The output signal is lowered when coupled to the next amp stage.

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DaniKowa

Joined Jul 1, 2021
48
Hi, I have a problem I need to understand. Probably the question will be simple If explained in a simple way, and for this reason I chose this section. Very often it happens that the output voltage of the signal of one amplification stage is lowered once hooked to the next amp stage, even with a coupling capacitor. It often happens to me with discrete components like JFETs or transistors, and a little less with opamps. It is probably a physiological behavior of the signal chain, but I would like to understand why, otherwise I can not carry out some of my projects. Thank you for your contribution!
 

Ramussons

Joined May 3, 2013
1,100
An amplifier stage has an Input impedence (Zi) and an Output impedence (Zo). For a voltage amplifier, it is best to have an Infinite input Z and a Zero output Z. But, in practice, this is not always possible.

If the amplifier is a stand alone unit, we design to make Zi high and Zo low.

For each stage in an amplifier, the design will take into account the Input and Output Impedences, Power, Voltage.....
 

MrChips

Joined Oct 2, 2009
24,650
The output impedance Z1 and input impedance Z2 together constitute a voltage divider.
If you match the impedances, i.e. Z1 = Z2, the voltage gain is -6dB, i.e. Vout is half the input voltage Vin.

1626783613095.png
 

BobTPH

Joined Jun 5, 2013
4,054
You do not see this effect with opamps because the gain is set by negative feedback, which will keep the output constant even if the load varies (within reason). Effectively, opamp circuits with negative feedback have very low output impedance.

Bob
 

dcbingaman

Joined Jun 30, 2021
498
For a standard transistor common emitter stage the output impedance is the collector resistor value. You will see a very high impedance looking back into the collector because it is a constant current source to the output impedance for a BJT amplifier is the collector resistor value. If you want to eliminate all loading problems, simply have any of your amplifiers followed by an op amp unity gain buffer. You will not have this problem with op amps.
 

dcbingaman

Joined Jun 30, 2021
498
Just want to make sure you understand amplifiers somewhat. What is the input impedance of the left circuit? What is the output impedance of the left circuit. What is the input impedance of the right circuit? What is the output impedance of the right circuit? Consider the op-amp to be an ideal op-amp.

ImpedanceExample1.PNG
 
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dcbingaman

Joined Jun 30, 2021
498
What is the output voltage of this circuit?

Example2.PNG
Hi, I have a problem I need to understand. Probably the question will be simple If explained in a simple way, and for this reason I chose this section. Very often it happens that the output voltage of the signal of one amplification stage is lowered once hooked to the next amp stage, even with a coupling capacitor. It often happens to me with discrete components like JFETs or transistors, and a little less with opamps. It is probably a physiological behavior of the signal chain, but I would like to understand why, otherwise I can not carry out some of my projects. Thank you for your contribution!
One last question. What is the output voltage of this circuit. Assume the transistor gain is 100:
Example3.PNG
 

ericgibbs

Joined Jan 29, 2010
14,245
hi D,
Until 'dc' logs in ,if these are the two circuits, why do you need the input stimlus to determine in Zin.??
Rem ideal OPA's
E

Check the IN and NI on your version
A 09 Jul. 21 13.54.pngESP_ 616 Jul. 21 14.02.png
 
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MrChips

Joined Oct 2, 2009
24,650
dcb is providing you with important guidance.

All amplifiers are not equal. There are two ways to connect the input signal to an op-amp circuit.
You can configure the op-amp with the signal going into the inverting input (-IN) or the non-inverting input (+IN or NI).

The input impedance is different for the two configurations.

Learn how these differ.

https://www.electronics-notes.com/a...rational-amplifier-op-amp/input-impedance.php
 

MrChips

Joined Oct 2, 2009
24,650
No, no.
Negative feedback requires the feedback resistor to go from output to inverting input (-IN). It does not matter if you want to invert the op-amp symbol.
 

dcbingaman

Joined Jun 30, 2021
498
Hi @dcbingaman
what the signal parameter used for both circuits on post #7? I get unclear results with 1volt/1k input.
Sorry about the confusion. If you look at the original ASC sim file I have inverted the op-amps to that the non-inverting input is at the top. That is why your sim is not working. You can accomplish that by moving the op-amp. Press Control-E will mirror it and Control-R will rotate it, and pasting it back at the original position.
 
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dcbingaman

Joined Jun 30, 2021
498
No, no.
Negative feedback requires the feedback resistor to go from output to inverting input (-IN). It does not matter if you want to invert the op-amp symbol.
Here is your asc with corrections. Also, I made a mistake on R9 that is 1K. At 10K gain is to high and rails the op-amp.
CktUpdate.PNGI think it is fine to simulate them. My goal was can you figure out the answers by using circuit analysis. And understand how loading is going on. If you are struggling with that, we are here to help provide guidance.
 

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dcbingaman

Joined Jun 30, 2021
498
No, no.
Negative feedback requires the feedback resistor to go from output to inverting input (-IN). It does not matter if you want to invert the op-amp symbol.
Sorry, I was not clear on what I said. I should have said, move the op-amp out of the circuit, invert it so non-inverting input is at the top, then paste it back into the circuit. The wires will line up correctly so the inverting input is connected to the feedback resistor. You will also have to change the power supply voltages (swap positive input and negative input) being that the symbol is now inverted.
 
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MrChips

Joined Oct 2, 2009
24,650
Inverting the op amp symbol places the feedback resistor from the output to the inverting input? Not sure what you are trying to say? See schematic correction above?
I am trying to say that the feedback resistor goes from the output to the inverting input, just like I said.
I don't care how you draw your circuit. Just make sure that the above statement is satisfied.
 
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