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Designing a
Linear Power Supply
Bill Naylor, Electronworks Ltd
The article explains how to design a mains powered
linear power supply. You can use one of our
electronic kits to evaluate the theory.
If you have any suggestions for improving this
application note, please drop us a line at:
enquiries@electronworks.co.uk
It is worth reading the articles on Capacitors and
Diodes before proceeding.
In our article on Capacitors, we learned that one of
their applications is the storage of a dc voltage.
We will now examine how they are used in power
supply circuits.
The circuit of FIG 1 is used in most mains based
equipment from industrial electronics to TVs.
FIG 1
The 240V mains is applied to the input of a
transformer with a turns ratio of 20:1. This means
that for every 20 turns on the input side (the
primary winding) there is 1 turn on the output side
(the secondary winding).
Now, 240V is a measure of the main’s ‘rms’ voltage.
The rms voltage is equal to 0.707 of the peak
voltage, so in truth the peak voltage of the mains
is actually
This has important consequences as we shall see
later.
The output voltage is
 V
due to the turns ratio of the transformer.
Similarly, this is an rms voltage, so its peak is
So we have an ac waveform with a peak voltage being
applied to the diodes D1 – D4. (This configuration
of diodes is known as a bridge rectifier). Referring
to FIG 1, when the upper connection of the
transformer output is positive, diode D1 conducts
passing current to the output capacitor, C, and out
to the load. The current flows through the capacitor
and load and back through diode D2. Since the lower
connection of the transformer is negative, diode D2
is biased into conduction and so the current returns
to the transformer.
When the bottom connection of the transformer is
more positive, the same happens, but with diode D3
conducting a positive voltage to the top of the
capacitor. The current flows through the capacitor
and out through diode D4. Since the top connection
of the transformer is more negative, diode D4 is
biased on and conducts the current back to the
transformer
FIG 2 shows the waveforms that are present at each
part of the circuit.
FIG 2
The voltage across the capacitor is shown as the red
dotted line in FIG 2. If the capacitor is removed,
we will get the output voltage of the bridge
rectifier (shown in blue), so we can see that we
have transformed the waveform from a sinewave at the
input to the circuit to a pulsed waveform (shown in
blue - coming out of the diode bridge), then to a
smoothed voltage using a capacitor on the output (to
produce the red waveform).
So what output voltage do we get? The transformer
outputs a peak voltage of 16.9V, but its voltage is
then dropped by 0.6V as it passes through a diode
(D1). It is then dropped by another 0.6V as it
passes through another diode (D2), so the highest
voltage we will get across the capacitor is 16.9 –
1.2 = 15.7V. Referring back to our article on
capacitors, we now have to choose our capacitor to
have a working voltage of at least 15.7V. Good
engineering practice states you should not take a
component to within 80% of its maximum rating, so a
capacitor with a voltage of 20V should allow enough
‘headroom’.
So why isn’t the voltage across the capacitor flat
dc? Well, let’s examine the red waveform. When the
rectified output is at its maximum, the capacitor is
fully charged. As the rectified voltage drops the
capacitor holds charge, keeping the voltage high at
the output. However, during this time, the load
takes current from the capacitor, discharging it.
When the rectified voltage comes back up again, it
recharges the capacitor voltage and the cycle starts
again. This change in dc voltage across the
capacitor is called the ripple voltage.
The amount of droop on the capacitor is determined
by the current drawn from the capacitor (the load
current) and the capacitor value. (as a reminder, in
our article on capacitors we mentioned that the
discharge of a capacitor is dependent on the current
being drawn from it).
There is an equation that describes this discharge:
I is the current (in Amps) being taken from the
capacitor, C is the capacitor value (in Farads), dv
is the change in voltage and dt is the
change in time.
Now for some numbers…
Our load current is normally known. Let’s assume it
is 100mA (=0.1A). If our output voltage is a maximum
of 15.7V, we should normally design for a ripple
voltage of no more than 10%, although 5% is better.
I am going to design for a ripple voltage of 5%, so
this gives me a ‘dv’ of 0.785V. This is illustrated
in FIG 3.
FIG 3
The only other unknown is the value of ‘dt’. Looking
at FIG 2, we see that the input voltage is at mains
frequency (50Hz). The rectified voltage is at twice
that frequency (100Hz). This has a period of 10ms
(0.01s). We can see that our red waveform droops for
nearly 10ms (but not quite), but it is good enough
to approximate the droop time to be 10ms.
We now have all the information needed to work out
what capacitor value will give us no more than
0.785V with a load current of 100mA. Plugging these
numbers into the above equation gives:
Implying that C is 1273uF. Looking at my components
catalogue, the next value above 1273uF is 2200uF.
There are two versions – one with a working voltage
of 16V and one with a working voltage of 35V. I will
play safe and use the 35V version.
Putting this capacitance value back into the above
equation implies my ripple (dv) should be about
454mV.
And so my power supply is designed!
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