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Old 9th September 2006, 07:40 AM   #21
Progg70 is offline Progg70  Sweden
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Sorry now it's there
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Old 9th September 2006, 07:59 AM   #22
myanmar is offline myanmar  Myanmar
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Hello AndrewT

Quote:
Max 50V p to p.....50v peak / sqroot of 2 = about 35V RMS over 4 ohms is 35 squared / 4 is 306.25 W RMS.Is that true
Sorry for my mistake not 50V p to p.
Max +/- 50V swing saw in oscilloscope.
That will be 100V p to p

100Vpp=50Vpk=50/(root2)Vrms or Vac = 35Vrms

Power delivered into a 4r load is
P=Vpk*Vpk/2/Load=50*50/2/4=312.5W

Maximum current when delivering this power is
Ipk=Vpk/Load=50/4=12.5Apk

That power & I peak cannot supply from my power supply.



I average saw in oscilloscope is +/- 30V
30Vpk=50/(root2)Vrms or Vac = 21.21Vrms

P=Vpk*Vpk/2/Load=30*30/2/4=112.5W

That may be possible for my power supply.

I think I must use sine wave generator for measure true power output.

Is that true ?

Can Output power vary because of frequency ?
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Old 9th September 2006, 08:40 AM   #23
Progg70 is offline Progg70  Sweden
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Thanks guys!

Every one said that the current was to low. To have the correct current, 3 Amp for the amp is it the same as – 1,5 amp negative side sin wave and + 1,5 amp on the positive side or …? (see picture with red questions.)

How is it with the class B amps? is it -1,5 over the PNP (– side) transistor and 1,5 over the NPN transistor (+ side), (see the picture). How is it with Single end where the whole signal is on one transistor and you have a current source? Should the current source was on the whole current 3 Amp or… (see picture.)
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Old 9th September 2006, 09:38 AM   #24
AndrewT is offline AndrewT  Scotland
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This was not requested but turned out to be usefull!
Quote:
Adding pk (zero to peak) or pp (peak to peak) as a suffix usually clarifies and avoids ambiguity.
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Old 9th September 2006, 09:52 AM   #25
AndrewT is offline AndrewT  Scotland
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Hi Prog,
firstly, get rid of your proposed 4ohm load.

8ohm speakers are a difficult enough load for an amplifier. Designing for true 4ohm ability just makes the job that much harder.

Each supply rail will see alternate peak currents of Ipk. That was the purpose of my presentation style.
The smoothing caps and any decoupling along the route supply most of the current spikes required by the load. The long term DC current comes from the transformer through the rectifier.
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Old 9th September 2006, 09:57 AM   #26
AndrewT is offline AndrewT  Scotland
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Hi Prog,
the ccs shown on your single ended amp carries constant current.
The load current is correctly shown as peaking at Ipk. But the arrow is pointing to the transistor, it should be pointing to the load.
The extra current, to supply the load, comes from the working output transistor.
It's current is Iq +-ILoad and can vary from zero (or almost zero as distortion increases badly as zero is approached) to two times Iq (again almost twice).

So much for all the erroneous references to constant current in a single-ended ClassA amplifier.
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Old 9th September 2006, 10:13 AM   #27
MikeB is offline MikeB  Germany
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Hi Progg70, sorry if i sound rude, your questions are too fundamental, you really need to work out them on yourself. Without understanding how all these currents sum up, you will never be able to understand any electronics/electricity. You must learn that by understanding.

A few hints:

- Q2 is reversed, the emitter needs to show up.
- The current through a ccs is constant, that's what makes it a current source.
- Get perfectly familiar with ohms law, I = V/R, V = I*R, R = V/I
- Think of currents like flowing water. You know what current is flowing through the 4ohms. This current needs to be a sum of the currents from both transistors, but don't forget the sign ! (The direction of the current flowing)

An example for the single ended:
- outputvoltage +1v, current through 4ohm is 0.25a, current through upper transistor is 1.25a
- outputvoltage -1v, current through 4ohm is -0.25a, current through upper transistor is 0.75a

Mike
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Old 9th September 2006, 10:48 AM   #28
kvholio is offline kvholio  Netherlands
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Kirchhoff's Current law:
http://www.physics.uoguelph.ca/tutor...Q.ohm.KCL.html

With kind regards,

Klaas
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Old 9th September 2006, 11:03 AM   #29
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Default You are a Jerk...as you can see, there are multiple visions about the subject

There are many variables that can fool the basic calculations, and even those calculations produce different results ..many results are provided by experienced people here.

The Watt meter will give the correct answer...we can go close with calculations...but as you see...where is the rigth answer?

I receive critics because made errors, if only perfect people could write in the forum you probable will se nobody...if we acept that there's a single correct answer, there are many making errors.

There are simple questions that have not a simple answer, as depends of this and that.

This show us that we have the need to exercise to be humble, as this simple question is showing me, and i hope showing to you to, that the correct answer may depends from other factors, and to colect all them is not so easy.

My vote goes for Michael, as the one has a lot of practice, not only theories...the one know the differences between calculations and real world data resultant of meterings....The humble factor in my point of view.

But, there are in this thread another practice monster...Anatech, this one had reparied more amplifiers than many of this thread could ever see in their lives...and one incredible good engineer, Janemann, as he is one of the forum references... and there are others that i did not mention.

regards,

Carlos
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Old 9th September 2006, 11:33 AM   #30
Progg70 is offline Progg70  Sweden
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Thanks everyone! ;-)

I think I have to build and test the circurit to get the right values for the current, watch it in real time and make the curve looking good.
A lot of speakers go down to 4ohm and below in different frequency areas.
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