>
> 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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