Re: message repost

On Fri, 5 Sep 1997, Zonn wrote:

> At 10:41 PM 9/4/97 -0700, Anders wrote:
> >If you look at the P314 schematic and "remove" the pre-drive transistors,
> >zeners, and resistors, you can see that this effectively "disconnects" the
> >unregulated power from the base of the power transistors. So all I did was
> >put my LV PCB in the "gap". Now for a visual...
> >Being inspired by your fine ascii art, Zonn...here
> >is my attempt at it. My LV PCB is wired up basically as follows:
> >
> >(+33V unregulated) ---------*----------------------+
> > | /
> > | | / Q102
> > | |/ 2N3716
> > +--[LM317]---------| (NPN)
> > | |\
> > | | \
> > | V---- to + HV and
> > | + deflection BJTs
> > [adjust parts] adjust to +25V
> >
> >
> >(-33V unregulated) ---------*----------------------+
> > | /
> > | | / Q103
> > | |/ 2N3792
> > +--[LM337]---------| (PNP)
> > | |^
> > | | \
> > | \---- to - HV and
> > | - deflection BJTs
> > [adjust parts] adjust to -25V
> >
> >So as you can see, the regulators keep the base voltage constant, thus
> >causing the power transistors to "follow" the output of the regulators.
> >The output voltage of the power transistors goes to the HV and to the
> >deflection BJTs, and is only one diode drop less than the output voltage
> >of the regulators.
> >I designed the adjust circuitry for maximum ripple rejection (80db if you
> >read the data sheet). So the DC power generated at the emmiters of the
> >power transistors is *very* clean/flat, and constant. And, since the base
> >current is approximately two orders of magnitude less than the output
> >current, the three-terminal regulators "don't get overworked".
> >
> >An area of improvement that has cropped up in this discussion is to add
> >some form of current limiting to the unregulated power -- i.e., limiting
> >the current into the collector of the power transistors.
>
> What you have done is what WG *tried* to do using a zener diode as the
> reference into the base of the power followers.

Yes, but the BJT follower configuration is a much, much more robust way to
go. I will explain...
 
> Obviously using the LM3x7 is a much nicer way to go! The only nitpicky
> thing I see is that your not actually regulating the output voltage. By
> regulating the base of the transistor you leave the voltage drop across the
> BE unregulated. My understanding is that while this is right around .6v it
> can vary based on temperature and current draw, though in this case it
> should be pretty insignificant. (What's a few millivolts compared to 27 volts?)

OK. Here we go! Good observation that the output voltage is actually less
than the regulator's output. This is true. The output voltage of the
follower is almost the output voltage of the regulator --
i.e., Vout = Vin - Vf; where Vout is the voltage at the BJT emitter, Vin
is the voltage at the BJT base, and Vf = Vbe which is at worst case 0.7
volts.
Without going into the entire math proof (very long), the Vout = Vin - Vf
equation is good as long as Rb (the base resistance) is very small. You
will notice that my design has no external base resistor. With a very
small Rb, we can assume Rb = 0, the equation for the follower, Vout = Vin
- Vf, is exact and not an approximation of the follower output.
So, say you want +25 volts at the follower output. You just adjust the
regulator to outpu +25.7 volts, or with a DMM on the follower output
adjust the regulator until you get +25 volts. Very user friendly.

> A bigger problem could be adding current limiting to the design. I haven't
> really thought it through for the above design, but when placing the pass
> transistor inside the regulation loop it's a simple matter of adding a
> resistor and another transistor to allow for current limiting.

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.
 
> (Nice ASCII art!)

Thanks!

-Anders

+-----------------------------------------
| Anders Knudsen
| ASIC Design Engineer
| Adaptec, Inc., Boulder Technology Center
+-----------------------------------------
Received on Sun Sep 7 21:26:11 1997