Re: current regulator -- explanation?
- From: Chris <cfoley1064@xxxxxxxxx>
- Date: Thu, 6 Dec 2007 03:23:43 -0800 (PST)
On Dec 6, 2:56 am, Rex <r...@xxxxxxxxxxxxx> wrote:
I found my way to this page today:http://www.gizmology.net/LEDs.htm
looking for information about light output.
As I was reading, I found a circuit for regulating the current through
LEDs. It has two transistors with one having the base-collector shorted.
The circuit is about 2/3 down the page and has the caption,
"LEDs driven by a simple current regulator".
I don't see exactly how this senses the current through the LEDs and
regulates it. Does it work? If so, can someone explain or give a link to
how the circuit works.
Posted here because maybe this is basic stuff I have missed or am just
too dumb to see.
Assuming it does work (or even if not), is this a good implementation
(function/parts)? Is there a better way?
Hi, Rex. Current mirrors are a good choice if you're making and IC,
because you can match the transistors well, and temperature tracking
between transistors is easy.
If you want an easy current source that has pretty good compliance
over a fairly wide range, you could do worse than this (view in fixed
font or M$ Notepad):
|
| VCC
| +
| |
| .-.
| | |
| | |
| '-'
| |
| |
| Vb |<
| <------| PNP
| |\
| |
| |
| |
| |
| ^
| Isource
|
| |
| ===
| GND
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
By controlling Vb, you can achieve pretty good control of current over
a fairly high voltage range with only one transistor and one resistor.
The emitter voltage will be one diode drop higher than the base
voltage. Let's say you wanted a current source of 20mA, and you had a
power supply that varied from 8 to 16VDC. You might do something like
this (view in fixed font or M$ Notepad):
|
| VCC VCC
| + +
| | |
| V .-.
| - | |36 ohm
| |1N914X2 | |
| | '-'
| V |
| - |
| | |<
| o-------| 2N3906
| | |\
| | |
| | |
| | |LED
| |3.3K |
| .-. V ~
| | | - ~
| | | |
| '-' |
| | |
| === ===
| GND GND
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
By adding the two diodes and the 3.3K resistor, you're ensuring that
Vb will be two diode drops below Vcc. That means Ve will be one diode
drop below Vcc. A 36 ohm resistor is chosen using Ohms Law:
0.7V / 0.02A = 35 ohms
This will not work perfectly, but it is an acceptable current source
for an LED.
Cheers
Chris
.
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