0
\$\begingroup\$

Also is it actually zero or just very low?

\$\endgroup\$
1
  • \$\begingroup\$ 1) what circuit are you asking about ? Depending on the circuit I can make the output impedance low (not zero!!) or much higher than low. 2) I can make a low output impedance using an opamp that has a high output impedance !!! Yes read that again. How? By using feedback. Also, read the free Ebook: "Opamps for everyone": web.mit.edu/6.101/www/reference/op_amps_everyone.pdf \$\endgroup\$ – Bimpelrekkie Oct 6 '20 at 7:52
3
\$\begingroup\$

Why is the output impedance of a non-inverting op amp zero?

It is about the so-called "differential output impedance" that is the ratio of voltage and current changes - dRout = dVout/dIout.

There is a voltage-type negative feedback in this circuit that makes the op-amp keep a constant output voltage when the output current changes (the op-amp output behaves as an "ideal" voltage source). So the ratio is almost zero.

\$\endgroup\$
2
\$\begingroup\$

At DC its just very low. For a voltage follower at DC its is approximately Z / G, where G is the open loop gain, and Z is the open loop output impedance. And for most op amps G is very high (10,000 or more) and Z is low (usually a few ohms).

For example take an op amp having an open loop gain of 10,000 and an output open loop output impedance of 10 ohms. The closed loop output impedance would be close to 0.001 ohms.

As an example, suppose you configure the op amp as a voltage follower (closed loop gain = 1).

If the output current changes by amount di then op-amp output stage must increase the output voltage by Z * di to keep the output voltage at the correct level. But to do that the difference between the + and - input terminals must have changed by Z * di / G. Therefore the closed loop output impedance dv/di = di / (Z * di / G), which is just Z/G.

At higher frequencies output impedance increases due to finite gain bandwidth product.

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

A bipolar-technology opamp (UA741 complete schematic is online, as are other bipolar opamp schematics) will have the open_loop Rout set by the idling current of the output bipolar transistors. We can estimate that.

In an opamp using a total Iddq of 1mA, we might guess at 0.2mA in the output (class A/B) stage. In that case, each of the 2 (pullup, pulldown) devices will have

  • Rout = 0.026 volts / 0.2mA = 0.026v / 0.0002 amps = 5 * 26 = 130 ohms

Given a pullup (presumed NPN common_emitter) and a pulldown (presumed PNP common_emitter), the Rout at DC can has a MINIMUM of

  • R_up || R_down = 130 || 130 = 65 ohms, as a MINIMUM

The "MINIMUM" applies because the base circuitry of each output device will also affect the Rout. Without analysis (requiring operating currents and Early Effect parameters and Beta, etc), we merely suggest 65 OHMS AS A MINIMUM.

If you visit some Analog Devices Inc high_speed opamp datasheets, you will find plots of Rout (actually Z_out) over frequency, even up at 100 or 500MHz. At such speeds, the bondwires in the package start to matter, and one particular ADI datasheet does show anomalies somewhat near 1GHz.

As other answers suggest, the open loop gain comes into the maths.

And the open loop gain has a phase shift. That phase shift, used properly in equations, shows the Zout appears as an INDUCTOR.

And when you pair an inductor with a capacitor, you encounter ringing .

Thus capacitive loading on opamps ------ is asking for ringing or oscillation.

\$\endgroup\$

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