Very basic transistor usage

[Jason Richard]

First off, please excuse my ignorance as I know very little about this stuff
and searching google has only helped to confuse me more.
What I would like to accomplish is to drive 12 bi-color LEDs (red/blue) from
two seperate inputs. I would like the red potion of the LEDs to light when
there is network activity and the blue LEDs to light when there is hard
drive activity. This seems simple enough, but the bi-color LED is common
cathode, and judging by the readings I took, the network and hard drive LEDs
toggle the cathode to turn on or off the LEDs. So, what I was thinking is
that I could use a transistor to drive the anodes of the 12 LEDs, but thats
about as far as I got. The specs of the LEDs are Blue: 3.2v at 20mA and
Red: 2.2v at 20mA. Also, the readings I took showed the hard drive activity
LED cathode swinging from 5v (off) to 0v (on) and the network LED swinging
from 3.3v (off) to 0v (on). Any drawing of a suitable circuit would be
greatly appreciated and any explanation of how the circuit works would be
even better!
Thanks for any and all help!
-Jason

 

[CFoley1064]

Subject: Very basic transistor usage

From: "Jason Richard" pcjason*SPAMTHIS*@sbcglobal.net
Date: 10/9/2004 5:21 PM Central Daylight Time
Message-id: <[email protected]>

First off, please excuse my ignorance as I know very little about this stuff
and searching google has only helped to confuse me more.
What I would like to accomplish is to drive 12 bi-color LEDs (red/blue) from
two seperate inputs. I would like the red potion of the LEDs to light when
there is network activity and the blue LEDs to light when there is hard
drive activity. This seems simple enough, but the bi-color LED is common
cathode, and judging by the readings I took, the network and hard drive LEDs
toggle the cathode to turn on or off the LEDs. So, what I was thinking is
that I could use a transistor to drive the anodes of the 12 LEDs, but thats
about as far as I got. The specs of the LEDs are Blue: 3.2v at 20mA and
Red: 2.2v at 20mA. Also, the readings I took showed the hard drive activity
LED cathode swinging from 5v (off) to 0v (on) and the network LED swinging
from 3.3v (off) to 0v (on). Any drawing of a suitable circuit would be
greatly appreciated and any explanation of how the circuit works would be
even better!
Thanks for any and all help!
-Jason

Hi, Jason. You've got two jobs here -- change both logic signals to a common
+5V/0V, and then drive the LEDs.

Here's one possible way to do this (view in fixed font or M$ Notepad):

VCC
+ VCC VCC VCC
| + + +
.-. | | |
2.2K| | | | .-.
| | | | | |2.2K
'-' | | | |
___ | |/ | '-'
.--|___|-o-| 2N3906 >| | ___
| 2.2K |< 2N3906 |-o-|___|--.
| | /| 2.2K |
| | | |
| .-. .-. |
| 75 ohm| | | | |
| | | | |120 ohm |
| '-' '-' |
| | | |
| Blue V ~ Red V ~ |
| - ~ - ~ |
| | | |
| | | |
| '---------o--------' |
| | |
Netwk|\ | === |
-| >O--' GND |
|/ |
2/8 74HC244 |
|\ |
-| >O-----------------------------------------------'
HD |/
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

For your 12 bi-color LEDs, you'll need three 74HC244s and 24 small signal PNP
transistors like the 2N3906. The HC244 is a non-inverting buffer, and each
logic gate will provide a solid 5.0V output for any input significantly above
1/2Vcc (2.5V if you use a 5V supply for the ICs). For each IC, be sure to make
the two enable inputs logic low (0V), or the outputs won't work at all.

Now, when the logic output goes low, it will pull current from the base of the
PNP transistors through the 2.2K resistors (around 2 mA). That will turn on
the transistors, so they will conduct through the emitter to the collector.
This is called using transistors as switches. For a PNP current-sourcing
switch, a logic "1" is off, and a logic "0" is on. As long as your base
current is around 1/10th of what you want to switch, it will work fine. You
notice that I wanted to switch 20 mA, so I made it so the base drive would be
about 2 mA.

Now you have to do the math on the series resistors that limit the current
going through the LEDs when the transistors are "on". You add the expected
voltage across the LED and the voltage across the switching transistor, and
find out what's left. For the red LED, let's figure 2.2V for the LED plus
about 0.3V for the "on" transistor, which will leave you 2.5V. You need to
figure out what value of resistor is necessary to have 2.5V dropped across it
when 20mA is going through it. For that, you use Ohm's Law. Sir Georg Ohm
said:

V = I * R, or

R = V / I, so

R = 2.5V / 0.02Amps = 125 ohms

Choose 120 ohms as the nearest value. Do the same calculation for a 3.2V LED
to get 75 ohms for that series resistor.

Here's the datasheet to give you the pinout for the 74HC244:

http://www.fairchildsemi.com/pf/MM/MM74HC244.html

For a 2N3906, with the pins down and the flat of the plastic TO-92 case facing
you, the pinout from left to right is E-B-C.

Good luck
Chris

 

[Robert Monsen]

First off, please excuse my ignorance as I know very little about this stuff
and searching google has only helped to confuse me more.
What I would like to accomplish is to drive 12 bi-color LEDs (red/blue) from
two seperate inputs. I would like the red potion of the LEDs to light when
there is network activity and the blue LEDs to light when there is hard
drive activity. This seems simple enough, but the bi-color LED is common
cathode, and judging by the readings I took, the network and hard drive LEDs
toggle the cathode to turn on or off the LEDs. So, what I was thinking is
that I could use a transistor to drive the anodes of the 12 LEDs, but thats
about as far as I got. The specs of the LEDs are Blue: 3.2v at 20mA and
Red: 2.2v at 20mA. Also, the readings I took showed the hard drive activity
LED cathode swinging from 5v (off) to 0v (on) and the network LED swinging
from 3.3v (off) to 0v (on). Any drawing of a suitable circuit would be
greatly appreciated and any explanation of how the circuit works would be
even better!
Thanks for any and all help!
-Jason

5V
-----------o----------o----------------,
| | |
\ \ |
10k/ 10k / |
\ \ |
/ / |
| | |e
| o---/\/\/-----b| PNP
| | 220 |c
| | | 130
| |c o----\/\/\- RED ANODE
o--------| NPN |
| |e o----\/\/\- RED ANODE
| | |
10k |c | o----\/\/\- RED ANODE
IN -/\/\/--| NPN | |
|e | '----\/\/\- etc
| |
| |
----------o----------o------------------------
GND
created by Andy´s ASCII-Circuit v1.25.250804 www.tech-chat.de

Separate but equal circuit for BLUE anodes, except you use 82 ohm
resistors to limit the current through them.

I'd use 2N3904s for the NPNs, and 2N4403s for the PNPs, but just because
I have them laying around. Make sure the PNP can pass at least 300mA.
Its a switch, so its not dissipating much power (ie, it won't get hot)

One odd thing is that if this circuit ends up always being ON, they may
be controlling the LEDs by using a high impedance for off, rather than
just bringing the voltage to the Vcc rail (your measurements say they
probably aren't doing this, but you never know). If so, you'll need a
10k resistor from IN to 5V as well.

 

[Wong]

Hi, Jason. You've got two jobs here -- change both logic signals to a common
+5V/0V, and then drive the LEDs.

Here's one possible way to do this (view in fixed font or M$ Notepad):

VCC
+ VCC VCC VCC
| + + +
.-. | | |
2.2K| | | | .-.
| | | | | |2.2K
'-' | | | |
___ | |/ | '-'
.--|___|-o-| 2N3906 >| | ___
| 2.2K |< 2N3906 |-o-|___|--.
| | /| 2.2K |
| | | |
| .-. .-. |
| 75 ohm| | | | |
| | | | |120 ohm |
| '-' '-' |
| | | |
| Blue V ~ Red V ~ |
| - ~ - ~ |
| | | |
| | | |
| '---------o--------' |
| | |
Netwk|\ | === |
-| >O--' GND |
|/ |
2/8 74HC244 |
|\ |
-| >O-----------------------------------------------'
HD |/
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

For your 12 bi-color LEDs, you'll need three 74HC244s and 24 small signal PNP
transistors like the 2N3906. The HC244 is a non-inverting buffer, and each
logic gate will provide a solid 5.0V output for any input significantly above
1/2Vcc (2.5V if you use a 5V supply for the ICs). For each IC, be sure to make
the two enable inputs logic low (0V), or the outputs won't work at all.

Now, when the logic output goes low, it will pull current from the base of the
PNP transistors through the 2.2K resistors (around 2 mA). That will turn on
the transistors, so they will conduct through the emitter to the collector.
This is called using transistors as switches. For a PNP current-sourcing
switch, a logic "1" is off, and a logic "0" is on. As long as your base
current is around 1/10th of what you want to switch, it will work fine. You
notice that I wanted to switch 20 mA, so I made it so the base drive would be
about 2 mA.

Now you have to do the math on the series resistors that limit the current
going through the LEDs when the transistors are "on". You add the expected
voltage across the LED and the voltage across the switching transistor, and
find out what's left. For the red LED, let's figure 2.2V for the LED plus
about 0.3V for the "on" transistor, which will leave you 2.5V. You need to
figure out what value of resistor is necessary to have 2.5V dropped across it
when 20mA is going through it. For that, you use Ohm's Law. Sir Georg Ohm
said:

V = I * R, or

R = V / I, so

R = 2.5V / 0.02Amps = 125 ohms

Choose 120 ohms as the nearest value. Do the same calculation for a 3.2V LED
to get 75 ohms for that series resistor.

Here's the datasheet to give you the pinout for the 74HC244:

http://www.fairchildsemi.com/pf/MM/MM74HC244.html

For a 2N3906, with the pins down and the flat of the plastic TO-92 case facing
you, the pinout from left to right is E-B-C.

Good luck
Chris

Hi Chris,
Something that I don't understand. Why is there a "voltage divider"
between the node of a 2.2K resistor and the transistor base?
Or it just the pullup resistor because of 74HC244 ?
Think that you wont get 2mA when Netwk is "1" and 244 output is "0",
I think you will get less than that, right ?

 

[CFoley1064]

Subject: Re: Very basic transistor usage

From: [email protected] (Wong)
Date: 10/11/04 3:00 AM Central Daylight Time
Message-id: <[email protected]>

Hi Chris,
Something that I don't understand. Why is there a "voltage divider"
between the node of a 2.2K resistor and the transistor base?
Or it just the pullup resistor because of 74HC244 ?
Think that you wont get 2mA when Netwk is "1" and 244 output is "0",
I think you will get less than that, right ?

Hi, Jason. It's common practice when using a transistor as a switch to have a
pulldown (NPN) or pullup (PNP) resistor from the base to the emitter to keep
things well-behaved. On is really on, and off is really off. In fact,
"digital transistors" are made these days which have the series resistor and
pullup/pulldown from base to emitter built in, and I'll usually use one of
those when using a discrete transistor as a switch. Fewer components to place,
less hassle. But if you're driving the transistors from HCMOS, which has logic
levels pretty much at the supply rails (unless you're driving some current),
the pullups technically wouldn't be necessary. Back in days of yore you'd need
these extra resistors because TTL logic levels would lead to fuzzy turn-on and
turn-off states. Times have changed. Good observation.

Good luck with your project
Chris