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21st April 2019, 18:03 #1
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Output Resistance of Amplifier with Feedback ( Series  shunt) Loading

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21st April 2019, 19:59 #2
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Re: Output Resistance of Amplifier with Feedback ( Series  shunt) Loading
Hi,
R= V / I.
If you try to move the voltage of the common node between both resistors (which is the output node)....
You need to fight against both resistors.
Let's assume the output voltage is 3V, Rl = 3 ohms and ro = 1 ohms.
Then initially the voltage across Rl = 3V and the current according ohm's law is 3A.
The voltage across ro = 1V and obviously 1A, too.
The internal generator voltage is 4V.
If you now pull down the ouptput voltage by 1V to 2V..
Then the voltage across ro increases and the voltage across RL decreases.
Obviously the current of ro increases and the current of RL decreases.
To increase the current of ro you need to draw (sink) current externally.
To decrease the current of RL you need to draw (sink) current externally.
Both currents are "sink" direction, thus they add.
Now let's calculate:
Rs = delta_V / delta_I = delta_V / ( delta_V / ro + delta_V / RL) = ...all deta_V ... cancel out:
Rs = 1 / (1/ro + 1/RL) ... which is the formula for parallel connection of resistors.
KlausPlease donīt contact me via PM, because there is no time to respond to them. No friend requests. Thank you.

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22nd April 2019, 02:13 #3
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Re: Output Resistance of Amplifier with Feedback ( Series  shunt) Loading
KlausST explanation is quite relevant, of course. A more formal explanation is as follows:
This relates to the two port approach of analyzing feedback loops. Since it is a seriesshunt feedback we need to find the loading from the feedback network at the input and output of the amplifier and this way we can isolate the forward path and the feedback path. The loading at the input side is found by shorting the right terminal of R2 to ground, because there the feedback is shunt. Looking into R1 we see R1R2. For the loading at the output of the opamp, we disconnect the connection from R2 to the opamp input (because it is series feedback on this side) and look into R2 from right to left into the feedback network  we see R1+R2, which is the output loading.
To find the output resistance of the so composed forward path we eliminate the input voltage source Vs by making it short. This makes V1=0 and hence the controlled voltage source uV1=0. Then, ro is also connected from output to ground.
It is then obvious that output resistance will be ro(R1+R2)RL.

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22nd April 2019, 06:54 #4
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Re: Output Resistance of Amplifier with Feedback ( Series  shunt) Loading
Hi,
I didn't take R1 and R2 into account because of the "open loop" statement.
But for sure to get "open loop" you may use the R1and R2, but disconnect their center node from the Opamp input..
Then (R1 + R2) is another parallel path and is leading to formula of post#1.
KlausPlease donīt contact me via PM, because there is no time to respond to them. No friend requests. Thank you.

23rd April 2019, 11:08 #5
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Re: Output Resistance of Amplifier with Feedback ( Series  shunt) Loading
The explanation is extrenmely simple: It depends from where you are looking into the resistor network. Looking into the output node we see three parallel pathes to ground. Hence, the resistors of these pathes are to be considered in parallel.
Another example: In your circuit, take the two resistors R1 and R2  without all the other parts .
Are they connected in series? Answer : Not possible without additional information.
It depends from where you are looking into this small "network".
(1) Looking from the left or from the right (the other end grounded)  they are, indeed, in series.
(2) Lookinng into the midpoint (both ends grounded), they are in parallel.
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