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#1
06-26-2009, 03:14 PM
 howartthou Member Join Date: Apr 2009 Location: Melbourne, Australia Posts: 96

Question:
A certain amplifier has a no-load output of 5 V. A 1KΩ load resistor is connected to the amplifier and the output is reduced to 3V. Whats the output impendance of the amplifier?

My attempt:
Hmmm. This is probably simple. But no formula comes to mind. I can only blame my poor textbook because I know I have some brains.

I know that:
I=E/R
So: I = 5/1000 = 0.005 A

I know that:
R=E/I
So: 3/0.005 = 600 Ω

What the hell am I missing here?
#2
06-26-2009, 04:13 PM
 Jony130 Senior Member Join Date: Feb 2009 Location: Poland/Wroclaw Posts: 2,512

Well,
maybe you can solve this circuit

Find R1 if you know
Vin=5V; Vout=3V and R2=1KΩ
#3
06-26-2009, 11:42 PM
 R!f@@ Senior Member Join Date: Apr 2009 Location: Euphoria.., The One & Only {GMT +5} Posts: 7,055 Blog Entries: 3

The out put of an amplifier depends on the out put stage design.
And keep in mind that an amplifier will deliver it's maximum power only when the load is same as the output impedance of the amp. If the load is different the power is reflected back in to the amp output stage and dissipated as more heat and may over heat and blow.
try to vary the load and try to find the maximum voltage and current delivered to the load. the max power will tell you the output Z practically.

Rifaa

__________________
R!f@@
$\bullet$ Do not get angry and do not be harsh because you will never regret being kind $\bullet$

#4
06-27-2009, 02:13 AM
 Audioguru Banned Join Date: Dec 2007 Location: Ontario, Canada Posts: 9,411

Amplifiers made in the last 50 years have an output impedance that is much less than the speaker's impedance. An output impedance of 0.04 ohms and less is common. It provides very good damping of the resonances in a speaker and provides max output power instead of one-quarter power.

Old vacuum tube amplifiers used an output transformer with an output impedance that was the same as the speaker's impedance. They provided almost no damping.
#5
06-30-2009, 12:34 PM
 howartthou Member Join Date: Apr 2009 Location: Melbourne, Australia Posts: 96

Quote:
 Originally Posted by Jony130 Find R1 if you know Vin=5V; Vout=3V and R2=1KΩ
Hi Jony

Well, Found a formula for a voltage divider: Vo = ViR2 / R1 + R2.

Using algebra to find R1 you end up with: R1 = (ViR2/Vo) - R2.

So: R1 = (5 x 1000 / 3) - 1000 =667Ω

I would like to know:

1. How do you know R2=1k and not R1.

2. The formula I used is for a voltage divider circuit which isn't exactly an amplifier now is it? So how did you know to treat this as a voltage divider circuit?

3. Why isnt it obvious to me that this question is about a voltage divider

Audioguru and Rifaa, I really appreciate your feedback but I am struggling here just to understand the basics here.

Audioguru, ummm, your answer is beyond me right now, although I am sure its insightful. I think I need to learn how to do addition and you are trying to teach me long multiplication - excuse the analogy

Rifaa, umm, output impedance of an amp? How can an amp have an output impedance? Surely its the load connected to the amp that has the impedance?? The amp itself may have an impedance but is does not "output" an impedance, does it
#6
06-30-2009, 02:50 PM
 R!f@@ Senior Member Join Date: Apr 2009 Location: Euphoria.., The One & Only {GMT +5} Posts: 7,055 Blog Entries: 3

The thing about amp is that it produces a voltage at the speaker thereby creates current.
So In theory Any voltage source has an Internal resistance. And if the Source produces AC then the force opposing AC is called Impedance, thus Out put Impedance and this Impedance is the thing that dissipates heat at the out put stage. Which totally means the out put stage has an internal resistance

Rifaa
__________________
R!f@@
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Last edited by R!f@@; 06-30-2009 at 03:07 PM.
#7
06-30-2009, 10:02 PM
 Audioguru Banned Join Date: Dec 2007 Location: Ontario, Canada Posts: 9,411

A modern half-decent amplifier has a very low output impedance. If its output is 5.0V without a load then its output is 4.9999V with an 8 ohm load.

It has plenty of negative feedback. When its output voltage drops a little due to load current then it increases its output to be almost correct.

The output impedance has nothing to do with heat. The output transistors get hot because they are resistors in series with the power supply and the load.
#8
06-30-2009, 11:40 PM
 Jony130 Senior Member Join Date: Feb 2009 Location: Poland/Wroclaw Posts: 2,512

Quote:
 Originally Posted by howartthou 1. How do you know R2=1k and not R1.
"A 1KΩ load resistor is connected to the amplifier"
And in this case, R2 from voltage divider is a RL for the amplifier.
So we can solve output impedance:
IL=3V/1K=3mA
and when RL=∞ Vo=5V
Rout=(5V-3V)/3mA=666Ω

Quote:
 Originally Posted by howartthou 2. The formula I used is for a voltage divider circuit which isn't exactly an amplifier now is it? So how did you know to treat this as a voltage divider circuit?
In electronic circuits, we always hooking the output of something (output of a amplifier, output of a voltage input source...) to the input of something else (load resistor, input stage of a next amplifier...).
So in our case, the signal source (Vin=5V) is the output of a amplifier with series internal impedance Rout, driving the load resistance R_Load=1K ( or next stage input impedance).
After we draw this in a simplified schematics we see our old friend the voltage divider.

And you have to remember that output impedance of a amplifier isn't a "real" resistor.
It is a model that help as analyses the circuits.
The amplifier is a source of a voltage and current to load resistor.
So when RL is disconnect form the amplifier, we get the maximum voltage form the amplifier.
Connecting RL to the amplifier, and the output voltage of a amplifier drops.
Engineers knows that resistor can do a voltage drop.
And that why they model the voltage drop of a amplifier by a internal resistance.
It is the simplest way to do so.

Quote:
 Originally Posted by howartthou 3. Why isnt it obvious to me that this question is about a voltage divider
Because you just start you're adventure with electronic circuits.

Last edited by Jony130; 07-01-2009 at 12:06 AM.
#9
07-01-2009, 07:41 AM
 R!f@@ Senior Member Join Date: Apr 2009 Location: Euphoria.., The One & Only {GMT +5} Posts: 7,055 Blog Entries: 3

Audioguru
Quote:
 The output impedance has nothing to do with heat. The output transistors get hot because they are resistors in series with the power supply and the load.
Well. If the transistor is creating resistance for a typical AC signal. and it dissipates heat, then I'll say it is the Impedance that is responsible. Prove me wrong.

Rifaa
__________________
R!f@@
$\bullet$ Do not get angry and do not be harsh because you will never regret being kind $\bullet$

#10
07-01-2009, 10:12 AM
 Jony130 Senior Member Join Date: Feb 2009 Location: Poland/Wroclaw Posts: 2,512

For me you are wrong.
The output impedance is only a "theoretical creature" (more or less).
To show (model) the voltage drop for small-signal analysis, and has nothing to do with power dissipation in amplifier and his output current capacity.
And equation for power dissipation in push-pull emitter follower (SEPP):
Ptot=(Vcc^2)/(∏^2*RL) ≈0.1*(Vcc^2)/RL
As you can see Ptot has nothing to do with Rout

Last edited by Jony130; 07-01-2009 at 10:24 AM.

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