Re: message repost

At 10:10 AM 9/8/97 -0500, you wrote:
>
>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.

Yes, I have thought about this. If something in the HV shorts, or just one
of the deflection transistors shorts, then the load will tend to look like
Re = 0, and POW, the LV power transistors are gone! I am looking at a
couple of possiblilities for current limiting on my LV PCB. As soon as I
have gotten around to analyzing them, I will post up what I have found. In
the mean time, I am open to any and all suggestions...

-Anders.

 -----------------------------------------
| Anders Knudsen
| ASIC Design Engineer
| Adaptec, Inc., Boulder Technology Center
| anders_knudsen@btc.adaptec.com
| http://www.adaptec.com
 =========================================
Received on Mon Sep 8 08:20:12 1997