zoidberg said:
I need to link two green leds to one of the outputs of a 7432, but the
output current is insufficient. What can I do? I thought that a cmos
buffer (4010) put between the 7432 output and the leds can be a good
solution. Could it work?
Otherwise would a 2n3904 npn bjt transistor work?
http://snipurl.com/wwsv
TTL outputs can only source a small amount of current, but they can sink
16mA current, which can drive an LED or two. I think you want the LED's
to turn on when the OR gate output is high, so the gate can't supply the
LED current directly. Also the high level output voltage is typically
around 3.4V, so stacking the LEDs in series pretty much guarantees there
won't be enough forward voltage to turn them on.
I suppose you could use the CD4010 if you did not need much LED forward
current. A CD4010 with a 5V supply is only guaranteed to source 3mA at
room temperature.
The following should work with higher LED currents. View in a fixed
width font like Courier:
Vcc Vcc
| |
.-. .-.
300 | | | | 300
| | | |
'-' '-'
| |
Vcc | |
| V -> V ->
.-. - -
1K | | | |
| | o------'
__ '-' 1K |
-\ \ | ___ |/
| |-o-|___|--|
-/__/ |>
|
Gnd
(created by AACircuit v1.28.6 beta 04/19/05
www.tech-chat.de)
The transistor is acting as a saturated mode switch. The base current
is supplied by the pullup resistor to Vcc at the gate output. The LEDs
need individual current setting resistors. In parallel one LED may hog
the current. If the LEDs were in series there would not be enough
voltage left across the resistors to reliably set the LED current.
Assuming a saturated transistor Vce of 0.1V and LED forward voltage of
2V this leaves 5V-0.2V-2V=2.9V across the current setting resistor. By
Ohms law a 300 ohm resistor would produce 2.9V/300ohm or a bit less
than 10mA of forward current in each LED.
Jack Rouse
