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John Fields
 
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Default Dropping 1V from a Regulated 6V Wall Wart

On Tue, 30 Dec 2003 08:28:53 -0800, Watson A.Name - "Watt Sun, Dark
Remover" wrote:

In article ,
mentioned...
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


Yeah, I was thinking something really low like that.

But these approaches have been to establish a new regulated voltage,
dependent on the value of the zener. There's nothing wrong with this,
but it would be simpler to just use a LDO 5V regulator as someone
suggested.


---
I agree.

Plus, you won't have the Zener tempco and delta VZ/delta IZ (small
though it might be since the change in IZ will only be caused by the
change in the transistor's Ib as the load current goes from 0mA to
200mA) to contend with. And, a BIG plus for the LDO, you won't be
dissipating the power the Zener will be _all the time_ , regardless of
the load current.

--
John Fields