Driving LEDs with higher current or from a higher positive voltage.

[Manj]

Hi,

I am novice!! can any one help me out? for a project I want to drive
series leds using led driver.The led output driver is programmed to be
either open-drain with a 25 mA current sink capability at 5 V, and 10
mA source capability at 5 V.To drive leds in series for higher current
and higher voltage they have suggested to usea n-p-n transistor. I want
current sink of 60mA and vcc applied to the led anode is 12v,Can any
one help me with the parameters for the transistor, does this requier a
biasing resistor? can any one suggest part no for a npn transistor to
meet this requierment.Thanks

Copy this into a .txt file
+ 12VCC
| |
| .-.
| 1 Leds | |
| | |
| '-'
| |
|
|
|
| .-.
| 2 Leds | |
| | |
| '-'
| |
| __ |
| -| \ ___ |/
| | )o-------|___|-o-|
| -|__/ R ?? | |>
| .-. |
| R?? | | ===
| | | GND
| '-'
| |
| ===
| GND

 

[Chris]

Hi,

I am novice!! can any one help me out? for a project I want to drive
series leds using led driver.The led output driver is programmed to be
either open-drain with a 25 mA current sink capability at 5 V, and 10
mA source capability at 5 V.To drive leds in series for higher current
and higher voltage they have suggested to usea n-p-n transistor. I want
current sink of 60mA and vcc applied to the led anode is 12v,Can any
one help me with the parameters for the transistor, does this requier a
biasing resistor? can any one suggest part no for a npn transistor to
meet this requierment.Thanks

Copy this into a .txt file
+ 12VCC
| |
| .-.
| 1 Leds | |
| | |
| '-'
| |
|
|
|
| .-.
| 2 Leds | |
| | |
| '-'
| |
| __ |
| -| \ ___ |/
| | )o-------|___|-o-|
| -|__/ R ?? | |>
| .-. |
| R?? | | ===
| | | GND
| '-'
| |
| ===
| GND

Hi, Manj. Just about any small signal NPN transistor will do here. If
you're using through hole parts instead of SMT, and you're looking for
something inexpensive, you could do worse than the venerable 2N3904.
The 2N4401 is another good, very commonly available choice.

The idea behind using a transistor as a switch is that you want to
deliberately overdrive the base to ensure the transistor as a switch
turns fully ON (and has a very low saturation voltage). That means
putting about 1/10th of the expected collector current into the base.
You've said your logic output can source 6mA, so use a 680 ohm resistor
for both Rs. That should get you close enough.

Good luck
Chris

 

[Chris]

Manj wrote:

+ 12VCC
| |
| .-.
| 1 Leds | |
| | |
| '-'
| |
|
|
|
| .-.
| 2 Leds | |
| | |
| '-'
| |
| __ |
| -| \ ___ |/
| | )o-------|___|-o-|
| -|__/ R ?? | |>
| .-. |
| R?? | | ===
| | | GND
| '-'
| |
| ===
| GND

By the way, from your diagram, you'll need a series resistance with
your LEDs to limit current, too. If you're driving two LEDs in series,
you could do worse than put a 150 ohm 1/2 watt resistor in there like
this (view in fixed font or M$ Notepad):

|
| +12V
| +
| |
| V ~
| - ~
| |
| V ~
| - ~
| |
| .-.
| | |150
| | | 1/2 watt
| '-'
| |
| ___ |/
| o-|___|-o-| 2N3904
| 680 | |>
| .-. |
| 680| | ===
| | | GND
| '-'
| |
| ===
| GND
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Good luck
Chris

 

[Rich Grise]

Hi,

I am novice!! can any one help me out? for a project I want to drive
series leds using led driver.The led output driver is programmed to be
either open-drain with a 25 mA current sink capability at 5 V, and 10
mA source capability at 5 V.To drive leds in series for higher current
and higher voltage they have suggested to usea n-p-n transistor. I want
current sink of 60mA and vcc applied to the led anode is 12v,Can any
one help me with the parameters for the transistor, does this requier a
biasing resistor? can any one suggest part no for a npn transistor to
meet this requierment.Thanks

I agree about the transistors - the R values aren't critical - I usually
use about 1K from output to base, and 4K7 from base to ground.

But don't forget to include the current limiting resistor for the LEDs!:

+ 12VCC
| |
|
[R] = (Vcc - Vled) / Iled (better check this -
| it's off the top
| of my head.)
|
|
| .-.
| n Leds | |
| | |
| '-'
| |
| __ |
| -| \ ___ |/
| | )o-------|___|-o-|
| -|__/ R 1K | |>
| .-. |
| R4K7 | | ===
| | | GND
| '-'
| |
| ===
| GND

Good Luck!
Rich

 

[Manj]

Thanks Chris!!!Can you explain me why do we put a resistor from base to
ground?Is it necessary?Is there a N-P-N transistor array (preferably
surface mount)avilable, becas I need to construct a led matrix of 8X8
and placing single transistor and 2 res will requier more pcb space.

In some of the websites,I refered to,they used a resistor pulled up to
vcc(5v) at the base of the transistor.Is this also one of the method to
ensure transistor is fully turned ON?Is this a good solution?can you
explain this?

Even if I drive 2 leds with 12v and minimum current of 20mA ,then
according to the current flow in the leds (Ic collector current)i will
change the resistor values for the appropriate base current and to
ensure transistor is ON,did I get this right?

 

[Peter Bennett]

Thanks Chris!!!Can you explain me why do we put a resistor from base to
ground?Is it necessary?Is there a N-P-N transistor array (preferably
surface mount)avilable, becas I need to construct a led matrix of 8X8
and placing single transistor and 2 res will requier more pcb space.

Use a ULN2803 - it is 8 NPN darlington transistors in an 18 pin DIP,
with the inputs arranged to accept TTL logic signals.

 

[Bill Bowden]

Thanks Chris!!!Can you explain me why do we put a resistor from base to
ground?Is it necessary?Is there a N-P-N transistor array (preferably
surface mount)avilable, becas I need to construct a led matrix of 8X8
and placing single transistor and 2 res will requier more pcb space.

The resistor from base to ground is not needed. It just wastes space.

In some of the websites,I refered to,they used a resistor pulled up to
vcc(5v) at the base of the transistor.Is this also one of the method to
ensure transistor is fully turned ON?Is this a good solution?can you
explain this?

You only need a pullup resistor if the logic output is open collector.
If the output supplies a voltage in the high state, a pullup resistor
is not needed.