Re: PMOS in parallel with NMOS

On May 8, 6:57 pm, rickman <gnu...@xxxxxxxxx> wrote:
On May 8, 12:58 pm, Tomás Ó hÉilidhe <t...@xxxxxxxxxxx> wrote:

On May 8, 4:20 am, rickman <gnu...@xxxxxxxxx> wrote:

Before I bother to address the design issues you present, I want you
to go back and reread the last couple of messages you posted and tell
me exactly what circuit you have described.  I only see described a
pair of transistors with their B-E junctions connected directly to the
power supply.  I have no idea if you are talking about an emitter
follower configuration or a common emitter configuration.

Take a PNP bi-polar transistor.
Connect 0 volts directly to the base.
Connect 5 volts directly to the emitter.
Connect the anode of the LED to the collector, and the cathode of the
LED to ground.

What power supply are you using?


The 0 volts on the base is provided by a microcontroller pin.
The 5 volts on the emitter is a direct connection to Vcc (which
happens to come from an LM7805 voltage regulator. The voltage
regulator is powered by a 9 volt square battery).


When you turn this circuit on, how
long does it operate before the smoke comes out?


Nothing "bad" happens. Also, the green LED's don't glow yellow.


 Do you actually have
5 volts on the emitter?


Yes, the emitter goes straight to the 5 volts coming out of the
LM7805.


What is the voltage on the base and
collector?


The base is getting 0 V from a microcontroller pin.
The collector goes to the anode of an LED, and the cathode of this LED
goes straight to ground.


 There is something wrong with this.  Either the circuit is
not what you expect, or your supply can't drive more than a few mA and
the voltage is very low.


I know, it's a bit funky, I realise. If I use a power supply of better
quality then I've to turn down the pulse width of the LED's to stop
the board from dying. I realise this is dodgy, which is why I'm
currently going over the design.


 Even then, the voltage across the B-E has to
be equal to the sum of the voltage across the E-C added to the voltage
across the LED.  The LED will not turn on without at least 1.5 volts
or so.  The B-E will not support more than about 0.9 volts without
burning up.  So which is it, the transistor going up in smoke or the
LED not lighting?


The B-E voltage *should* normally be about 700 mV. Since I have 5 V
applied across it though, there must be 4.3 V dropped *somewhere*.
Maybe it's dropped in the internal resistance of the microcontroller
pin, maybe it's dropped inside the transistor some how. I don't know.
I'll be going over it.


If the circuit does not burn and the LED lights, then explain to me
what is wrong with what I have said above.


I can only conclude that there's non-negligible voltages being dropped
internally in the components. (Yes, I realise this is a bad thing).


In your description above, where is the narrow pulse width???  As
others have told you, if you read the data *** for the LED they tell
you that there is a maximum current regardless of the duty cycle.
That is because at sufficiently high currents damage is done to the
LED.  It may not be apparent for awhile, but it is cumulative.  If you
don't believe this, then why does the data *** give you a max
current spec?


I've only ever heard things like "maximum current rating of 25 mA".
I've never seen any mention of "max max maxium" for use in display
multiplexing. Before I order more LED's I'll go over the data***.


It seems quite conceivable to me that I'm experiencing the same thing
with the transistor, i.e. I'm putting a massive current thru it but
it's OK because the pulse width and duty cycle are low enough that it
won't get damaged.

Yes, if you keep them low enough.  But if you connect the base and
emitter of a bipolar transistor across a 5 volt supply it will burn
up, period, full stop.


Well firstly there's no such thing as a perfect power supply for two
reasons:
1) You never have just a voltage. It's always "a voltage behind a
resistance". Many times, the internal resistance is negligible, but
other times you'll find that a 5 volt supply is only giving you 4 V
because there's a voltage being dropped inside the power supply.
2) You're assuming that the voltage supply is linear in terms of
resistance versus current. That is, if you halve the resistance across
it, the current flow should double. Irrespective of the internal
resistance of the supply, there's other things that make resistance
versus current flow non-linear. (Admittedly, I don't know of any of
these things, but I'm pretty sure an LM7805 wll start acting weird if
you try to draw higher and higher currents from it).

Anyway, the point I'm making is: Just because you put 5 V across B-E,
that doesn't mean that the whole 5 V is going across the B-E junction.
You could have some volts dropping in the supply, or even in the metal
legs of the transistor.

Anyway, I know the design isn't great, but it definitely does work. I
can show you my design schematic in Protel which clearly shows 5 volts
being put directly across B-E.


I can assure you that pulling too much current from an I/O pin on an
MCU is a bad thing.  You think that by trying it with one part on one
day that you have proven something.  Yes, you proved that one part
will suffer your abuse for one day.  But again, you are creating
cumulative damage to the I/O pin and the chip is likely to fail
prematurely.  Even if only one in a hundred chips won't take your
abuse, that is a *huge* failure rate for semiconductor devices.


Again I'll have to look into this also. First an foremost though, I
want to play around with using one pin to both clock and reset my
counter :-D


However, if you
think the output capacitance of the MCU pin is limited the current to
the transistor, whew! you have a lot to learn.

I'm sure I could find a sufficiently low pulse width and duty cycle
that would make ths happen. Whether this pulse width and duty cycle is
within the MCU's capability, I don't know.

Make what happen?


(Make the output capacitance of the MCU pin limit the current, as
mentioned in the quote within a quote above). Of course, realise that
it would take a ridiculously short pulse width for the output
capacitance of the MCU pin to be non-negligible.


BTW, what is the speed
of your multiplexing?  Is it above 100 MHz?

There's 16 columns. Each column stays lit for 220 microseconds. The
shifting time is neglibile compared to the 220 microseconds, so it
takes about 3.5 milliseconds to perform one full flash. That gives a
frequency of about 284 Hz.

In the old days they used test equipment that would drive pulses into
the traces on a board regardless of whether a driver was already
driving the trace.  One of the vendors showed us a video of a bond
wire glowing red from the excess current something like 100 times a
second.  The resulting mechanical stress weakened the bond wire and
the parts would fail in the field.

Does this sound at all familiar?  High current, low duty cycle, latent
defects.  Are you beginning to see the picture?


I do see the picture. If an LED can take a constant current of 25 mA
though, I still think it should be able to take a bit more current if
it's being multiplexed. How much more though, I don't know.
.