Wagner Lipnharski wrote:

> > +6v (battery)
> >  ---------------------*-------------------------------
> >                       |
> >                BC327  |        L1=470uH
> >  ----,                E        ( <1ohm )
> >     C|----R---------B    Q1
> >      |   2k2          C
> >      |                |
> >     S|----R-----------|-----------------,
> >      |   270 ohm      |                 |
> >      |                |            A    |
> >      |                *----L1------*----|---*-----,
> > PIC  |                |            |    |   |     |
> >      |     ,----------|------------*    |   -    SOL COIL
> >      |     |          |            |    |   ^     |
> >      |     R          |            |    |   |     |
> >      |     |          |            CS   |   '-----*
> >     V|-----*          -       470uF|    |         |
> >      |     |          ^            |    |         C
> >  ----'     R   1N5819 |            |    '-------B    Q2
> >            |    scho  |            |              E  BC337
> >            |          |            |              |
> >  ----------*----------*------------*--------------*---
> >  GND


> What is the PIC VCC in the above circuit?
> If it is +5V, then Q1 will be always active, no matter what. It would only
> cut if VBE would be lower than 0.6V, with +5V at VCC, the most the PIC can
> deliver via the 2k2R is 4.95V, so, at least 1.05V will be Q1 VBE.

Yes I should have been more specific, the "6v battery"
is the PIC Vdd voltage, probably two 3v cells and a
series diode to make it about 5.5v.


>the 2k2R suggests 2mA as Base current, at best
> Beta=80, Q1 will be conducting around 180mA maximum.

Not so, i've recently done a lot of testing of BC327
in buck regulators, and even at Ib=1mA Vce=0.2v@100mA
and even better at Ib=2mA of course.
This buck cap charger is *designed* to use a small
current from the battery, if the pulse on period is
correctly short enough the current drawn from the
battery (Q1 Ic) should be 30mA or even less. So 1mA
to 2mA Ib is about right for hard saturation.


> If its load is lower
> than 33 ohms (L1, CS, COIL), then Q1 will not saturate and its VCE will
> heat it up.

No, because it is a buck switch and the Ic current
is a function of L1 inductance and pulse length.

> The 270 ohms resistor suggests that 16mA will be injected at Q2 base, is
> that right?

10mA+ is about right, BC337 Ib=10mA Vce=0.1v@800mA,
he said the solenoid coil is about 500mA so I allowed
a decent margin.


> what is the reason for a buck charge to the cap to +6V?
> A simple resistor woudn't do the same job?

His original requirement was for a way to PULSE a solenoid
(drawer latch) and the whole circuit to run from a very
small battery that had limited current ability and a
small total capacity. The buck circuit should be close
to 90% efficient to charge the cap, a resistor is
simpler but much less efficient. This may equate to
3x more drawer openings before the battery is flat and
allows more flexibility to reduce battery current drain
and reduce total cap charging time. Since the PIC is
already available the cost is only one inductor, a
diode and two resistors more than using a resistor to
charge the cap.

> I guess the battery voltage is not +6V, isn't it?

Correct, the voltage is two 3v cells and whatever needed
to run the PIC, I assumed a drop diode as this is a
micropower circuit.
-Roman

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