# Input Impedance OpAmp Circuit

• posted

Hello,

I'm trying to figure out what the input impedance to this op-amp circuit is. I have 2 ways I've been thinking about this, and I'm not sure which one is right, or if they're both wrong. I'm looking for the input impedance as seen by the AC voltage source, V1.

I posted an image of the circuit in question here:

Way 1: It's the impedances of C1, C2, and R3 in series. Done

Way 2: Thinking about a virtual short, it's C1, C2, R3, and R2 in series. Done

Basically, I'm wondering if R2 plays a role in the input impedance (as defined above) in this circuit.

much thanks!

• posted

Yep. Done.

No "virtual short". The inverting input of the OpAmp is a "virtual _ground_".

R1 and R2 might influence REALLY high frequency input Z. You should bypass the R1-R2 juncture to ground to minimize the power supply noise path.

...Jim Thompson

```--
| James E.Thompson, CTO                            |    mens     |
• posted

Kinda what Jim said: at reasonably low frequencies the inverting input to the op-amp is a virtual ground, by virtue of the really high gain of the op-amp and the feedback through R4.

But once you approach the corner frequency of the circuit, which depends on the gain and the op-amp GBW product, then your input impedance will go up as the virtual ground starts looking like a virtual lossy inductor due to the feedback through the op-amp's low-pass nature.

Just where you have to start worrying about this depends on what you're doing with the circuit: if you need 16 bit precision then you need to start worrying at frequencies over 250 times lower than the circuit's corner frequency; if you aren't that desperate for precision then it's not that bad.

```--
Tim Wescott
Wescott Design Services```
• posted

=A0 =A0 ...Jim Thompson

=A0 =A0| =A0 =A0mens =A0 =A0 |

=A0 | =A0 =A0 et =A0 =A0 =A0|

=A0|

=A0 =A0 =A0 |

Why is there no virtual short? This is what I think is meant by the two terms

Virtual Short: the inverting and non-inverting inputs are both at the same voltage potential, essentially behaving as if they were shorted together.

Virtual Ground: if one input is tied to ground, then the other will also have a 0V with respect to ground, so even if it is not physically tied to ground it's still virtually tied to ground because of the other input.

neither input is tied to ground in my circuit. there's a DC voltage at the non-inverting and the AC signal at the inverting.

• posted

Since the non-inverting input only has a DC source on it, for any AC signal the voltage at the non-inverting signal is zero... hence the virtual ground. (...granted, there's a finite resistance between the "real" ground -- R1||R2 -- and the non-inverting input, but since the ideal op-amp input draws no current, there's no voltage drop and hence the non-inverting input is still a virtual ground.)

(In case this isn't clear: If you tell your multimeter to measure AC and hook it up to a DC power supply, the steady-state response is 0V.)

---Joel

• posted

The feedback forces the two nodes to be at the same potential... OpAmp in equilibrium. Same potential doesn't necessarily mean "short".

Potential-wise but NOT current conduction wise.

Virtual ground... behaves like ground for _AC_ signals.

...Jim Thompson

```--
| James E.Thompson, CTO                            |    mens     |
• posted

Think of it this way; the output of the opamp will do anything in it's power (i.e. be at whatever voltage within its rails) to make the voltage across its inputs zero. Since the '+' input is at a fixed voltage (not affected by the output voltage or the '-' input), it falls out of the AC equation. It's as if a battery were put at this point equal to the voltage at the '+' input.

It's not a short, rather a controlled voltage. V- is forced to be the same voltage as V+.

There is nothing magical about ground. Erase all your ground symbols (and connect those points together) and draw a ground at the '+' input. You've changed nothing.

Again, ground is a relative thing. You can define it to be anywhere.

• posted

The way you describe your "virtual short", current would have to flow from the non-inverting terminal, making it more like a _real_ short.

What really happens is that the op-amp does it's level best to make the inverting terminal voltage equal to the noninverting terminal voltage. It's _not_ a short in the sense that current going into the junction at the inverting terminal comes out the noninverting terminal. In a circuit like yours it becomes a virtual AC ground, or a virtual voltage source, or whatever other "virtual this or that" makes sense to you and still yanks the inverting terminal around to match the noninverting terminal voltage.

```--
Tim Wescott
Wescott Design Services```
• posted

ot

s

Thanks for the responses, I have a follow up question

why is it that the feedback resistor and the load resistor don't play a role in the input impedance?

• posted

Crank your way thru the feedback (math) equations and you'll see.

To amuse yourself, then consider what happens when the OpAmp rails :-) ...Jim Thompson

```--
| James E.Thompson, CTO                            |    mens     |
• posted

And when the input frequency gets high enough, even if everything is staying linear.

It's amazing how much this stuff gets passed over in casual discussions and even in textbooks, but then suddenly gets important when you're doing something like feeding a 16-bit ADC.

```--
Tim Wescott
Wescott Design Services```
• posted

No fundamentals are being taught in the schools anymore :-( ...Jim Thompson

```--
| James E.Thompson, CTO                            |    mens     |
• posted

It can have an impact on the input impedance. When the thing gets above

200C and an amperage smell wafts through the room ... :-)

Seriously, I had a Harris opamp where every pin combination measured under 1ohm. It was a ceramic package which is probably why its lid hung on.

```--
Regards, Joerg

http://www.analogconsultants.com/```
• posted

"Phut" disease ?:-) ...Jim Thompson

```--
| James E.Thompson, CTO                            |    mens     |
• posted

In my case (mid '80's) they taught plenty of fundamentals, just not much practical application. So there was lots of focus on filter bandwidth, but it was rare to have an exercise that let you figure out just how far down from a 3dB corner you had to be if you wanted less than 5 degrees phase shift, or accuracy within 0.01%, etc.

It _was_ easy to figure out once a senior guy pointed it out, but one did spend the first few years out of school smacking one's forehead and saying "d'oh".

```--
Tim Wescott
Wescott Design Services```
• posted

I was fortunate all the way around...

```--
| James E.Thompson, CTO                            |    mens     |
• posted

"Jim Thompson" wrote in message news: snipped-for-privacy@4ax.com...

I have a suspicion that if you and Larkin had shared a cube back some 40 years ago, by now you'd be best buddies. :-)

Had you talked to anyone at Dickson about "possible employment opportunities" prior to actually quitting Motorla?

Nice history there, Jim; thanks...

• posted

Perhaps. Larkin takes poorly to criticism... can never admit error. Tom and I took it more as a learning experience... some of my best schemes came from Tom betting me I couldn't do it... like that NPN+lateral_PNP gimmick to make an essentially beta-independent PNP mirror... a cup of coffee bet ;-)

Then, one Friday, Tom claimed he could get 5-Watts out of an MC1554 Audio Amplifier (TO-5). I said no way. Monday I arrive to find that Tom had taken a block of Aluminum, machined a TO-5 sized hole with a clamp, drilled holes thru which he piped CO2... 5-Watts? No problem ;-)

No. But I knew Bob Rutherford who was VP... one call and I was hired.

...Jim Thompson

```--
| James E.Thompson, CTO                            |    mens     |
• posted

On 16Jul2010, in the Inverse Marx Generator thread, you wrote "I got a free ride thru MIT way back when scholarships were awarded only on merit. " Which was it? Work or free ride for 4 years? Art

• posted

message

Learn to read all the details. Tuition, room and board... no book or miscellaneous expense coverage. But it was pretty much a free ride... my "take-home" was about \$20/week. ...Jim Thompson

```--
| James E.Thompson, CTO                            |    mens     |