View Single Post
  #40   Report Post  
Active8
 
Posts: n/a
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" said...
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 figured that since the 6V input was already well regulated, that it
would be easy to just drop a single volt. My subject: line kind of
said that. And if I use a 1N4002 inline with the input, it gives me
5.4 to 5.2V or so, over most of the current range, which is not all
that bad. And I did say I wanted to keep it simple.


I'll trade a pot for a zener and R anyday, but I haven't looked at
what this thing will do over the full current range. It's so close
to what you started with, I thought it would be a snap to check
out.

It's not that
I'm unappreciative of all the input others have given, it's been a
good learning experience seeing these neat little circuits. I've
saved many of them to disk, to experiment with later. So I thank all
who contributed.

I'm just glad to find an electronics post or 2 lately. Lots of long
OT stuff I'm guilty! I feed it, but it's fun.

--
Best Regards,
Mike