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Old 17th July 2007, 11:19 AM   #1
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Default Pure Class A Single End Amplifier Idea!

Hi all,

After my sucessfull Class A headphone project (by the way I had some modifications on it and it works very good now)
DC Offset on Class A headphone amp.
I decided to make a speaker amplifier based on it.
Click the image to open in full size.
I've been inspired from Rod Eliott's "Class A Op-Amp" project to make it a pure Class A design.
The bias current is 2 A or so. And it can give about 20W RMS into 8 ohm.
I havent tested it yet. But on 5spice software it works very very good. With 0 phase drift and 250Khz bandwidth. However I cannot make FFT and THD analyses.
Anyway, may you have a look and give some critics on it pls.
Thanks in advance a lot.
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Old 17th July 2007, 11:39 AM   #2
AndrewT is offline AndrewT  Scotland
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Hi,
the LTP input pair are not balanced.
R1 (22k) with near 24V will pass about 1.1mA.
Half of this should go through each of Q1 & 2.
Q5 passes about 9mA and it's base current will be about 20 to 40uA.
This leaves about 0.5mA through R4. It's value should be about 1k1 to 1k2.
You could add a trimming pot across a slightly higher value of R4. Try 1k3//(6k8+10kpot). Then set up the trimmer to balance the Vre on each side of the LTP. Oh, no Re, try adding emitter degeneration resistors of 22r to 200r. This will reduce the gain slightly but may sound better.

Why did you choose a & b grades of bc559/549?
I would have chosen c grade bc560/550 throughout.
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Old 17th July 2007, 04:20 PM   #3
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Lightbulb Some thoughts

You don't want any possible DC from the signal input to Q1 base.
Solution: Put a capacitor between the input and Q1 base.

You want to minimize any DC offset that your amplifier could produce.
Solution: Connect a big series capacitor with R9. (I mean big capacitance.)

You want to maximize linearity of the differential amplifier, reduce gain and thus also reduce possible stability problems.
Solution: Put small equal emitter resistances to Q1 and Q2.

While the Zener diode D1 is a voltage source, it is also a noise generator. The noise is amplified in Q7 and goes to the speaker.
Solution: Connect a capacitor parallel with D1 to suck the produced noise.

If your amplifier has stability issues, you must compensate.
One possible solution: Try a small capacitor between Q5 collector and base.

Instead of the current source (R2, R3, Q3 and Q4) maybe you want to use something simple.
Solution: Use a bootsrap connection from the amplifier output. You need one capacitor and two resistors for that.
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Old 17th July 2007, 06:07 PM   #4
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Regarding output stage, I have already seen something like this

The input LTP might be improved, and the output stage should be driven from low impedance, rather than from Q5 collector.
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Old 18th July 2007, 07:24 AM   #5
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Quote:
The input LTP might be improved
Right.

Quote:
the output stage should be driven from low impedance, rather than from Q5 collector
I disagree...
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Old 18th July 2007, 07:55 AM   #6
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Input LTP now improved I guess...
Click the image to open in full size.
The 15pF on NFB line to kill 2,2Mhz oscillation which I recently noticed.
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Old 18th July 2007, 08:05 AM   #7
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Is there any specific reason for using a Darlington instead of a Mosfet as a current source?
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Old 18th July 2007, 08:11 AM   #8
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No there is no special reason to use darlington. In fact I use only a power resistor for this position in my first design. (Headphone amp). After then I though to put a 5A regulator IC (like LM338 or LT1083) as a constant current source but I faced that power dissipation limits exceeded. Then I found that solution on a web design. Its not special, and not my design.
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Old 18th July 2007, 08:22 AM   #9
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ok,thank you very much indeed. may be PMA can explain it more.
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Old 18th July 2007, 08:50 AM   #10
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Hi Dxv,
that mirror on the input LTP is tried by many and many more citicise it for poor sound quality.
In addition the current into the base of Q5 unbalances the LTP. Is that why is sounds bad?

Go back to resistor loading.
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