On Tue, 30 Dec 2003 12:03:49 -0600,
said...
On Tue, 30 Dec 2003 16:54:34 GMT, Active8
,invalid wrote:

On Tue, 30 Dec 2003 08:57:09 -0600,
said...
On Tue, 30 Dec 2003 12:36:45 GMT, Active8
,invalid wrote:


Rough guess...



---+------------------+
| |
| 10 |
| ___ |/
+---|___|---+----|
| |
| |
- |
5.6 500mW ^ +---------
|
-------------+----------------

---
Closer guess:

Ib ~ Ie/ß = 200mA/100 = 2mA

IZt = 20mA

R = (Vin-VZ)/(IZt+Ib) = 0.4V/0.022A = 18.18...R ~ 20R


Ok, I incorrectly guessed Beta (Chaos Master told me how to get
special characters and it doesn't work here) of 200 and set Iz at
40mA (a waste) to get the 500mW zener to operate at about half it's
rating. I know, I only needed 10mA if Ib really were 10mA.

---
Any common 5.6V 500mW Zener will have its Zener voltage specified at a
particular test current; 20mA. This current is what _must_ flow through
the Zener in order for the reverse voltage it drops to fall within the
bounds specified.

I glanced at a generic curve. Not even sure where the Zener specs
(if any) *are* on this box.

40mA will not only waste 20mA, it could cause the
Zener voltage to be out of spec. The beta of the transistor (more
properly the alpha) is almost unimportant in this case since the base
current will be very small compared to the collector/emitter current and
its change will have very little impact on the Zener current from
no-load to full-load.

Yes, Beta's just for rule of thumb stuff. Base current is
negligible which is the idea of the pass trans, anyway. It allows a
practical resistor for such a small drop, whereas not using it
forces the resistor to handle most of the current and you know...

--
Best Regards,
Mike