# resistor transistor basics

Discussion in 'Electronic Basics' started by Jan Nielsen, May 13, 2007.

1. ### Jan NielsenGuest

I have several LEDs that use the same voltage, they will be on and off
at different times, so the current usage will not be the same, can they
still share a resistor, when they are in parallel to the resistor ?

Also can a transistor be used to switch a lower voltage than the control?
If not would a simple resistor before the base (green line) help ?

I have a drawing of my idea
http://codebin.dk/stuff/resistor_transistor_basics.jpg
The 100R matches 0.02A (each led) with a 2V drop from supply to the LED.

Thanks,
Jan

2. ### EeyoreGuest

If you turn on and off each LED with a transistor in parallel with it that
'shorts out' the LED to turn it off, you can run several LEDs in series from a
single current source. The number you can run this way depends on the supply
voltage available of course. I have run up to 16 LEDS this way for example
(audio signal level meter).

This is an excellent way of saving power.

Graham

3. ### Jan NielsenGuest

Eeyore skrev:
Oh, the transistor goes to the other led connector ofcause.
But will the voltage after the transistor be 3V(plus a little) or 5V ?
If 5V then I need the resistor before the transsistor base instead ?
I cant run at the LED power because I have 2 colors that need a
different voltage each.

I am just wondering if the 5LEDs with the same voltage can share a
resistor, when they will be varying in current draw.

/Jan

4. ### John PopelishGuest

If you turn on more than 1 led at the same time, how they
divide the current through the series resistor is a bit
undefined, depending on how well they and their switch
match. but even with good matching, having two on at the
same time, they will be dimmer than when only one is on. I
would give each its own current limiting resistor unless I
was sure that only one would be on at a time.

Transistors take at least a diode drop (about .6 to .7 volts
base to emitter to switch on. The collector to emitter
voltage is fairly independent of the turn on process.

I think I would put the three NPN transistors in the ground
side of the LED circuits, emitter to ground, collector to
LED, with a separate resistor between each LED and the +5
supply. Is the transistor control signal also a +5 or
ground voltage? If so, a base resistor of about 4.7k would
of the collector current. That should be enough to turn the
transistors on to less than a diode drop, collector to emitter.

This assumes that the control signal is positive when you
want the LED to be on.

You could also drive the LEDs with the transistor acting as
a voltage follower (that copies the control voltage except
for a diode drop in the signal) with the base connected
directly to the control voltage, the collector to +5 and the
emitter to the resistor and LED in series to ground. The
saves 3 base resistors, but wastes a good fraction of a volt
more across the transistor, so you may need to lower the LED
resistor a bit to compensate.

5. ### John PopelishGuest

Oh, another thing, you will have to make use of both LED
pins. Your drawing shows connections to only one.

6. ### EeyoreGuest

Yes each (npn) transistor's emitter and collector go to the LED's cathode and anode
respectively..

I don't know what you mean by 'after the transistor'.

They aren't actually that different (apart from blue LEDs).

The suggestion I'm making is that you have the LEDs in *series* not parallel.

What's the maximum number of LEDs that will illuminated at one time ?

Graham

7. ### Jan NielsenGuest

John Popelish skrev:
[SNIP]

Thanks, that explains it.
The drawing is wrong, I use both legs ofacause.

/Jan

8. ### JasenGuest

They can but brightness will be lower (and probably uneven) with two
(or more) lit.

your circuit puts no current through the led so they won't light at
all.
FETs can.
for bipolar transistors that would be essential.

Bye.
Jasen

9. ### ChrisGuest

Hi, Jan. Resistors are really inexpensive (you can get them for a
penny or two each in quantity from just about any mail order source).
Running two LEDs of even slightly different forward voltages isn't a
good idea, because the higher forward voltage LED will get starved for
current by the lower.

If your power supply voltage was high in relation to the LED forward
voltage (say, 12VDC), the difference of a few tenths of a volt
wouldn't matter very much as long as you chose the series resistor
accordingly, and only turned on one LED at a time. But if you turn on
two or more, one will almost always hog the current.

If you happen to be driving the LEDs with a microcontroller, you might
be able to work your plan by multiplexing the LEDs you want on. Let's
say you want all three LEDs in your diagram to appear to be on. You
might turn each of the LEDs on for 1/3 of the time at a switching rate
of 1KHz or so. The human eye can't follow switching that fast, so
persistence of vision will result in the appearance that all three are
on simultaneously, even though only one is actually on at any
instant. Most larger LED displays work on this basic principle, to
save power and make physical wiring easier.

Good luck
Chris

10. ### Jan NielsenGuest

Chris skrev:
Indeed, the reason to share a resistor was merely to reduce the amount
of components, totally it will be 30resistors just for the leds on that
circuit when they cant share, but its ok.
I payed 5\$ for 2000resistors on ebay some time ago

/Jan

11. ### Rich GriseGuest

No. They never have exactly the same forward voltage drop, so one
LED will hog all of the current until it blows, then the next, and
so on. The current rises exponentially with the voltage.

Use a resistor (or current regulator) in series with each LED, or in
series with each series string of LEDs.

Good Luck!
Rich

12. ### Anthony FremontGuest

Yes, that's the theory which apparently assumes that Vf is some unchanging
value. Now you should go out and get some LEDs and actually try it for fun.
Provided they are the same type, I'm betting that they will all light. I'm
not saying that they will all be the same brightness or they will all pass
the same current, but they will all most likely light. Vf rises with If,
they will seek some equilibrium point.
Thanks for setting me straight on that, I was thinking that current rose
linearly with applied voltage. ;-)

13. ### Rich GriseGuest

That's the thing - they do the exact opposite of "seek some equilibrium
point." The one that gets the more current gets hotter, which lowers its
Vf even more - it's called "thermal runaway."

I'm sure you can get away with it, but I would never, ever, ever try to
foist it off in any kind of commercial design that has to be reliable.

Good Luck!
Rich

14. ### Anthony FremontGuest

No way to measure Tj, so I can't comment on that aspect of it, but IME
forward voltage drop increases with forward current. At least when keeping
things below MAX specs. Check it for yourself.

15. ### Paul E. SchoenGuest

Some of the high power white XLamp LEDs (www.Cree.com) I have been working
with have equivalent series resistance as high as 2 ohms (3_7090), while
others are as low as 0.7 ohm (XR-E). That's probably how most of the cheap
multi-LED flashlights just put them all in parallel, and often don't even
use a dropping resistor. However, I've also seen a lot of them fail,
probably because of current hogging (thermal runaway) at higher
temperatures.

The temperature coefficients of Vf are typically -2.8 to -3.2 mV/Deg C for
the 2 ohm version (3_7090), and -4.0 mV/Deg C for the 0.7 ohm (XR-E).

Paul