Hi Byron, re your posted comments below about the
parts values, my simple circuit is just an example,
and not really designed for 8 amps!

It was designed for absolute simplicity, not as
a working model for your high power needs. ;o)

If you are expecting 2A as in your calcs I would
suggest some simple circuit changes. Some caps
in the right spots, a resistor network which doesn't
need as much current, some thought about bad things
like overcurrent and spike protection, etc.

It's probably going to cost another transistor
to fix the resistor network issue and a few discrete
parts to get this circuit looking realistic at 2A.

I do think your calcs are very conservative, i've
been using low sat transistors a bit lately and
would suggest a 15A transistor beta at 2A will
saturate quite nicely at about 60-80mA. I do advise
testing the sat voltage yourself because it varies
from type to type and some are better than others.

Is there any reason your ruled out the FET??
At 2A the FET sat volts will be very low, and at
12v in and 7805 5v you still get close to 12-4.4
=7.6v to turn on your p channel FET. Then your
resistor ladder will be quite cool. ;o)

Then you just need 2 of these regulators, one from
the vehicle battery to your gell cell, keeping it
charged to optimum, and the other regulator from
gell cell to the PC 12v rail.
-Roman




Byron A Jeff wrote:
>
> On Fri, Jan 11, 2002 at 05:03:04AM +1100, Roman Black wrote:
> > Byron A Jeff wrote:
> >
> > > Now we're getting somewhere. This is the issue I want to address. Please
> > > take a stab at explaining the controller using the 7805. I know that it's
> > > possible to use the GND pin as an adjust pin. However I just don't see how
> > > to get over the dropout voltage issue. I just took a look at the National
> > > Semi LM340 datasheet, and even with an output current of 0A it starts to
> > > drop out about 1.5V above the regulation voltage. So how can it be used as a
> > > control element when Vin is in the dropout region?
> >
> > Wow! That was a long post. :o)
> > Here's my views on building a low dropout series
> > regulator as simple as possible.
> >
> > First thing is separate the thing into two parts:
> > * power (obviously a low sat power device)
> > * control
> >
> > ONLY the power device needs to be low dropout. The
> > control circuit can be anywhere, and you put it in
> > the best place, referenced to the zero rail.
> >
> >
> >                       Low sat PNP
> > Vin -------------------------E C-----------
> > Vout
> > (13.8v)     |           |     B     |     (Reg 12v)
> >             |           R     |     |
> >             |           |     |     |
> >             |           |-----|     R1
> >             |                 |     |
> >             |  7805           C     |
> >             |--I  O---------B       |
> >                 G             E-----|
> >                 |        NPN        |
> >                 |                   R2
> >                 |                   |
> >                 |                   |
> > Gnd
> > ---------------------------------------
> >
> >
> > Our power stage is just a low sat PNP like a BD204
> > or better part. The entire control system is referenced
> > to Gnd, and a 7805 is used as a cheap easy precision
> > zener, giving exactly 5v no matter what the input
> > voltage does.
> >
> > R1 and R2 are a voltage divider tied to Vout. They
> > set the NPN emitter at 5v-0.6 =4.4v whenever the
> > output is at the correct voltage. Any less than the
> > correct Vout, and the transistors will turn on as hard
> > as needed to cause regulation or saturation, whatever
> > comes first.
>
> So that's the "magic" formula!! The R1/R2 ratio is just like the divider on
> an LM317. So if I calculate R1/R2 so that the midpoint gives 4.4V when the
> PNP collector is 12V. The transistors will lock in to put the PNP collector
> to 12V.
>
> Young grasshoppa finally understands. Thanks.
>
> So I did a quick calc using Vo = (R2/(R1+R2))*Vin with 12.0 for Vin and 4.4
> for Vout and got a ratio of R1 = 1.73*R2.
>
> >
> > Make sure the R1/R2 divider passes more than 10x
> > the max expected PNP base current to keep regulation
> > fairly accurate. If you keep this current substantially
> > higher than the PNP base current the regulation will
> > be close to perfect, at all voltages right down to
> > saturation.
>
> And only at the cost of the base current which is the one item I have no
> concern over in this circuit. And just to verify the base current will be
> 10 * (Iout/beta) right?
>
> So let's put it all together. I went poking around the Philips NTE/ECG
> replacements parts site, because there are several Atlanta Distributors.
> I picked the NTE2314. 15A max, Vce(sat) of 0.25 typ, 0.5 max at 8A draw.
> Min beta of 30 at 8A.
>
> R2 would need to draw 10 * (2/30) -> 0.6A of current @ 4.4V. That works out
> to 7 ohms. If I use R1=12 ohms and R2=7 ohms it'll set the output voltage
> to 11.92V.
>
> Only one missing piece: what's the purpose of the emitter to base resistor
> on the PNP. A pullup to somehow force complete cutoff of the transistor?
> What value might it need to be?
>
> Another question: can R1/R2 be fed into a voltage follower? At that low a
> value the resistors will burn more than a half amp all the time.
>
> > -Roman
> >
> > PS. It will work just as good with a P type FET
> > as the power part, if Vin is 12v or higher.
>
> Then the Rds determines the dropout.
>
> This is a winner! I'll start working on it immediately.
>
> Thanks for all your help.
>
> BAJ
>
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