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From: Mike Monett
X-Mailer: Mozilla 2.02 (Win16; I)
Subject: Re: Amplifying stage with negative Vgain??
References: <firstname.lastname@example.org> <3DEF7AA3.email@example.com>
Date: Thu, 05 Dec 2002 11:39:50 -0500
NNTP-Posting-Date: Thu, 05 Dec 2002 11:39:53 EST
Organization: Bell Sympatico
Fred Bloggs wrote:
> > The circuit is a dead simple single stage C/C amp. R1 and R2 (both 3k)
> > split 12v supply in half to bias the base of Q1 at 6v; there's no
> > collector resistor (straight to Vcc) and RE is 1,080 ohms. The input
> > and output coupling caps are both 1n. There was no load when the
> > measurement was made - other than my scope of course.
> > This is *not* a simulation as you may well have gathered by now.
> If the BC337 pinout is such that you have connected E to the input
> junction of the 3k's, B to 12V, and C to the 1080 output resistor, then
> you end up with a configuration that gets 2Vp-p across that 1080 for
> 8Vp-p at the input. You have almost certainly done something like this.
> You can figure this out if you measure the quiescent DC voltages at the
> three nodes- what are they?
Your hookup sounds strange. If E=emitter, B=base and C=collector, you have
invented a most wonderful emitter follower. I don't think it will work tho.
With the base connected to +12, both junctions are forward biased and will
be about 0.7V below +12, or 11.3V. You would have to drive it pretty hard to
get 2V p-p out. With 8V in, you wouldn't see anything at the output.
Normally, the emitter follower is hooked up like this:
B = input
C = +12V
E = 1080 ohm resistor
Since you are getting some signal out, you may have reversed the emitter and
collector leads as Tony suggested. This will function but at greatly reduced
beta. Try swapping them. Here's the DC operating parameters from SPICE:
DC Operating Point Voltages
Bipolar Junction Transistors
The nodes are shown on the schematic at
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