Re: Very basic transistor usage

From: Wong (tatto0_2000_at_yahoo.com)
Date: 10/11/04 

Date: 11 Oct 2004 01:00:10 -0700

cfoley1064@aol.com (CFoley1064) wrote in message news:<20041009190520.15331.00001406@mb-m22.aol.com>...
> >Subject: Very basic transistor usage
> >From: "Jason Richard" pcjason*SPAMTHIS*@sbcglobal.net
> >Date: 10/9/2004 5:21 PM Central Daylight Time
> >Message-id: <JvZ9d.27219$QJ3.21317@newssvr21.news.prodigy.com>
> >
> >First off, please excuse my ignorance as I know very little about this stuff
> >and searching google has only helped to confuse me more.
> >What I would like to accomplish is to drive 12 bi-color LEDs (red/blue) from
> >two seperate inputs. I would like the red potion of the LEDs to light when
> >there is network activity and the blue LEDs to light when there is hard
> >drive activity. This seems simple enough, but the bi-color LED is common
> >cathode, and judging by the readings I took, the network and hard drive LEDs
> >toggle the cathode to turn on or off the LEDs. So, what I was thinking is
> >that I could use a transistor to drive the anodes of the 12 LEDs, but thats
> >about as far as I got. The specs of the LEDs are Blue: 3.2v at 20mA and
> >Red: 2.2v at 20mA. Also, the readings I took showed the hard drive activity
> >LED cathode swinging from 5v (off) to 0v (on) and the network LED swinging
> >from 3.3v (off) to 0v (on). Any drawing of a suitable circuit would be
> >greatly appreciated and any explanation of how the circuit works would be
> >even better!
> >Thanks for any and all help!
> >-Jason
>
> Hi, Jason. You've got two jobs here -- change both logic signals to a common
> +5V/0V, and then drive the LEDs.
>
> Here's one possible way to do this (view in fixed font or M$ 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 data*** 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 ?