The schematic is incorrect. As wired the switch will not affect the voltage in any way.
The correct verison would be as follows
_______
| |
| |
|_1_2_3_|
| | |_____From transformer/rectifier
| |
| |_______Output
| | |
| / ---
| \ ---
| /390 |
| \ |
|_| |
|
/
\ 10R
/
\
|
/
\ 10R
/
\
| To switch and variable resistor
Note that a 2200uF capacitor on the output is really way too much. This amount of capacitance does not help the regulator work better, it slugs the transient response. A single 100nF will suffice in most cases with maybe a 10uF cap if you are expecting big load spikes.
Also note that when the switch is inbetween positions, i.e. when you switch it, the output voltage will briefly rise to almost the unregulated voltage, probably destroying any sensitive equipement on the output. You can use a make before break switch to stop this, or you could change the whole scheme and switch the 390R resistor to other values. This way the voltage will fall to 1.25 volts during switching.
Regards
Mike Rigby-Jones
-----Original Message-----
From: Mike M [SMTP:elektrikman@DYNAMITEMAIL.COM]
Sent: Tuesday, November 09, 1999 2:50 AM
To: PICLIST@MITVMA.MIT.EDU
Subject: Adjustable Power Supply HELP
Hey guys, i built a real (cheep) quick adjustable power supply using an LM317...i basically followed the schematic at:
http://expert.cc.purdue.edu/~bymaster/power_supply.gif
anyway, when i check the output voltage on the 317, no matter what values of resistance i put the adjust pin and the output pin always read 16.8v. There is a 18v input. The 5v 7805 regulator works fine... Now i know i probly blew out the chip because while testing i turned the 5k pot to 0, there should be 20ohm in series wit the 390ohm resistor and when u do the math
1.25(1+ 20/390) = 1.31v
In my circuit..the two 10ohm resistors started to smoke and turned black. Why?? They are 1/4 wat resistors (probably why but all i had on hand)..???
mikE
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