AVR-Chat

Thanks Graham,

I ended up opening the supply to have a look inside. There is a current
mode PWM chip on the underside of the board - M62281FP (datasheet
http://www.ortodoxism.ro/datasheets/MitsubishiElectricCorporation/mXrttz
r.pdf ).

Unfortunately my electronics knowledge is thin. Would anyone know
whether I can adjust the output voltage of the psu by playing with
inputs of this chip? Maybe fool the feedback circuit?

Off topic I know.

Many thanks
Dean

-----Original Message-----
From: AVR-Chat@yahoogroups.com [mailto:AVR-Chat@yahoogroups.com] On
Behalf Of Graham Davies
Sent: Friday, 20 May 2005 8:03 AM
To: AVR-Chat@yahoogroups.com
Subject: [AVR-Chat] Re: A little off topic, but PSU question....

--- In AVR-Chat@yahoogroups.com, "Claxton, Dean J" 
<dean.j.claxton@t...> wrote:

> ... I placed a pair of 1N5404
> diodes in series for ... drop
> of around 1.4V (to be on the
> safe side ... the diodes get
> rather toasty ...

This seems to be a reasonable solution to me, assuming you can't find 
an obvious way to turn down the voltage.

The problem with putting diodes in parallel is that they may not 
share the current particularly equally. You can get around that by 
just using more than are strictly necessary, say five instead of 
three.

Another approach would be to use a resistor and a diode in series 
instead of two diodes. Suppose, for example, the current really is 
three amps and you decide to put one amp through each of three 
diodes. If you measure the voltage and it is still high by, say, 0.3 
volts, then just put a 0.3 ohm resistor in series with each diode. 
This will drop the additional voltage and ensure you get even sharing 
of the current.

One thing to remember, though, is that when you heatshrink the whole 
lot together, there's the same amount of heat coming out of your 
bumdle whether you have two or six diodes in there. Things may run 
cooler while you have the diodes separated, but when you mash them 
together they could be almost as hot once again.

Graham.





 
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--- In AVR-Chat@yahoogroups.com, "Claxton, Dean J" 
<dean.j.claxton@t...> wrote:

> ... There is a current mode PWM
> chip on ... the board - M62281FP
> ... adjust the output voltage 
> ... by playing ...

What you've probably got here is something like the application circuit 
at the end of your data sheet.

You need to know first of all that there are voltages in here that are 
way more hazardous than what comes out of your power outlet. Even if 
you've had a jolt from 110 volts AC, do not assume you will survive 
electrocution by the higher DC voltages inside this power supply. Don't 
touch anything while it's on. Plug it into its load and wait for it to 
discharge even after you've turned it off and disconnected it from the 
outlet before you touch it. Wear safety glasses. I'm serious.

What you need to find is the circuit that senses the output voltage and 
feeds it back to the chip. Look for the optoisolator and the voltage 
reference shown at the bottom right of the application circuit. If 
there are two optoisolators, the one connected to CT(OVP) is over-
voltage protection. It's the one connected to EA OUT that you want. 
Trace this part of the circuit and draw out the schematic. Assuming it 
looks like the application circuit, what you want to do is to make 
small adjustments to the ratio of the two resistors at the right that 
form a voltage divider across the power supply output. To reduce the 
output by 10%, try putting a resistor in parallel with the upper of 
these two that is ten times the value of the one that's already there. 
Hands off, stand back, safety glasses on and see what you get. Power 
off, disconnect, discharge and adjust accordingly. Don't go far from 
the original design voltage. You're playing with the voltage at the VCC 
pin of the chip as well.

Graham (no liability accepted).

(think the quoting attribution got messed up due to poor reply
formatting)

> I ended up opening the supply to have a look inside.

Why not look inside the LCD monitor instead. If there is a regulator in
its input then its likely all this fuss is moot.

> The problem with putting diodes in parallel is that they may not 
> share the current particularly equally. You can get around that by 
> just using more than are strictly necessary, say five instead of 
> three.

Not really. Need series resistor in line with each parallel diode to
help balance things out.

Silicon diode gets hot its resistance drops drawing more current and
making more heat.

-- 
David Kelly N4HHE, dkelly@HiWAAY.net
========================================================================
Whom computers would destroy, they must first drive mad.