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question about switch mode supply

L

Louie

Jan 1, 1970
0
can someone explain to me something about how switch mode power supplies
work? i read an article that states:

"The duty cycle of chopped DC will effect the AC voltage level generated
on the transformer's secondary. A long duty cycle means a larger output
voltage (for heavy loads) and a short duty cycle means lower output
voltage (for light loads) (for heavy loads) and a short duty cycle means
lower output voltage (for light loads)"

what i don't get is the "long duty cycle means a larger output voltage"
part. i figure that the secondary voltage is determined by the primary
voltage and the ratio of windings. how does this work?

thanx, Louie
 
J

John G

Jan 1, 1970
0
I would say the article was written by someone who does not know much.
If it is anys smart the output voltage will be controlled by the circuitry
which controls the duty cycle of the transformer so that the rectified
filtered voltage is constant. And as you think this means wider pulses not
higher pulses from the transformer.
 
J

John Popelish

Jan 1, 1970
0
Louie said:
can someone explain to me something about how switch mode power supplies
work? i read an article that states:

"The duty cycle of chopped DC will effect the AC voltage level generated
on the transformer's secondary. A long duty cycle means a larger output
voltage (for heavy loads) and a short duty cycle means lower output
voltage (for light loads) (for heavy loads) and a short duty cycle means
lower output voltage (for light loads)"

what i don't get is the "long duty cycle means a larger output voltage"
part. i figure that the secondary voltage is determined by the primary
voltage and the ratio of windings. how does this work?

thanx, Louie

This description is so simplified that it no longer makes sense.
There are lots of different topologies with different waveforms and
concepts so I am guessing about which one this is trying to describe.

I think the above description is about a version that produces pulses
in alternating directions from some DC source into a transformer, and
then rectifies the pulses out of the secondary, before passing those
pulses through some low pass filter. Duty cycle refers to the percent
on time of the pulses. Since the output, here, is rectified AC, when
the primary sees a 50% of cycle pulse in each direction, the rectified
output is on 100% of the time, and the DC out is at maximum. If the
primary pulses are shorter than 50% of a cycle, and the rest of the
half cycles consist of zero volts, the rectified output will be a
pulse train (the pulses being the same peak voltage as the first case)
with some zero volt sections between them. The low pass filter after
the rectifier will average this pulse waveform as a lower value of DC
than the full duty cycle version.

The aim of most switching supplies is to keep the DC output voltage
constant, in spite of both source voltage variations (height of pulses
varying) and load current variations (changes in the resistive voltage
drops of the switches, transformer windings, rectifiers and filter
caused by changes in current).

Here is a good general introduction to switching regulator concepts
and circuits:

http://www.national.com/appinfo/power/files/f5.pdf

found on this list of papers:

http://www.national.com/appinfo/power/0,1768,95,00.html
 
B

Bill Bowden

Jan 1, 1970
0
Louie said:
can someone explain to me something about how switch mode power supplies
work? i read an article that states:

"The duty cycle of chopped DC will effect the AC voltage level generated
on the transformer's secondary. A long duty cycle means a larger output
voltage (for heavy loads) and a short duty cycle means lower output
voltage (for light loads) (for heavy loads) and a short duty cycle means
lower output voltage (for light loads)"

what i don't get is the "long duty cycle means a larger output voltage"
part. i figure that the secondary voltage is determined by the primary
voltage and the ratio of windings. how does this work?

thanx, Louie

I think it works more like an inductor than a transformer.
Say the primary inductance is 1mH and the applied voltage
is 12 and the pulse width (time on for applied voltage) is
100uS. This means the current will ramp from 0 to about 1 amp
in 100uS and the average current is 1/2 amp. If the duty cycle
is 50%, the average current will be 1/4 amp and the average
power is 12*.25 or 3 watts. Assuming 100% efficiency, and
a load resistance of 100 ohms, the output should be around
17 volts DC. A longer duty cycle will increase input current
and output voltage on the load.

But this doesn't account for the transformer turns ratio.
I can't quite see how the turns ratio effects the operation except
maybe to reduce losses. If you put 3 watts in, and 3 watts goes
into a 100 load, the voltage must be 17, regardless of turns ratio.

What am I missing?

-Bill
 
L

Louie

Jan 1, 1970
0
yeah, i sorta understand now how they work. but i have to test this PS
unit. how can i trick it into thinking its in circuit so i can measure
its output?

louie
 
L

Louie

Jan 1, 1970
0
thanks, John, that helps alot. it wasn't easy, but i read through the
reference you cited, so now i have some understanding of how they work.
now what i have to do is troubleshoot this PS that came out of a
projection TV.
if i could figure out how to load the output so it thinks its in
circuit, then i could just measure the voltages and see if its working
or not.
problem is, i have no documentation telling me what the voltages are
supposed to be.
however, there are several zeners in the control circuit, which is on
the output side of the board, seperated from the input side by the
multi-tap trans and the opto coupler. maybe the zeners can tell me what
the levels are supposed to be.
or maybe i just don't have the slightest clue what i'm doing.
and, yes, i do have a resistor to drain the caps.

Louie
 
L

Louie

Jan 1, 1970
0
i believe the reason for the high frequency of switching is so that you
can have a smaller transformer. with the unit i'm working on the caps
and switching regulator are almost as big as the transformer.

as for the ratio, i think you can be pretty loose with it as long as the
secondary has comparable inductance with the primary. as far as i can
tell, that is.

louie
 
U

uuuuu

Jan 1, 1970
0
Regarding the first link, why does the current in Fig 30 (the current
through the L) rise and fall linearly ?

Anoher question:
There is a law that current change through an inductor cannot happen
instantly, so what happen in the switch is opened in one go ?
 
J

John Popelish

Jan 1, 1970
0
uuuuu said:
Regarding the first link, why does the current in Fig 30 (the current
through the L) rise and fall linearly ?

Anoher question:
There is a law that current change through an inductor cannot happen
instantly, so what happen in the switch is opened in one go ?

The answer is the same for both questions. The rate of change of
current through an inductance is proportional to the voltage across
it. This property defines inductance.

In figure 30, the ramp up rate of change is proportional to the input
voltage minus the output voltage, because this is the voltage across
the inductor when the switch is on. The ramp down slope is
proportional to - the output voltage and a diode forward drop, because
the inductor can produce no more reverse voltage than this without
forming a conduction loop to the output through the diode, which then
carries the inductor current as it winds down its stored energy.
Since the voltage in both directions changes little during that part
of the cycle, the current rate of change is similarly constant for
those two ramps.

There is a discontinuity in the inductor voltage as the switch opens,
and the inductor voltage changes polarity. This voltage rate of
change is limited only by the speed of the switch and the stray
capacitance of the inductance and the other circuit components tied to
it. But the magnitude of total voltage change is limited by the turn
on of the diode.
 
J

John Popelish

Jan 1, 1970
0
uuuuu said:
Are you saying that the reverse voltage which is induced on L can be
greater than the forward voltage that was there just before the
switching occured, i.e. greater by 0.6V (diode drop) ?

Sure. For example, if the inductor has 7 volts switched into it, and
5 volts out, there is only 2 volts causing current ot increase when
the switch is on, but when the switch turns off, there is -(5 +.6) or
so across it as the inductor produces enough induced voltage to
forward bias the diode.
Is it possible for the induced voltage to be arbitrarily high ?

Yes. The induced voltage is limited only by the speed that the switch
turns off, and the capacitive load across the inductance and the
parallel resistance across the inductance (that represents the current
induced in the core). If the inductor and switch were perfect and the
stray capacitance approached zero, the induced voltage would approach
infinity (but would last for a time approaching zero) and the inductor
current would quench to zero almost instantaneously.
Also, can you please explain the formula in fig. 28, i.e. how is this
formula arrived at ?

This equation just describes the averaging effect of a low pass filter
with no losses. The average voltage over the time of one pulse cycle
(Tp) of a pulse that spends Ton time at Vpk, and (Tp-Ton) at zero
volts is:

Vout = ((Ton*Vpk + 0*(Tp-Ton))/Tp = (Ton*Vpk)/Tp

Of course, it a real circuit, the off state voltage is one diode drop
negative, and the inductor has some resistive voltage drop, so this
formula overestimates the output voltage.
Also, I believe you did not say why the voltage on L increases and decreases
linearly and produces a saw-tooth wave form, which formula are you using ?

The voltage across L is assumed to have one of two fixed values during
a cycle. when the switch is on, it is assumed to be the unregulated
voltage supply, and when the switch is off, it is assumed to be the
negative of the output voltage plus a diode drop.
 
U

uuuuu

Jan 1, 1970
0
John Popelish said:
Sure. For example, if the inductor has 7 volts switched into it, and
5 volts out, there is only 2 volts causing current ot increase when
the switch is on, but when the switch turns off, there is -(5 +.6) or
so across it as the inductor produces enough induced voltage to
forward bias the diode.


Yes. The induced voltage is limited only by the speed that the switch
turns off, and the capacitive load across the inductance and the
parallel resistance across the inductance (that represents the current
induced in the core). If the inductor and switch were perfect and the
stray capacitance approached zero, the induced voltage would approach
infinity (but would last for a time approaching zero) and the inductor
current would quench to zero almost instantaneously.


This equation just describes the averaging effect of a low pass filter
with no losses. The average voltage over the time of one pulse cycle
(Tp) of a pulse that spends Ton time at Vpk, and (Tp-Ton) at zero
volts is:

Vout = ((Ton*Vpk + 0*(Tp-Ton))/Tp = (Ton*Vpk)/Tp

Of course, it a real circuit, the off state voltage is one diode drop
negative, and the inductor has some resistive voltage drop, so this
formula overestimates the output voltage.


The voltage across L is assumed to have one of two fixed values during
a cycle. when the switch is on, it is assumed to be the unregulated
voltage supply, and when the switch is off, it is assumed to be the
negative of the output voltage plus a diode drop.

OK, amlost done :) now how about the current,
I meant to ask why the current is rising and falling in a triangle
wavefrom (figure 30). Is it derived from integrating di/dt in the
V=Ldi/dt formula ?
Thanks!!
 
J

John Popelish

Jan 1, 1970
0
uuuuu said:
OK, amlost done :) now how about the current,
I meant to ask why the current is rising and falling in a triangle
wavefrom (figure 30). Is it derived from integrating di/dt in the
V=Ldi/dt formula ?
Thanks!!

Precisely. The voltage and the rate of change of current (the slope
or di/dt or the current waveform) are proportional to each other. The
proportionality constant that relates them is the inductance, L.
 
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