Newbie transistor question

[realexander]

Hi,

I'm having trouble getting a transistor to work the way I thought it
should, and I hope someone could help me.

I'm using a PIC to drive common anode 14-segment LEDs. Since the PIC
I/O line can't supply enough current, I'm trying to use a 2N2222
transistor instead. The PIC directly drives the 2N2222's base (no
intervening resistor), the collector is tied to +5 and the emitter to
the common anode. The problem is that, while the LEDs light, they are
very, very dim - only visible in a dark room.

I thought maybe that the PIC wasn't saturating the transistor, so I put
in a pull-up resistor on the base. No difference. I connected the base
directly to +5, and the LEDs were still dim. The emitter shows +4V (5
volts in the collector, 4 volts out the emitter??!!) If I tie the LEDs
common anode to +5, they light up nicely, but of course I can't
multiplex a bunch of LEDs with the anodes tied to +5.

Am I using an inappropriate transistor? Am I using it wrong?

Thanks,
Bob Alexander

 

[martin griffith]

Hi,

I'm having trouble getting a transistor to work the way I thought it
should, and I hope someone could help me.

I'm using a PIC to drive common anode 14-segment LEDs. Since the PIC
I/O line can't supply enough current, I'm trying to use a 2N2222
transistor instead. The PIC directly drives the 2N2222's base (no
intervening resistor), the collector is tied to +5 and the emitter to
the common anode. The problem is that, while the LEDs light, they are
very, very dim - only visible in a dark room.

I thought maybe that the PIC wasn't saturating the transistor, so I put
in a pull-up resistor on the base. No difference. I connected the base
directly to +5, and the LEDs were still dim. The emitter shows +4V (5
volts in the collector, 4 volts out the emitter??!!) If I tie the LEDs
common anode to +5, they light up nicely, but of course I can't
multiplex a bunch of LEDs with the anodes tied to +5.

Am I using an inappropriate transistor? Am I using it wrong?

Thanks,
Bob Alexander

Do you have series resistors to limit the led current? They are
essential

Check the pinout of the 2N2222, might be upside down


martin

 

[Leon]

You should have a series resistor in the transistor base to limit the
current.

Are you sure the PIC pin you are using is actually configured as an
output? Disconnect it and check that it is actually being driven high
and low.

Leon

 

[Tim Wescott]

Hi,

I'm having trouble getting a transistor to work the way I thought it
should, and I hope someone could help me.

I'm using a PIC to drive common anode 14-segment LEDs. Since the PIC
I/O line can't supply enough current, I'm trying to use a 2N2222
transistor instead. The PIC directly drives the 2N2222's base (no
intervening resistor), the collector is tied to +5 and the emitter to
the common anode. The problem is that, while the LEDs light, they are
very, very dim - only visible in a dark room.

I thought maybe that the PIC wasn't saturating the transistor, so I put
in a pull-up resistor on the base. No difference. I connected the base
directly to +5, and the LEDs were still dim. The emitter shows +4V (5
volts in the collector, 4 volts out the emitter??!!) If I tie the LEDs
common anode to +5, they light up nicely, but of course I can't
multiplex a bunch of LEDs with the anodes tied to +5.

Am I using an inappropriate transistor? Am I using it wrong?

Thanks,
Bob Alexander

A 0.6 or 0.7V drop from base to emitter is to be expected -- if you're
seeing 4.3 or 4.4 volts on the emitter (assuming 5V on the base) and
rounding down then your transistor is working as expected.

What are you doing to limit the current of your LEDs, and how are you
switching the cathodes of the LEDs? I suspect that with the voltage
drop on the transistor your current limiting resistor is now too large,
and your LED current is insufficient. You can probably get your current
back up either by using lower resistance current limiting resistors, or
by using a PNP transistor in common emitter. The PNP in common emitter
will only drop 0.2V or so -- connect it emitter to +5V, collector to the
LED, resistor between PIC and base, and remember that pulling the base
_down_ will turn the transistor _on_.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Posting from Google? See http://cfaj.freeshell.org/google/

"Applied Control Theory for Embedded Systems" came out in April.
See details at http://www.wescottdesign.com/actfes/actfes.html

 

[Bob]

Hi,

I'm having trouble getting a transistor to work the way I thought it
should, and I hope someone could help me.

I'm using a PIC to drive common anode 14-segment LEDs. Since the PIC
I/O line can't supply enough current, I'm trying to use a 2N2222
transistor instead. The PIC directly drives the 2N2222's base (no
intervening resistor), the collector is tied to +5 and the emitter to
the common anode. The problem is that, while the LEDs light, they are
very, very dim - only visible in a dark room.

I thought maybe that the PIC wasn't saturating the transistor, so I put
in a pull-up resistor on the base. No difference. I connected the base
directly to +5, and the LEDs were still dim. The emitter shows +4V (5
volts in the collector, 4 volts out the emitter??!!) If I tie the LEDs
common anode to +5, they light up nicely, but of course I can't
multiplex a bunch of LEDs with the anodes tied to +5.

Am I using an inappropriate transistor? Am I using it wrong?

Thanks,
Bob Alexander

The emitter being at 4V is about right. It can never reach the collector's
voltage in the configuration you've got it in (common collector).

If 4V isn't enough to light the LEDs then it may mean that either the LED's
current-limiting resistor is too big, or (somehow) there are two LEDs (in
series) per segment.

If you rearrange your circuit to be common emitter configuration then you
can get the emitter and collector voltage to be closer together, during
saturation. If you use an NPN, in the common emitter config, then you'll
have to make sure that the PIC can source enough base current (through an
appropriate series resistor). If the PIC can't source enough current, but
can sink enough current, then you'll need to use a PNP device.

It seems that you do understand the basics. You just need to dig a little
deeper to find the root of the problem.

Bob

 

[colin]

You should have a series resistor in the transistor base to limit the
current.

I think hes using an emitter follower so the base current should be self
limited.

Are you sure the PIC pin you are using is actually configured as an
output? Disconnect it and check that it is actually being driven high
and low.

It sounds like your cathode drivers are needing more volts to work than
available with the voltage drop accros the transistor (1v) And the leds
(~2v) might leave them 3v wich should be enough, but if you are using an
emitter follower for the cathode drivers too it be you have too few volts
accros the current limiting resistor, you may need to adjust this
accordingly.

alternativly you can use a pnp common emitter for the anode driver with base
limiting resistor, or a P chanel FET, of course the logic output will have
to be inverted, this should reduce the volt drop wich is inevetable with an
emitter follower.

Colin =^.^=

 

[John Popelish]

Hi,

I'm having trouble getting a transistor to work the way I thought it
should, and I hope someone could help me.

I'm using a PIC to drive common anode 14-segment LEDs. Since the PIC
I/O line can't supply enough current, I'm trying to use a 2N2222
transistor instead. The PIC directly drives the 2N2222's base (no
intervening resistor), the collector is tied to +5 and the emitter to
the common anode. The problem is that, while the LEDs light, they are
very, very dim - only visible in a dark room.

I thought maybe that the PIC wasn't saturating the transistor, so I put
in a pull-up resistor on the base. No difference. I connected the base
directly to +5, and the LEDs were still dim. The emitter shows +4V (5
volts in the collector, 4 volts out the emitter??!!) If I tie the LEDs
common anode to +5, they light up nicely, but of course I can't
multiplex a bunch of LEDs with the anodes tied to +5.

Am I using an inappropriate transistor? Am I using it wrong?

(discussing NPN transistors, like the 2N2222)
If the transistor were big enough to carry the current and still have
significant current gain, the emitter would have a voltage about .6 to
..7 volts (one forward biased silicon diode drop) below the base
voltage. The problem is that transistors lose gain rapidly as the
collector voltage gets close to or below the base voltage. In this
common collector application, you expect operation with both collector
and base voltage at 5 volts (no extra positive collector voltage to
keep the gain up). So the current gain falls and the base to emitter
drop rises. I don't know what the total emitter current is with your
load (but it can't be very high if all the LEDs are dim), so this may
or may not be the problem. You may have collector and emitter
reversed (the emitter is the lead near the tab on the can).

All this can be extracted from the data sheet:
http://www.fairchildsemi.com/ds/MM/MMBT2222A.pdf
Note figure 3 and 4. Figure 3 shows the base to emitter drop with the
collector voltage so low that the current gain is reduced to 10.
Figure 4 shows the base to emitter drop when the collector voltage is
5 volts more positive than the emitter.

You could use a PNP transistor, such as a 2N4403, but you will need a
base resistor to limit the base current to a reasonable value (say,
1/20th of the worst case LED load current), if the PIC can pull down
that much. If the required base current is more than the PIC can
reasonable supply, you need two stages of PNP. The first connected as
a follower (similar to what you have, now, except that the collector
connects to ground). But then you add a base resistor between its
emitter and the saturating switch transistor's base. The resistor's
value will have to take into account the two base emitter drops in
series that use up some of the 5 volt supply.

 

[realexander]

Thank you all for your responses. To answer the question most of you
had, no there are no resistors in the circuit. My impression was that
resistors are needed to prevent anything from frying. So far, neither
my LEDs nor my PIC have fried. Maybe the currents are low enough, or
maybe I haven't been running the circuit long enough.

I worked through your suggestions. Many of you mentioned that +4 coming
from the emitter might not be enough. Bob (not me, a different and

If 4V isn't enough to light the LEDs then it may mean that ... there are two LEDs (in
series) per segment.

He led me to the answer, though perhaps not the way he intended. The
display I'm using is a dual 14-segment alphanumeric display, with two
common anodes (one for the left digit, one for the right). I initially
had +5 wired to both anodes. When I was ready to add transistors, I
added one transistor to one of the anodes, but left the other anode at
+5. I assumed they were working independently, so why not leave one
digit at +5? The LEDs controlled by the transistor were dim, while the
ones wired to +5 were bright. Bob made me stop and think, and I
realized that the two digits could affect each other, since segment x
of one digit has the same cathode as segment x of the other. I
disconnected the common anode of the other digit, and the
transistor-controlled digit became bright!

Even a newbie like me understands what was happening now. The LEDs need
a voltage drop of 2.6V. The LEDs being fed directly by +5 had (5 - 2.6
= 2.4V) at their cathode. The LEDs fed by transistor were getting +4 in
their anode, and saw a (4 - 2.4 = 1.6) difference between their anode
and their cathode, which was not enough to light - at least not very
brightly.

Once again, thank you all!

- Bob Alexander

 

[Rich Grise]

I thought maybe that the PIC wasn't saturating the transistor, so I put
in a pull-up resistor on the base. No difference. I connected the base
directly to +5, and the LEDs were still dim. The emitter shows +4V (5
volts in the collector, 4 volts out the emitter??!!) If I tie the LEDs
common anode to +5, they light up nicely, but of course I can't
multiplex a bunch of LEDs with the anodes tied to +5.

If you do, in fact, have a common anode, then that's the only way you
can possibly do it. From each individual cathode, run a 120 or 220 ohm
resistor to the collector of a 2N2222, and ground all of their emitters.
I'd put a resistor (maybe 1K ~ 4.7K) between the chip output and the
base, just to limit the current.

If you want to mux more than one display, then drive the common anode
with the collector of a PNP.

If the LEDs have separate anodes, then that's not common anode, it's
common cathode, and you have to drive it with PNPs, and control the
whole display with an NPN at the common cathode.

Good Luck!
Rich

 

[[email protected]]

Hi,

I'm having trouble getting a transistor to work the way I thought it
should, and I hope someone could help me.

I'm using a PIC to drive common anode 14-segment LEDs. Since the PIC
I/O line can't supply enough current, I'm trying to use a 2N2222
transistor instead. The PIC directly drives the 2N2222's base (no
intervening resistor), the collector is tied to +5 and the emitter to
the common anode. The problem is that, while the LEDs light, they are
very, very dim - only visible in a dark room.

I thought maybe that the PIC wasn't saturating the transistor, so I put
in a pull-up resistor on the base. No difference. I connected the base
directly to +5, and the LEDs were still dim. The emitter shows +4V (5
volts in the collector, 4 volts out the emitter??!!) If I tie the LEDs
common anode to +5, they light up nicely, but of course I can't
multiplex a bunch of LEDs with the anodes tied to +5.

Am I using an inappropriate transistor? Am I using it wrong?

Thanks,
Bob Alexander

eh?

it depends what sort of current your LED requires, but the correctly
configured outputs of PICS are definately capable of directly sinking
and sourcing 25mA - more than enough for most LEDs. The datasheet I
have in front of me (16F84) lists "High current sink/source for direct
LED drive" as one of the peripheral features on the over-view page
(with the pin-out diagram of the chip)

Obviously if you are driving something that requires more than 25mA
then you will need to buffer the signal, but not for "normal"
operations.

I drive LEDs direct from PICs with a 150R resistor which gives the LED
20mA - plenty bright enough for most indicators. As a newbie, this is
calculated as supply voltage-2 (the LED will roughly "use" 2 volts
from your supply) divided by current required...
which gives 3V / 0.02A = 150R Ohm's Law.

Why are you driving the anode through the transistor? You should
connect the anode directly to (in this case) +5V and then each segment
should be connected to the PIC pin via a 150R (or so) resistor.
Remember that with a common anode display, you need to drive the pins
in a reverse logic, so set a pin low to light the segment. So in any
output, 1 is seg off, 0 is seg on.

You need to re-do your drive circuit then go back to your code with a
single LED and make sure you can switch it on and off correctly. Then
connect up a segment of your display - with the same results - then
finish off by replicating your control code for all segments - it has
to work. Are you sure you are TRISing the port pins correctly? also,
make sure you have turned off any special features of the pins, i.e.
analogue input ports, PWM etc... Keep it really simple until you have
got control.

Might I also suggest that you take this thread to
sci.electronics.basics
as this group is populated by engineers that like to eat noobs )

Best of luck

H