Re: message repost

On Sun, 7 Sep 1997, Anders Knudsen wrote:

> OK, now comes a very interesting "feature" to the BJT follower
> configuration. If you believe the math (and all this shit is based on
> math!), then the follower has the power to deliver *just* the right amount
> of current to the load -- in our case the load is the HV unit and the
> deflection transistors. Now all this math gets pretty hairy, and kinda
> makes me sick to look at it! (That's why I got into digital design ;-) But
> here is the "mechanism" by which the BJT follower delivers just the right
> amount of current. The collector current (Ic), which is the current
> supplied to our load, is related to Vbe by the v-i characteristics of the
> base-emitter junction. When voltage is applied to the base (greater than
> Vf or 0.7V) current starts to flow throught the base (Ib) and the
> collector. If we call the load (HV and deflection transistors) Re, this
> value is of course not constant, the voltage drop across Re limits the
> value of Vbe. An equilibrium condition is then reached where Ic and the
> drop across Re estabish the exact Vbe needed to proved Ic -- we have
> feedback! The equilibrium value of Vbe will always be close to Vf (0.7V)
> for any Ib, because the relationship between Ib and Vbe is an exponential
> one. The current Ib will be much much less than the current Ic. The
> current through the load will be Ib + Ic (Kirchoff's current law). Ib is
> much smaller than Ic, since Ic = (Bf)(Ib); where Bf is the current gain of
> the BJT, probably around 100 for the power transistors used in the WG6100.
> So say Ic = 3 amps, then Ib is 30 milliamps. Thus the power wasted in the
> base-emitter is Pbe = (Vbe)(Ib) = (0.7)(.03) = 21 milliwats.
> So what the follower is doing here is providing current gain because the
> load current is much larger than Ib.

        The problem is limiting the current through the transistors, or
adding "short circuit" protection. If the load resistance, Re tends
toward 0, obviously, Ie gets VERY large, and Ib = Ie/(Bf + 1) gets pretty
large too. You need to add at least current limiting resistors to teh
base and the emitter to stop that circuit from first saturating (whereupon
all of the equations that you mentioned up above break down) and then
destroying themselves.

        Like I said, current limiting could be as easy as finding the peak
current draw through the existing LV supply, and adding current limiting
resistors to Anders's design.

        If I missed something like this, previously, let me know. It's
"crunch-time" here at work, and I haven't been following this thread very
closely...

Joe

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Joseph J. Welser jwelser@ccwf.cc.utexas.edu
Design Engineer -- Crystal Semiconductor Corporation
Ph.D. Student in E.E. -- University of Texas at Austin
Work: jwelser_at_crystal.cirrus.com http://www.crystal.com
P.O. Box 17847; Austin, TX 78760
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Received on Mon Sep 8 08:10:44 1997