Discussion in 'Homework Help' started by howartthou, Jun 26, 2009.

1. ### howartthou Thread Starter Active Member

Apr 18, 2009
96
0
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. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,662
1,311
Well,
maybe you can solve this circuit

Find R1 if you know
Vin=5V; Vout=3V and R2=1KΩ

3. ### R!f@@ AAC Fanatic!

Apr 2, 2009
9,652
1,110
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

4. ### Audioguru Expert

Dec 20, 2007
10,787
1,212
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. ### howartthou Thread Starter Active Member

Apr 18, 2009
96
0
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. ### R!f@@ AAC Fanatic!

Apr 2, 2009
9,652
1,110
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

Last edited: Jun 30, 2009
7. ### Audioguru Expert

Dec 20, 2007
10,787
1,212
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. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,662
1,311
"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

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.

Because you just start you're adventure with electronic circuits.

Last edited: Jun 30, 2009
9. ### R!f@@ AAC Fanatic!

Apr 2, 2009
9,652
1,110
Audioguru
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

10. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,662
1,311
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: Jul 1, 2009
11. ### Audioguru Expert

Dec 20, 2007
10,787
1,212
Most class-AB amplifiers have an efficiency of 50% when the output level is close to clipping. A typical audio amplifier has a damping factor of 200 so its output impedance is 8 ohms/200= 0.04 ohms at low frequencies.
If the output power into 8 ohms is 100W then for the 0.04 ohms to dissipate 100W its current would need to be 50A RMS which is impossible.

12. ### R!f@@ AAC Fanatic!

Apr 2, 2009
9,652
1,110
hee hee hee what's the use.
I don't think we can prove what we leaned are wrong. Still When I built an amplifier, All I think of is how well the out put stage can withstand low frequencies without too much heat. Even though Impedance and all the stuffs are there, I do not think about them. If it works, that's good enuf for me.

Rifaa

13. ### howartthou Thread Starter Active Member

Apr 18, 2009
96
0
Jony130
Thank you for your answers, they help a lot, especially the diagrams, wish I could see things as clearly as you do, but all I have time for is to study on paper, I havent had time to build hardly any ciruits yet. Really appreciate your assistance, thanks again.

Riffa and Audioguru
I enjoy reading your responses too although I barely understand them. Still confused about how an amp has an output impendance with no load. And why the output impendance is meant to match the impedance of a load once connected?

In an amp circuit nothing should really happen until a load is connect right? So when you say ourput impendance do you mean a resistance like say a lighy globe has a certain resistance? A 60 watt x 240v light globe for example.

P = E x I
60 = 240 x I
I = 250mA

R=E/I
R=240/0.25=960Ω

So the light globe has an R of 960 but the R is the load.

So how does an amp have an output impedance without a RL? Do you mean it is designed with a certain output impedance in mind and that a load to be attached should have a matching impedance?

Last edited: Jul 4, 2009
14. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,662
1,311
Light globe has a resistance (strongly nonlinear resistance)
And this resistance is a load resistance for the mains voltage 240V.
In audio circuit we don't wont to Rout=RL.
Because in theory amplifier should act as a ideal voltage source EMF whit 0Ω internal resistance.

15. ### hobbyist AAC Fanatic!

Aug 10, 2008
885
85
To understand output impedance look at these diagrams
The output impedance is the resistor marked RC.
the resistor in the box represents a transistor.

1. shows the output voltage across the transistor which becomes the output voltage fed to a load. This is at no load voltage.

2. shows voltage output to a load resistance equal to RC.

3. shows volt. out when load resistance is much smaller than RC.

4. finally voltage to the load when you change output impedance RC, to be much smaller than the load resistance.

From this you can see why it is important to try to keep the RC (output impedance of amp stage) as low as possible with respect to the load it's driving.

16. ### millwood Guest

that's almost too easy: if you have a mechanism to adjust the output devices (think of them as resistors for this analysis) so they produce the same voltage on the load, you have created an ideal voltage source from resistive devices!

output impedance of a (power) amplifier with negative feedback is typically very low in the audio band. However, they can be high or very high at very low frequencies (think of those capacitor-coupled output stage), or very high frequencies.

Apr 2, 2009
9,652
1,110

HUH!!

Rifaa

18. ### Beduin Member

May 20, 2009
16
0
Thanks man, you are 1337!
We should have karma system in the forum

19. ### Beduin Member

May 20, 2009
16
0
Hey there m8

You know what, am in your situation! I am a noob

I think it will be a good strategy if you start reading about the basics of opamps. I can sense that you are not quite clear about opamp characteristics/analysis as Ideal and none ideal situations? If not, do that my friend, I saw some excellent videos on youtube* and there are millions of sites about opamps.
I would love to help out, but I can see there are alot more skilled people here.. I don't even understand what they are saying

And my friend, I think guys like me and you have ALOT to look forward too hence abstraction needed (am going to build a robot next term).