Re: About some parts on logical circuits

On Wed, 23 May 2007 12:07:55 +0200, "Dr Gang" <dr.gang@xxxxxxx> wrote:

I didn't see the resistors that you were referring to. Was this by any
chance on some pushbutton switch logic? Often you will see a cap and
resistor in a debounce circuit.

Precisely, I wasn't aware of the pull-down resistor issue. They're the ones
I was thinking about.

The transistor part brings no interrogation (maybe it should...).

If you want to take a look at the schematic, here it is :
http://dr.gang.free.fr/test/Swtiching.pdf

The ICs are dual JK flip flops and a quad OR gates. The schematic shows
JFET
transistors but I'll use more common NPN ones.

There's a difference in the way that a bipolar loads the JK's output
pin as compared to a FET, but you probably won't have any problems.

That's something I can't figure out : what is the load seen by the JK's
output ? Since each transistor will have to feed a dozen of LEDs at 40mA
each, that makes a rather high current and I'm wondering if this won't suck
too much from the JK's output which will "see" the LED network through the
transistor's gate.

That's a lot of current! A dozen? You need to make sure that the
driver transistor can handle it. Let's say you're driving 500ma.
You'll need to account for dissipation by multiplying the voltage
across the transistor by .500A to get the wattage. Small heatsinks
would probably be a good idea.

The required source current can be approximated by dividing 500ma by
the beta of the transistor. Usually driver transistors don't have very
high betas. The way around that is often 'Darlington' transistors,
which are a piggy-backed configuration. But there's another problem
with that: You may not have enough voltage at the JK output to push
the darlington into conduction. The required drive voltage would have
to be the turn-on voltage for the LED + the forward breakover of a
darlington (1.2v) + some margin.

Here's another idea: Run the LEDs off the transistor's collector. That
would eliminate the voltage threshold problem, but you'll still need
to make sure the JK can source the current. You'll need to make sure
that the JK's output is not loaded, since the emitter will be going to
ground (base-emitter looks like a diode to the driving circuit), so
yse a resistor in series with the JK output going to the transistor
base.

You'll still need current limit resistors on the LEDs (I recommend
separate resistors for each LED).

Also, high-efficiency LEDs should not need 40ma. That will trim the
drive current and wattage requirements way down.

In the same idea, in a cascade of triode, what is the current going from the
first tube plate to the next's grid ? For me, a grid is a dead-end which
therefore can't draw current. I'm probably not seeing things the right way
but can't figure it out.

A tube grid draws almost 0 current. But if it's driven positive (with
respect to cathode), it will conduct. Normally, guitar amps don't
account for that in preamp stages but it can happen. The basic idea in
a preamp is to amplify voltage, so current is not a direct concern
(can't cover the subtleties here). Usually only output stages are
designed for current.

And change to individually drawn FF's and gates. Rather than drawing
the package outline, draw the bullet-shaped logic symbols for AND/OR
and individual boxes for the FFs. It makes your intent more clear.

Indeed the circuit is messy if you stick your head in it for the first time.
Luckily enough, there are competent people in this NG. :)

It's a pain to trace the logic. I didn't. <g>

Speaking of intent, what are you going for here?

I'm making a 3 channel amp. The switching scheme involves optocouplers. Each
channel has its own network of 5 to 8 optocouplers to turn on (while the
others would turn off) when switching to it. This schematics is the
footswitch pedal that'll allow me to change channels. There are mechanical
systems to switch between 2 channels but I've never seen anything fit for a
3 channel switch so I went logical.

If you decode your schematic as I suggested above, I'll take a look at
the logic. But you gotta do that.

.