From: Chuck Simmons
Organization: You jest.
X-Mailer: Mozilla 4.61 [en] (X11; U; Linux 2.0.33 i586)
Subject: Re: Help! Diode biasing of Class B amplifiers
Date: Sat, 05 Oct 2002 14:32:11 GMT
NNTP-Posting-Date: Sat, 05 Oct 2002 07:32:11 PDT
Paul Burridge wrote:
> I'm having serious difficulties getting my head around this
> I've seen several circuit diagrams illustrating the use of
> compensating diodes to establish the correct operating point for the
> output transistors; just bringing them to the verge of conduction to
> minimize x-over distortion. All fine, except that I can't see how the
> input signal is supposed to ignore the 'one-way-only' property of
> these diodes. For instance, across the +20v and -20v rails of the twin
> supply, there is shown, in series, a 4k resistor, two diodes and
> another 4k resistor. Both diodes point the same way with arrows
> towards the neg. rail and are thus forward biased. The input is shown
> as being applied in the middle between them. The bases of the
> complimentary output pair are connected to the point between each
> diode and its resistor.
> A positive-going part of the input cycle can only reach the base of
> the lower transistor, which is a PNP and therefore 'expects' a
> negative signal to function; whereas a negative-going input signal
> portion can only reach the top transistor (an NPN) which 'expects' a
> positive signal to its base. I was tempted to think the publishers had
> mixed up their polarities, but this circuit is so prevalent in so many
> different books both US and UK that seems impossible. Any ideas,
You need to look at this differently. Keep in mind that there is a bias
current through the diodes. If the preceeding stage raises the voltage
at the junction of the diodes toward the positive rail, the voltage at
the base of the NPN will be pulled toward the rail by the resistor. The
voltage drop across the diode will drop because there is less current
from the resistor. The current in the lower diode will increase but if
the load is capable of sinking current, the NPN will conduct current
into the load as a voltage follower and its base voltage will rise
making up for the drop in the voltage across the top diode. Thus the PNP
transistor current will not change much.
Sorry, that is really confusing.
For that type of circuit to work well, there is usually a boost voltage
for the diode bias. This is needed to get enough base drive for the
output transistors. In AC applications, the output can be rectified and
filtered to produce the needed boost. Drat! Even more confusing.
Oh, nevermind. The circuit works as advertised.
... The times have been,
That, when the brains were out,
the man would die. ... Macbeth
Chuck Simmons firstname.lastname@example.org