\endgroup\$ – Spehro Pefhany Jun 4 at 0:48
  • \$\begingroup\$ Tried calculating input and output impedance with the method of Input impedance=Zin=Vin/Iin Output impedance=Zout=Vout/Iout Not sure if I am not right track \$\endgroup\$ – Prajakta Bhutkar Jun 4 at 1:27
  • \$\begingroup\$ Probably \$\Delta V_{OUT}/\Delta I_{OUT}\$ is better, but how are you approaching that? \$\endgroup\$ – Spehro Pefhany Jun 4 at 1:30
  • \$\begingroup\$ I have connected two different load resistors and took readings of voltage and current flowing across them. To get Ξ”π‘‰π‘‚π‘ˆπ‘‡/Ξ”πΌπ‘‚π‘ˆπ‘‡ I found this article as well which I think is helpful if anyone else is also looking for answers to the above question. rohmfs.rohm.com/en/products/databook/applinote/ic/amp_linear/… \$\endgroup\$ – Prajakta Bhutkar Jun 5 at 23:24

The [closed loop] output impedance of an opamp is well modeled as a INDUCTOR, in series with some small resistance. We will predict values, or ranges of values, during this answer. To arrive at ClosedLoop values, we'll first understand the OpenLoop performance of a "typical" opamp.

The effect of Closed Loop Zout being INDUCTIVE plus a small Resistance, is crucial in understanding why some values of Capacitors will cause ringing or oscillation when shunting the OpAmp output pin. These values will vary from opamp to opamp, from technology to technology, from specific opamp internal circuit to other specific internal circuits; after reading this answer, you will understand this dependency.

The manufacturer may or may not give you a plot versus frequency, for Av=+1, Av = -1, Av = -10, etc.

The manufacturer may or may not give you a typical number in the parameter tables.

But here is how to proceed.

Output stages of complimentary bipolar (modern opamps) will have a class AB circuit that drives the Vout pin. Operating those 2 devices at 0.5mA produces an incremental Rout, of each bipolar device, of 52 ohms; with the 2 emitters driving Vout effectively in parallel, the Rout is halved to 26 ohms OPENLOOP.

Some high speed bipolar opamps run their output stages at higher currents, and you will see plots or values in the param tables, of 15 ohms or 10 ohms, again these values are *OPENLOOP.

Let us have an example here. Our opamp has DC_gain of 1e+6, F3dB of 10Hz, Unity Gain Band Width (UGBW) of 10MHz, and output loop Rout (at DC) of 10 ohms.

How does the circuit perform in a voltage_follower topology?

  • From DC to 10Hz, the Rout will be 10/1,000,000 = 10 microOhms.

  • As the open_loop gain rolls off with increasing frequency, the decreasing loop gain produces a RISING Zout with a phase_shift. This effect is exactly that of an inductor in series with the ClosedLoop circuit.

  • Let us predict that inductor. Once we have that value of Inductance, we may shunt the Vout pin with some capacitance, and we can predict the frequency of ringing or of oscillation. Knowing the Rout (in this case, its 10 ohms), we know the dampening resistance, and can predict the Q (-3 dB bandwidth will follow) of the R+L+C network.

  • What do we know? At the UGBW frequency, the opamp no longer has any control over the output voltage, and the full DC_Rout value appears on the output pin. You can see this effect in certain datasheets, if the Zout plot is extended to and past the UGBW. Analog Devices has some opamps thus characterized.

  • For our example opamp, the Rout (ZOUT) is 10 ohms at 10MHz, assuming 90 degree phase margin at that frequency.

  • What is the inductor value? Given Zl = 2 * PI * F * L, we simply do a division of Zl/( 6.28 * UGBW) == L == 10_ohms/ 62.8MegaHertz, and we find the inductor == 10/62 microHenries = 160 nanoHenries.

Now what happens when a Capacitor loads (shunts) this opamp voltage_follower circuit?

Depends on .... how close to UGBW is the resonant LC frequency? At worst, you may need to INSERT an external dampening discrete resistor. So let us run some maths.

Suppose the "inductor" is indeed 0.16uH. Place a capacitive load of 0.16uF.

With the classic resonance formula of

  • 1/[ 2 * pi * sqrt(L * C) ]

we expect exactly 1MHz F_ringing.

[ I used the "OutZ" button in Signal Chain Explorer to view the results.]

F_ring is indeed 1MHz. With 20 dB peaking.

I then inserted a resistor between the Unity Gain opamp and the 0.16uF capacitor.

  • 1 ohm external dampening ==> 1dB peaking [using SCE BODE PLOT]

  • 0.1 ohm external ==> 14 dB peaking


Regarding input resistance and capacitance:

Input capacitance will strongly depend on whether the input diffpair has been cascaded to minimize Miller Multiplication. Examine the parameter table, and the schematic, of the Fairchild UA715 to learn of this benefit.

Input resistance will be different from Input (bias or leakage) Current.

FET/CMOS input stages will have nano/pico/femto amps of current at room temperature. At 125 Β° C, the input current into dates of FETs or the necessary ESD circuitry, may have increased 1,000s or 1,000,000X. If you casually use 1MegOhm resistors, a surprise awaits.

Input resistance will be high for FET/CMOS inputs, and relatively LOW for bipolar inputs.

Beware the SPICE modeling of input capacitance as the opamp changes from small-signal to SlewLimited behaviors. The use of cascade devices (common base, common gate) ensures LOW INPUT CAPACITANCE at all modes of operation.

Without cascading, the Cin during slewing will be lower (indeed 10X or 100X lower) than during small-signal operation (the final settling ).

Will the SPICE model show this?

| improve this answer | |

Your Answer

By clicking β€œPost 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.