piclist 2002\10\20\165558a >
Thread: NPN LED current drive
face BY : Wagner Lipnharski email (remove spam text)

Russell McMahon wrote:
{Quote hidden}

That's 100% correct, where you need to force a constant current.
There is of course, some drawbacks with the emitter resistor technique,
when dealing with battery fading voltage.

Suppose you are using a 9V battery in a circuit fed directly by this
battery. This circuit will drives a transistor with a constant current
emitter resistor and a LED at collector.

Lets see what happens when the battery is 9V or 5.5V;

  Battery 5.5 or 9V
 |                |
 |                |
.-----.        D  _V_  _-->
|     |            |
| DRV |            C
|     |---R1-----B[NPN]
|     | A   B      E
'-----'            |
 |                R2
 |                |
_|_              _|_

NPN Beta = 100

ID = 20mA
IE = 20mA x (1+ -----)

IE = 20mA x 1.01 = 20.2mA

IB = ------ = 200uA

R2   = 47 Ohms
IR2  = 20.2mA
VR2  = 0.9494 V
VBE  = 0.6V
VB   = VR2B = 1.594 V
IB   = 200uA
VR2A = 5.4 or 8.9V

When VCC = 5.5V;
VR2A = 5.4V
VR2 = VR2A - VR2B = 5.4 - 1.594 = 3.806V
R2 = 3.806 / 200uA = 19030 ohms.

If R2 = 19030 ohms when VCC = 8.9V;
VR2A = 8.9V

       8.9V - Vbe
IB = ------------------
     19030 + 47 x Beta

IB =  7.5V / 23730 = 316uA
IE = 316uA x 100 = 31.6mA
VR2 = 47 x 31.6mA = 1.4852 V


It shows that constant current transistor configuration only works ok if
the base current is also constant, of course, it relies on that. The above
driver circuit is dependent of battery voltage (and it fades), doesn't
matter so much if using emitter resistor to fix current or not, the R2
voltage will change along with battery voltage.

The only way to keep the constant current would be to generate a fixed
driver voltage at the driver output, what seems not to be the case.  Some
drivers use an internal double diode to ground to clip the output driver
voltage to around 1.2V, even with a 3V battery application. It keeps VR2
fixed along with the LED current.

In a 3V application, as remote controls, VR2 should be as small as
possible, because when batteries fade out to less than 2.2V, less the
transistor VCE, a high VR2 can kill the IR signal not because the batteries
are low, but simple because there is not enough voltage around the IR LED.

In this cases, the VBE and the known BETA will be used as the constant
current controller.  With a knowing Beta of 100, driving 2V to a 7kohms
base resistor, will provide a 200uA base current, what will open 20mA
collector current for the LED. As a small NPN can goes down to a 0.3VCE
without saturating, it means the battery can fades down as much as
VLED+0.3V, with the same IR emission level.

It means what?  that if you have a constant base current, it will result in
a constant collector current, with a little more worries about making sure
all the transistor batch shows a relative small beta variations.

Using a very small REmitter (R2) will bend the constant current control to
the Beta and VBE factors, since now they can almost dictate IC, being VR2 a
small portion of it.  It means that using an emitter resistor to fix
current, it should develop a significative emitter-ground voltage, at least
higher than 0.6V, so variations in VBE will represent small variations at
the constant collector current.

The same effect can be seem on operation amplifiers as constant current
generators.  In the feedback network, if the compared voltage is very
small, the relation between the Compared Voltage and OPOffset (changes with
temperature) will be small, causing the constant current very instable and
sensible to temperature.

Control > 3V __|  |__
|                    .------------.
|         0.10909V   |            |
R 1k         |      / \           |
|            |     /   \         _V_ Laser
|   100k     V    /     \         |  21.4mA
o----R----o------(+)___(-)        |  Const
|         |             |         |  Current
_V_ 1.2    R 10k         '--R10k---o
| VREF    |           0.10909V    |
_|_       _|_                      R 5.1ohms

In the above (ideal) circuit, a probable offset of 5mV at the op-amp will
generate a change in the constant current, in the order of 5mV/109mV = 4.6%
or almost 1mA.  If by lack of VCC you need to reduce the compared voltage
to less than 50mV, then the ratio gets worse.


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