# Emitter-Follower Input And Output Impedance

Joined Jun 16, 2023
128
Hi:

I am looking for confirmation regarding the input and output impedance of this
class AB amplifier. See attached.

Assume IE1 = 42.261 mA and Rs = 2.2KΩ

 r'e1 = 25 mV / IE1 0.589​ Ω re1 = RE1║RL 0.008​ KΩ Rin(base) = βac1 * (r'e1 + re1) 0.133​ KΩ Rin(total) = Rin(base)║RB1║RB2 0.054​ KΩ

Assume βac1 = 223

 Rout = (Rs / βac1)║re1 0.006​ KΩ

Thanks

RS

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#### Bordodynov

Joined May 20, 2015
3,117
See

Joined Jun 16, 2023
128
What is this? This does not answer my query.

#### Audioguru again

Joined Oct 21, 2019
6,461
You are using little low power transistors that will be very overloaded in your circuit when it plays loudly.
The simulation uses transistor specs that are typical and are much higher than minimum spec that you might get.
With an 8V peak input, the simulated output shows much more transistor current and heating than is allowed.
The 180 ohm base resistors have a minimum of only 1.8V across them (10mA base current) for a collector-emitter current of 925mA which is far from the specs.

Your circuit is missing negative feedback that all amplifiers use that greatly changes input and output impedances.

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Joined Jun 16, 2023
128
You are using little low power transistors that will be very overloaded in your circuit when it plays loudly.
The simulation uses transistor specs that are typical and are much higher than minimum spec that you might get.
With an 8V peak input, the simulated output shows much more transistor current and heating than is allowed.
The 180 ohm base resistors have a minimum of only 1.8V across them (10mA base current) for a collector-emitter current of 925mA which is far from the specs.

Your circuit is missing negative feedback that all amplifiers use that greatly changes input and output impedances.
Again, no one is answering my query!

#### ericgibbs

Joined Jan 29, 2010
18,258
Hi RRR
asc file
E

Joined Jun 16, 2023
128

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#### ericgibbs

Joined Jan 29, 2010
18,258
Hi RRR.
This is what LTS shows for Zin and Zout for that amp.
Is this any help?
E

Posted a better plot!

Last edited:

Joined Jun 16, 2023
128
Hi RRR.
This is what LTS shows for Zin and Zout for that amp.
Is this any help?
E

Posted a better plot!

View attachment 303970View attachment 303971
Hi E:

Thanks. That's a big help.

Here are my plots. They could probably look better, especially the output impedance, but not sure what is needed.

Last edited:

#### Audioguru again

Joined Oct 21, 2019
6,461
Your simulation shows many DC voltages and DC currents but has no signal measurements.
Your input impedance is a little less than 0.1k ohms (why not show 200 ohms at the top of your graph?).
Your output impedance is zero for most frequencies so you should show 10 ohms or less at the top of your graph. The 100uF output capacitor should be 1000uF for all low frequencies.

#### Bordodynov

Joined May 20, 2015
3,117
You are using little low power transistors that will be very overloaded in your circuit when it plays loudly.
The simulation uses transistor specs that are typical and are much higher than minimum spec that you might get.
With an 8V peak input, the simulated output shows much more transistor current and heating than is allowed.
The 180 ohm base resistors have a minimum of only 1.8V across them (10mA base current) for a collector-emitter current of 925mA which is far from the specs.

Your circuit is missing negative feedback that all amplifiers use that greatly changes input and output impedances.
That answers your question. For probing, I applied a current of 1A to the output (in low-signal mode). The impedance shared with the load is equal to the load voltage. The joint impedance is the parallel connection between the load and the output impedance of the amplifier. I used the formula for this connection to calculate the amplifier output impedance.

#### ericgibbs

Joined Jan 29, 2010
18,258
Hi RRR.
If you limit the frequency range of the AC analysis, you can increase the plot resolution.

Also, to 'magnify' the plot display, place your cursor at the lower frequency you want to magnify, then keeping the mouse key held down, move the cursor to the higher frequency of interest.
So you now have the area of interest in the cursor drawn box, then let the mouse key up.

E

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#### ericgibbs

Joined Jan 29, 2010
18,258
Hi RRR,
You can also determine the Impedance of the two paths, that are in parallel, giving the total input impedance.
E

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