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EE1000, Lecture 11
Linear power supplies (Transformer, rectifier, regulator)
The ideal transformer
Current and magnetic flow in a transformer
Transformer circuit
Center-tapped transformer
See the animation of transformer (
Any load impedance ZL connected to the ideal transformer's secondary winding causes current to flow
without losses from primary to secondary circuits, the resulting input and output apparent power
therefore being equal as given by the equation
Combining the two equations yields the following ideal transformer identity
This formula is a reasonable approximation for the typical commercial transformer, with voltage ratio
and winding turns ratio both being inversely proportional to the corresponding current ratio.
Half-wave rectification
In half wave rectification of a single-phase supply, either the positive or negative half of the AC wave is
passed, while the other half is blocked. Because only one half of the input waveform reaches the output,
mean voltage is lower. Half-wave rectification requires a single diode in a single-phase supply, or three
in a three-phase supply. Rectifiers yield a unidirectional but pulsating direct current; half-wave rectifiers
produce far more ripple than full-wave rectifiers, and much more filtering is needed to eliminate
harmonics of the AC frequency from the output.
Full-wave rectification
A full-wave rectifier converts the whole of the input waveform to one of constant polarity (positive or
negative) at its output. Full-wave rectification converts both polarities of the input waveform to
pulsating DC (direct current), and yields a higher average output voltage. Two diodes and a center
tapped transformer, or four diodes in a bridge configuration and any AC source (including a transformer
without center tap), are needed. Single semiconductor diodes, double diodes with common cathode or
common anode, and four-diode bridges, are manufactured as single components.
For single-phase AC, if the transformer is center-tapped, then two diodes back-to-back (cathode-tocathode or anode-to-anode, depending upon output polarity required) can form a full-wave rectifier.
Twice as many turns are required on the transformer secondary to obtain the same output voltage than
for a bridge rectifier, but the power rating is unchanged.
Check out: and select diodes (half-wave rectifier and full-wave
Linear Regulator
A linear regulator is a system used to maintain a steady voltage. All linear regulators require an input
voltage at least some minimum amount higher than the desired output voltage. That minimum amount is
called the dropout voltage. For example, a common regulator such as the 7805 has an output voltage of
5V, but can only maintain this if the input voltage remains above about 7V, before the output voltage
begins sagging below the rated output.
"Fixed" three-terminal linear regulators are commonly available to generate fixed voltages of plus 3 V,
and plus or minus 5 V, 6V, 9 V, 12 V, or 15 V, when the load is less than 1.5 amperes.
The "78xx" series (7805, 7812, etc.) regulate positive voltages while the "79xx" series (7905, 7912, etc.)
regulate negative voltages. Often, the last two digits of the device number are the output voltage; e.g., a
7805 is a +5 V regulator, while a 7915 is a -15 V regulator.
The 7805 takes in a voltage between 7 and 30 volts and regulates it down to exactly 5 volts. The first
capacitor takes out any ripple coming from the transformer so that the 7805 is receiving a smooth input
voltage, and the second capacitor acts as a load balancer to ensure consistent output from the 7805.
Want to know what’s inside 7805 chip? Take a look at this link:
The reservoir capacitor (smoothing capacitor)
The output of rectifiers - although unidirectional - is not suitable to supply a radio because it contains
still a RIPPLE - an ac partition of the voltage. These ripples cause distortion to the radio. Therefore,
there are means needed which reduce these ripples and convert the output voltage of the rectifier to a
steady dc-supply.
The effect of the circuits doing this job is called SMOOTHING. The simplest form of smoothing is
obtained by connecting a capacitor in parallel to the output of the rectifier terminals.
During the first half of the ripple when the output voltage of the rectifier is increasing - the capacitor
will be charged. Therefore during this period, the capacitor stands for an additional load which means
the overall voltage at the output will be smaller than without the capacitor (depending on the internal
resistance of the voltage source). This effect will go on till the voltage has reached its peak.
During the second half of the ripple - when the voltage drops again - the capacitor is discharged again.
The capacitor stands now for a second energy source parallel to the original voltage source (the
rectifier). So the voltage at the output terminals will be higher than without the capacitor because the
current drawn from the rectifier is diminished. The capacitor will tend to fill the “valleys” between the
I am i
EE1000 HW #11
1. (true/false) Transformers work on direct current (DC).
2. A transformer has a turns-ratio of 5:1. If the input (primary) voltage is 120 Volts, what is the
output (secondary) voltage?
_________ ________________.
3. A transformer has a turns-ratio of 10:1. If the (output) current in the secondary winding is 1A,
what is the (input) current in primary winding?
___________ ______________.
4. Draw a full-wave rectifier using four diodes for the transformer shown below:
5. What device do we use to filter (smooth out) the output of a rectifier?
________ _________________.
6. Show how to use a 7805 linear regulator to produce a constant 5 Volts. You will need to show
two capacitors. (Do not give values for the capacitors; just draw them where they belong.)
From Rectifier