op amp basic : dc path

Thread Starter

Zeeus

Joined Apr 17, 2019
481
Seen somewhere to always provide dc path for function generator....Should assume the current flows into the + terminal? (can also assume for FET op amps?)

Doing lab soon but think placing 1M from "+" to ground will make no difference. Correct?
but ofcos dont want to use 4 resistors for simple amplifier when mates are using 4 for Howland

Best,
Mr Howland
 

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Ylli

Joined Nov 13, 2015
850
No. Because of the feedback, the op-amp negative input will be at the same potential as the input signal level - making the + op-amp input a near infinite impedance. Adding a 1 meg from the + to ground will create a voltage divider and reduce both the net gain and the input impedance.
 

Thread Starter

Zeeus

Joined Apr 17, 2019
481
No. Because of the feedback, the op-amp negative input will be at the same potential as the input signal level - making the + op-amp input a near infinite impedance. Adding a 1 meg from the + to ground will create a voltage divider and reduce both the net gain and the input impedance.
If add 1 meg, then the + input has almost all of source voltage, no? the input impedance will be 1 meg|| 2.7K so no much change
Missed something?

about 1st question : Thinking there has to be complete part for function generator current
The DC path issue is for allowing a path for input bias current to flow.

https://www.analog.com/media/en/technical-documentation/application-notes/AN-937.pdf


Regards, Dana.
Is this from a book? if yes, what's name?

Thanks
 

BR-549

Joined Sep 22, 2013
4,938
It's called an application note......AN .....and it's number is 937.

Many chip manufacturers(this one is analog devices) have application notes for their devices.

A great and proven, trustworthy source for study.

Edit: Think of them as a custom textbook.
 

AnalogKid

Joined Aug 1, 2013
8,214
Thinking there has to be complete part for function generator current
There already is. The output of the function generator almost certainly is a DC-coupled amplifier stage, and the ground or return connection from the FG to your circuit is the reference needed for R1 ro stabilize the DC operating point of the circuit.

To see more clearly what your circuit "sees", remove Vi and connect the left side of RB directly to GND. This is the equivalent DC circuit, equal to what happens when the FG output is turned down to zero V.

ak
 

Thread Starter

Zeeus

Joined Apr 17, 2019
481
There already is. The output of the function generator almost certainly is a DC-coupled amplifier stage, and the ground or return connection from the FG to your circuit is the reference needed for R1 ro stabilize the DC operating point of the circuit.

To see more clearly what your circuit "sees", remove Vi and connect the left side of RB directly to GND. This is the equivalent DC circuit, equal to what happens when the FG output is turned down to zero V.

ak


Understand your 2nd statement but not first..Anyways, too early to be concerned about simple thing

Please look (Learning basic op amp but "understood" this weeks ago but not working now for testing)

When the output goes past say 2v pk- pk, wrong to expect the LED on emitter of transistor to turn on?
At first the output from output op amp looking weird, changed the 470U cap (at first it was, 2.2U so changed to 470) and it was looking like sine wave...Frequency of slew rate for sine = rate/pi * Vout (does not work for this circuit but works for other circuits have tried)

the 22UF was at first 1UF then the LED was blinking..Anyways, dunno...will finish studying basic configurations soon to understand this better but right now, expecting the LED to turn on as soon as output passes a certain limit but not working as expected..What is wrong?

Thanks

output of 2nd op amp is low..strange, no?
 

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AnalogKid

Joined Aug 1, 2013
8,214
You don't say anything about what the circuit is supposed to do, what it is you are trying to achieve. It looks like an AGC circuit with a very short attack time and a very long release time.

Please add reference designators to ALL components.

ak
 

Thread Starter

Zeeus

Joined Apr 17, 2019
481
You don't say anything about what the circuit is supposed to do, what it is you are trying to achieve. It looks like an AGC circuit with a very short attack time and a very long release time.

Please add reference designators to ALL components.

ak
Yes AGC....Attenuated the input on first op amp then using output and NPN as voltage divider..
Follow diagram AK, you know what it is

So release time caused by the 22UF and 1meg? what is attack time?

Please explain..Yes it is AGC!

designators makes it look messy : hard for reader to follow
Thanks

Edit : THought explained earlier...Varying input signal...The NPN should not turn on unless base is about 3V - 4V? if it is not turned on then NPN is not attenuating signal so the voltage from output of 1st op can go to the 2nd..If the output from 2nd amp is about 2V, then the doubler increases to about 3. so NPN turns on to reduce then acts like divider with 470 to reduce input signal to 2nd op amp

Problem now is that, LED not turning on..

Actually, nvm..will figure out later

Thanks
 
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AnalogKid

Joined Aug 1, 2013
8,214
Follow diagram AK, you know what it is
No, *you* know what it is; I had to figure it out, and that is not the same thing. My point was that "what it is" is a kinda important thing to leave out of a post asking for free help.
So release time caused by the 22UF and 1meg? what is attack time?
The attack time is a logarithmic function of the dynamic impedance of a diode (not that diode, the other one), a capacitor (the one over there), and the output impedance of that thing to the right, forming a voltage divider with the other capacitor. Clear?
designators makes it look messy : hard for reader to follow
You could not possible be more wrong.

ak
 

Thread Starter

Zeeus

Joined Apr 17, 2019
481
No, *you* know what it is; I had to figure it out, and that is not the same thing. My point was that "what it is" is a kinda important thing to leave out of a post asking for free help.

The attack time is a logarithmic function of the dynamic impedance of a diode (not that diode, the other one), a capacitor (the one over there), and the output impedance of that thing to the right, forming a voltage divider with the other capacitor. Clear?

You could not possible be more wrong.

ak
Thanks
Not so clear though..Ts wishes clearer but will try again
 
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