line driver with automatic bias

[Ban]

features AB/operation of the output transistors with a bias current of 1.5mA
(set by 15k). The complementary pairs should have a similar beta. To have a
symmetrical clipping at around +/-11.5V a zener in the positive supply of
the opamp overcomes the bootstrapping action of the 100u capacitor.
Keeping the output transistors conducting reduces distortion by a whole
magnitude. There is no current limit, so an 100R output resistor protects
the transistors. Temperature compensation is intrinsic to the circuit, no
thermal runaway should occur.
.--------o------------o--o
| | | +15V
| .-. |
| | | |
| | |15k |
z '-' |
2.7V A | |
| .-----o |
| | \| |/
| --- PNP |--------| NPN
| --- >| |>
| |100u | |
___ | | | ___ | ___
o--|___|-o---)--)-----)---|___|----o-|___|-o
in 10k | | | | 10k | out
| | | | |
| | | \| |/
| | | NPN |--------| PNP
| |\| | <| |<
'-|-\ | | |
| >-o-----o |
o--------o-|+/ .-. |
GND | |/| OP27 | | |
| | | |15k |
=== | '-' |
GND | | | -15V
'--------o------------o--o
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Ciao
Ban
Apricale, Italy

 

[Martin Riddle]

What a pile of crapola. Did you get that from Larkin ?

...Jim Thompson
--

But, it has values

Cheers

 

[Ban]

This one is simpler and better in all ways:

http://www.abvolt.com/misc/line_drv.jpg

Sorry, but it has the same number of parts, and try to establish a stable
bias current from -25° to 125°. The variation is less than 5%.
The distortion at 10kHz 20Vpp 100R load is less than 0.025%. With some
4401/4403.
ciao Ban

 

[Ban]

This is nice and simple.

ftp://jjlarkin.lmi.net/OpampBoost.JPG

The resistor lets the opamp do some of the work at low level,
eliminating the deadband. Sort of. No thermal problems at all.

We could have a 500+ post thread about which "class" this is.

I don't understand Ban's circuit.

The resistor onthe upper left drives 1mA into the emitter of the PNP, from
which most will be taken by the collector, only 1/beta will flow out of the
base into the base of the NPN O/P transistor. This causes the same quiescent
current flowing in the output pair.
Temperature variation is minimal, since we do not have a bias voltage but
current drive, and the first transistor adjusts its base current
accordingly.

ciao Ban

 

[Ban]

What a pile of crapola. Did you get that from Larkin ?

Larkin will like it a lot after he has put it in the simulator and compared
to his circuit, maybe you should do it too?

 

[Tim Williams]

The PNPs are upside down.

Tim

 

[Grant]

The resistor onthe upper left drives 1mA into the emitter of the PNP, from
which most will be taken by the collector, only 1/beta will flow out of the
base into the base of the NPN O/P transistor. This causes the same quiescent
current flowing in the output pair.
Temperature variation is minimal, since we do not have a bias voltage but
current drive, and the first transistor adjusts its base current
accordingly.

So did you draw the PNPs upside down? Or operate them backwards on purpose?

Grant.

 

[Grant]

The PNPs are upside down.

I just asked that too, 'cos it don't make sense otherwise, and then it's
nothing new?

Grant.

 

[Ban]

I just asked that too, 'cos it don't make sense otherwise, and then
it's
nothing new?

The bases are connected together, a strange and unusual configuration. Put
it in a pspice simulator and see what happens.
Ban
Apricale, Italy

 

[[email protected]]

The resistor onthe upper left drives 1mA into the emitter of the PNP,

How? It's connected to the collector. Relying on reverse operation? ;-)

from
which most will be taken by the collector, only 1/beta will flow out of the
base into the base of the NPN O/P transistor. This causes the same quiescent
current flowing in the output pair.
Temperature variation is minimal, since we do not have a bias voltage but
current drive, and the first transistor adjusts its base current
accordingly.

....assuming your PNPs and NPNs are perfectly matched.

 

[[email protected]]

I think the PNP output transistor is drawn wrong, the emitter should
be connected to the output. And, either the PNP driver transistor
is acting as a zener diode, or its emitter is drawn on the wrong
end, too.

Both PNPs are drawn up-side-down, though it's still dumb if you flip them
over.

 

[[email protected]]

Is it really any different from this?

ftp://jjlarkin.lmi.net/SillyAmp.jpg

Sure. Your PNPs are drawn the right way. ;-) ...but you have no feedback;
crossover distortion is going to be *ugly*. ;-)

Even with feedback outside the followers, I think yours has worse crossover
distortion as the opamp slews.

 

[Ban]

Is it really any different from this?

ftp://jjlarkin.lmi.net/SillyAmp.jpg

John

uploaded my simulation file, has some beefier driver in sziklay
configuration.
The PNPs are in fact upside down.
http://i33.tinypic.com/20kx3lv.gif
Ban

 

[Grant]

....

uploaded my simulation file, has some beefier driver in sziklay
configuration.
The PNPs are in fact upside down.
http://i33.tinypic.com/20kx3lv.gif

Well at least your PNPs are right way up, but circuit looks too busy,
have you built this or just simulating? Looks to me as if you expect
some matching that's not going to happen in real life?

Thanks for putting up a clearer circuit!

Grant.

 

[Ban]

...

Well at least your PNPs are right way up, but circuit looks too busy,
have you built this or just simulating? Looks to me as if you expect
some matching that's not going to happen in real life?

Thanks for putting up a clearer circuit!

Grant.

Of course a simulated circuit looks a bit messy with a lot of additional
placeholders.
I have breadboarded the simple 4 transistor version and tried also
transistors with different betas. Best is to take the PNP from the C-bin and
the NPN B-suffix. The 4401/3 worked also.
The circuit needs very little compensation, since there is no dead-time. And
the transient behaviour is very good.
I want to try breadboarding a headphone amp, when I have tested different
output configurations in the simulation.

Ban

 

[[email protected]]

That's last one is not "mine"; I think it's equivalent to Ban's.

I know. I used it 40 years back. It sucked then, too (but was good enough
for the application). I don't think it's equivalent at all. I don't see
where the OpAmp goes open-loop for Vbe X 2. Maybe I'm missing something,
though.

The little arrow thingie that says "FB" is where the feedback goes.

You couldn't draw a couple of resistors?

But even with no feedback, it has no crossover distortion, at least at
low levels and low frequency. It's class A, beta biased. Horrible.

If there is no crossover, you don't need two transistors. A follower will
work.

You may as well follow the opamp with a single NPN emitter follower,
with a passive or constant-current pulldown. Works better, not
beta-funny, less parts.

More power.

 

[Tim Williams]

Thanks for the data. I guessed Ban's had better crossover performance
because I didn't see a way to reliably set the idle current in yours.
So, I assumed the idle current was effectively zero, causing a
deadband, which I guess was wrong.

Nahh, think current mirror.

Tim

 

[Ban]

On Sun, 29 Aug 2010 10:52:16 -0500, Vladimir Vassilevsky


Ban only has a 15K resistor to beta-bias the outputs. That's going to
limit the output swing capability. With a 600 ohm load, you may as
well just use an opamp all by itself.

One thing I like to do is use multiple opamps in parallel. For
example, two opamps each with a 100 ohm resistor to the load, net 50
ohm source. Or four opamps, same idea. If you want to go fast, into
the MHz or 10s of MHz range, discretes get messy.

The base-to-base thing is sort of cute.

John

With 1mA base current you can drive +/-10V into 50 Ohms, The O/P transistors
have a beta of 300+ You can also reduce the bias setting resistors easily.

 

[Tim Williams]

Why? VLV doesn't have a current mirror, just compound emitter
followers. Idle current is set by Vbe drops and matching.

Yes, in other words, a current mirror ;-)

Compare to the conventional two-diode-biased emitter follower. Assuming
the diodes are the same junction as the Vbe's (e.g., diode strapped 3904s,
or even better, one NPN and one PNP, don't use big fat 1N914s which have
much less drop), it's a current mirror.

Tim

 

[Martin Riddle]

The bases are connected together, a strange and unusual configuration.
Put it in a pspice simulator and see what happens.

Reversing the Collector and Emitter will guarantee that it will not or
may not always work.

Cheers