Re: About some parts on logical circuits


"flipper" <flipper@xxxxxxxx> a écrit dans le message de news:
1tod53lnh3af8buteo3o4i5tn8ifcsvm7l@xxxxxxxxxx
On Fri, 25 May 2007 12:35:25 +0200, "Dr Gang" <dr.gang@xxxxxxx> wrote:

Well, it depends on the whole design. but a 'will probably work no
matter what' choice might be a FQPF7N10L for 64 cents each at Mouser.

Given the money I spent so far on this project, I'm not counting the cents
anymore here :)

Ok. It's just instinct for me.


What transistor did you buy?

These are BD135, big packaging, can handle 1A.

I see. Well, that's a TO126 and while it says 1.5W to ambient that's
at 25C. I don't like putting more than 3/4 watt through those
unheatsinked. Conservative but I just don't like running them at the
boiling point.

That one is a good example for why they make different transistors.
While it says absolute max current is 1.5A if you look at the Hfe
numbers it's clearly intended for operation between, say, 20mA up to
150mA or so. That's the 90% Hfe range.

You get different minimums from different manufacturers (another
problem for product designers with part substitution) but the
Fairchild *** says Hfe min 40 at 150Ma and lower at .5A, down to 25.

Allright... as if it wasn't complex enough with a constant gain... I
understood that OAs were made precisely because the final product was much
more constant than transistor. Don't we have a solution there ?


That's a real problem for you because it would take 20mA into the base
for .5A of LEDs. CMOS gate won't do that. But, as I've said, a
Darlington front with a BC550 would take care of it.

It's high time for some stricter calculations. I've numbered the
optocouplers to turn on for each channel.
Clean requires 5 optocouplers.
Crunch requires 11 optocouplers (and I don't even play blues !!!)
and finally Lead requires 8 optocouplers. Some are common to 2 channels
(I'll use the not-Q signal of the flip flop corresponding to the 3rd channel
and therefore will need an additionnal transistor)

at a more reasonable 25mA for each, that'll bring us respectively to 125,
275 and 200mA. If settling for higher on-resistance makes the whole story
easier and safer, I'll do it.

I wondered about stealing filament power.

The problem with your 7 volt is that's filtered only and not a great
idea for digital circuits. It should be regulated and that'll get you
5 volts with a low dropout 5V regulator, depending on just what the
actual AC is when that winding only has a .5A load on it, and keep the
ripple real low. Hmm, but it might drop out on low line voltage. Not
good.

Better would be to diode double it to 14 and, since 8V 3 terminal
regulators are a dime a dozen, 8V out. You *will* need to heatsink
that thing, though, and secondary AC being high on high line voltage,
plus the low load bump, with aggravate it. Could easily hit 5 watts at
.5A (and I'm assuming you go to 20mA ea on the 21 LEDs) and that's not
a dinky 1 inch stamped metal heatsink. Need a bit of oomph.

With the precisions I brought (and should have brought sooner), it shouldn't
be that bad. What I don't understand here is why you don't want to use a 6V
or 7V regulator instead of a double the voltage output to run it through an
8V regulator. There are regulators for lower voltage than 14v.
It's the loss of 6V at 0.5A that would make the requirement for dissipating
3W with a heatsink right ?
With Duncan PSU designer, I quickly simulated how shaky would be the
rectified signal. With a 220µF cap, I get 0.05V of ripple, 1%. Can't CMOS
handle that ? Can't I just regulate the CMOS supply and stick to the normal
rectified signal for the transistors and LEDs ?
Now that I think about it, I think I forgot about the voltage drop in the
diode bridge... I'm good for ordering another batch of 20 LED resistors with
the right value...

Getting the 14V is like generating a +- supply but with the - end at
ground. I.E. One end of the winding goes to the junction of two
capacitors and the other end has two diodes. One positive going to the
'top' cap and the negative one going to the 'bottom' cap. When a +-
supply the junction of the two caps is ground and you get +- 7 volts.
Connect the -7 to ground and you get +14 on the top side, with +7 in
the middle.

I think I've seen that voltage doubler circuit once or twice before.

If the winding is putting out significantly more than 5 volts under
low load, like your .5A, you could cut it down a notch with some
series resistance, a power resistor. Or knock it down a notch,
regardless, to cut the heat load on the regulator because it's sort of
too little or too much either way.

I have to measure what comes out of there with my voltmeter. I measured the
6,3V filament winding and it is not above the specified 6,3V even without a
load so...

Or you cut put a diode or two in the ground leg of the regulator to
bump the output voltage up a bit. Goes up by the diode drop and
that'll reduce the voltage across it and, so, what it has to
dissipate. Two diodes would give you about 9.3 volts out. (I do that
to make 6.3VDC filament regulators from a 5V brick or 1 diode to get
12.6VDC from a 12V brick).

Like I said, lots of ways to skin the cat ;)

Well you sure know what you're talking about. It's comforting to have you
and others hanging around on this NG ! :)

Thx !


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