nouvelle electronique

[arex]

Hello everybody!

I found this diagram in the last issue of "Nouvelle Electronique". It is a simple class a amplifier, 10 watt output power, 10Hz to 100KHz. All resistors are 1/4watt. Nothing to tweak.

I am not sure on the wrightness of the diagram: R5=R6=18K???.
C2=220uF but C1 1000uF????. can anybody tell me, is this correct???.

Can I use a double power supply +- 12V by conecting C3 at 0 volt???.

And in general it is a useable/good diagram?

Thanks !

 

[Nelson Pass]

Looks a bit like the JLH with a constant current source
on top.

 

[arex]

I made a mistake: T1 is a pnp transistor.
The correct schematics joined:

 

[Tube_Dude]

I am not sure on the wrightness of the diagram: R5=R6=18K???.

Yes...correct! Is for the middle point at the output , to be at half the supply voltage.

C2=220uF but C1 1000uF????. can anybody tell me, is this correct???.

C2 must be at least 2.200 uF for good low frequency response , with a 8 Ohm load.

 

[Tube_Dude]

I made a mistake: T1 is a pnp transistor.

T1 in your second schematic must be connected with emitter to C3 and T3 must be a PNP transistor.

 

[darkfenriz]

I made a mistake: T1 is a pnp transistor.
The correct schematics joined:

No, no, no
T1 was OK, T3 wasn't

cheers

edit: Jorge was first, ignore

 

[arex]

WRONG again! T3 is a pnp too.
Excusez moi svp. THIS IS THE WRIGHT DRAWING.

Can I use a dual power supply by conecting C3 on 0 volt and eliminating C2??

Comparing with the JLH amp: the resistor on the Q2 colector (JLH) in a 7watt one.

Here (nouvelle electronique) the same resistor in the T4 emiter is a 1W ( 0.25 x 4). Why?

Thanks

 

[Tube_Dude]

Can I use a dual power supply by conecting C3 on 0 volt and eliminating C2??

No , because that way , the offset will not be controlled...

Here (nouvelle electronique) the same resistor in the T4 emiter is a 1W ( 0.25 x 4). Why

Because in your schematic the dissipation in that resistor is ~ 0,6W..so 1W is enough .

Attention in your last schematic T1 is still up side down

 

[moamps]

And something is wrong with amp's grounding.(-Ve and C3 on same potential?)
Also, quiescent current is too small to reach 10W/8ohm.

Regards,
Milan

 

[arex]

No , because that way , the offset will not be controlled...

I will join another diagram, hope the wrigt.

Please excuse my ignorance, how must I modify the schematic to eliminate C1 ??

 

[bobo1on1]

Dual supply version, thats what you want right?
Someone please comment on it.
I don't think the small 27 pF and 220 pF caps are necessary.

 

[eLarson]

Is there a benefit to having the 1711's emitter attached to nothing but the base of the output stage? (I'm thinking it looks a little naked without a resistor from emitter to the - rail, too, that's all.)

 

[arex]

Yes;

Thank you. This is what I wanted, but I don't realise by myself if your schematic is right. Comparing with JLH 1996 (see bellow), it seems to be a simplified version. Why did you changed the T1 base polarisation (R5=R6 in the initial drawing)?

Somebody else comments?

 

[wkopacz]

eLarson:
Tip142 is a darlington transistor which has internal on-chip resitors ie 8k, but if we add smaller value we speed-up amp...
I think that standard Miller compensation is good enough (C4), 47p is initial value...

An externally hosted image should be here but it was not working when we last tested it.

regards,wk.
for me current is too small...
R1=0.47R then I=0.7/0.47 =1.5A
1.5 x 24V = 36W per channel

 

[eLarson]

Ah! I hadn't thought to check the TIP142 data sheet. An internal path to ground (other than the base, that is). That would explain it.

Thanks,
Erik

 

[bobo1on1]

You could do without the cap in series with the 68R feedback resistor but any voltage drift from the base-emitter junction from T1 would be amplified 16 times.
You would have to test it, I'd rather not use a cap there at all.
Maybe a DC servo is a good solution.