🚀 go-pugleaf

From article <OTTO.90Oct15004800@tukki.jyu.fi>, by otto@tukki.jyu.fi (Otto J. Makela):
> that I don't know how a switching power supply works.  Could someone with the
> necessary knowledge give me the baby-talk -type explanation so that I won't
> get stuck in such an embarassing way the next time ;-) ?  Be gentle, I'm only
> a computer hacker with limited (and long-past) electronics experience.  Post
> only if you have iron-clad knowledge of this.  No "I thinks", please.

Switching power supllies work on the principle of storing energy on a magnetic
field. Let me digress (sp?) for a moment. Old linear regulation stepped down
120V to 6.3Vrms via transformer, rectified it to about 8.5 volts, and then
used a linear regulater (7805, etc) to produce 5V at 1 Amp. Unfortunately
this method dissapated (8.5V - 5V) * 1 Amp of heat and wasted it.
This produced, at best 60% effeciency. Ok for 5Watts out, not good for 220W.


1) Step down switching (most common) [sorry about the poor graphics]

                         /MOSFET
120V-[]-6.3Vrms-X-8.5Vdc-#-|--()()()()--5V
                           ^
                           \Schottky Diode to ground.

 - 120Vrms is stepped down via transformer to 6.3V (usually higher), where it
is rectified to 8.5V dc. A high power, low loss, semiconductor switch is used
to pulse this power through a coil (usually toriod-based) to the 5V output.
When the 5V supply gets to 5.1 V the semiconductor switch is turned off.
In order to keep current flowing in the coil, the magnetic field collapses,
and the input end is clamped to ground by the diode, as the output maintains
the required current. When the output voltage gets to 4.9 volts, the switch
is turned back on, and the cycle repeats. Losses are: Switch (.2V*1Amp),
and Diode (.3V*1 Amp). However, because the power is 'Switched' the switch
loss is only during the ON phase, and the diode loss during the OFF phase.
Additionally: power in is pulsed, as required, to produce over 95% effeciency.
As a rule of thumb, the duty cycle will tend to be [On:Off] 'Vout:(Vin-Vout)'.

Lost yet?
-At turn on time the coil will saturate, requiring that a differencial probe
measure input current, and turn off the switch when the current is too high.
This will also prevent damage if the output is shorted. Without this the
switch will pop sooner or later, and with it the power supply 'soft starts'.

-Most switchers are fixed frequency (50-200KHz), variable duty. These type
require a minimum load to operate. A better method, in my humble opinion,
is to run 'free running switchers'. These are, as the name suggests, variable
frequency and duty cycle, and adjust the best to changing loads.
They work by letting the over/under voltage, and over current circuits
adjust the output for themselves.

-If the output is constant, the coil can be 'tuned', for added effeciency.

2) Step up Switching

 8.5V------
          ()
          ()     Diode
          ------->|---12V output
          |
          # MOSFET

In this method the semiconductor switch pulses the input voltage, through
the coil to ground. When the switch is released, the coil will pass current
out to the 12V, with its other end tied to the input.
This method is usually the cleanest design, because everything is positive,
and referenced to ground (no diff probes).
This was the first common use of switchers, to step up the +5 to +12 for
NMOS chips.

2) Inverter

 8.5V ----# MOSFET
          |
          ----|<---  (-12V) output
          |
          ()
          ()
          ()
          ---Ground.

In this method, the semiconductor switch builds up a field in the coil,
and releases it. The coil swings in voltage, and is then used to supply the
inverted voltage. This circuit requires both differencial current probes
for input limiting and an inverter-type circuit to measure the (-12V)
output relative the the positive control circuit and is the most difficult
of the three to design.

X) how your PC power supply realy works......

The input 120V is rectified to 240Vdc on the hot (line) side. The control
circuit pulses a transformer* (not a coil). The low (isolated) side of the
transformer is rectified and filtered as +5V. An opto-isolator is used
to tell the control circuit the exact value of the +5V supply.
The +/- 12 Volt supplies are generally just rectified and filtered.

* Don't stick your fingers in here, please. It will hurt.

As with (1) above, the losses are in the switch, and the rectifier(s).
However the switch losses are a percentage of the input circuit and
the duty cycle. Since the input voltage is 240V, a lot less current flows
through the switch resulting in a lower voltage drop (5V). As a percentage
of 240V, this is only 2%. With a 50% duty this is only 1%.

Now that you're totally confused.....

Texas Instr., National, Motorola, Siliconix, etc all produce app notes
on switchers as a use for their high power mosfets.

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| \      / |     /=|     /=|      Tony Becker           MCRSYS Florida
 \ \    / /       ||      ||      uunet!mcrsys!tony     (813) 799-1836 (voice)
  \ \  / /        ||      ||
   \ \/ /         ||      ||      Opinions expressed are my own, and don't
    \  /          ||      ||      necessarily reflect those of any other
    /  \          ||      ||      members of my species.
   / /\ \         ||      ||        -It's nice to live in a society where this
  / /  \ \        ||      ||         is possible, and not a capital crime!
 / /    \ \       ||      ||
| /      \ |    |====|  |====|
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