RE: C64 power supply replacement

From: Baltissen, GJPAA (Ruud) (ruud.baltissen_at_abp.nl)
Date: 2004-01-16 09:57:08

```Hallo Groepaz,

And if you are not German speaking? Unfortunately I don't know the Englisch

The working is quite simple. Needed knowledge: if you raise or lower the
voltage at one pin of a capacitor, the other pin wants to follow it.

CR6

|\ |
+9V UNREG   ------| >|-----+
|/ |     |
|     CR5
|
| |          |    |\ |
9V AC --------| |----------X----| >|--------+---------  + 15V
| |               |/ |        |
A          B                       |
C90                       | ----- |
+-------+  C88
|
|
GND

The above is from the PRG. Assume we have no 9V AC connected, then at
crossing X we always measure 9 - 0.7 = 8.3 Volts. Let's now connect the 9V
AC. Assume we are at the bottom of the sinus. From now on the voltage level
at point A will rise about 13V. Point B wants to follow and C90 starts to
"suck current" through diode CR6 thus building up voltage. At the top of the
sinus, the voltage will drop again and B wants to get rid of its current.
But CR6 is one way only and the only way left is through diode CR5. C88
buffers the flow.

Another way to describe it:
+9V UNREG = normal air
C90 = piston + chamber
CR6 = one-way inlet valve
CR5 = one-way outlet valve
C88 = expansiontank.

..... and we have an aircompressor !!!

I hope this helpes :)

--
___
/ __|__
/ /  |_/     Groetjes, Ruud
\ \__|_\
\___|       URL: Ruud.C64.org

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