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Electronic Fundamentals I Page 6- The Basic Power Supply
The power supply is used to convert the AC energy provided by the wall outlet to dc energy. In most electronic equipment, the power cord supplies the ac energy at 120 V to the power supply.
The power supply then provides all the dc voltages needed to run the equipment.
The basic power supply is broken down into 4 elements as shown.
AC
_1) The Transformer
- The Rectifier
- The Filter
- The Voltage Regulator_
The Power Supply
The Transformer
The Rectifier
The Filter
- usually steps up or steps down the incoming line voltage depending on the needs of the power supply. This alternating voltage is then fed to the rectifier.
- is a diode circuit that converts the ac to pulsating dc. This pulsating dc is then applied to the filter.
- is a circuit that reduces the signal variations to become a smoother dc voltage. It can include one
Basic Power Supply -- Block Diagram Vin A^ Rectifier B^ Filter C RegulatorVoltage Vout
A
Vin
Vout^ V^ DC
120 V AC
B
C
Transformer
Figure 1
Page 6-2 Electronic Fundamentals I The Transformer components such as resistors, capacitors and inductors. We will study the capacitor as a filter. The filtered dc signal is then fed to a voltage regulator stage.
- is used to maintain a constant voltage at the power supply output. It also provides a further smoothing of the dc voltage.
We will be using a zener diode as a voltage regulator. Modern day circuits have superceded the zener diode regulator with more modern integrated circuits. Since the zener diode is the simplest of these circuits to understand, we will study it as a prerequisite to the more modern circuits available.
The basic schematic symbol for the transformer is shown in Figure 2. Note that it has two windings , the primary and the secondary. The input voltage is applied to the primary winding and the output voltage is taken from the secondary winding. The vertical lines
or several passive
between the windings represent an iron core transformer.
Figure 3 represents a simple transformer. Note that there is no electrical connection between the primary and secondary windings. AC is applied to the primary and this ac current creates a magnetic flux in the core of the transformer.
The Voltage Regulator
The Transformer (^) Primary (Input)
Secondary (Output)
Figure 2
V (^) P
N (^) P
V (^) S
N (^) S
Magnetic Flux
Load
2 : 1
Figure 3
Page 6-4 Electronic Fundamentals I
Ideally, transformers are 100% efficient. This means that the ideal transformer transfers 100% of its power to the secondary (The actual losses are small, so we ignore them).
If we assume that all the power that goes in is transferred to the output then:
Note: P = I*V
The Transformer
In the simple transformer shown in Figure 4, we have 12 turns creating the magnetic flux in the core. This same flux cuts the secondary windings. Since only 6 windings are here, the induced voltage is one half of that in the primary. Since 12 windings to 6 windings is a ratio of 2:1, we can calculate the secondary voltage if the turns ratio is known and the primary voltage is known.
This gives the formula:
In our case above, if V is 24Vac, Calculate Vs:
P
Calculating Secondary Voltage
V (^) S V (^) P
N (^) S N (^) P
V (^) S V (^) P
N (^) S N (^) P
= (^) V (^) S =
=_______ Vac
Figure 4
V (^) P
N (^) P
V (^) S
N (^) S
Magnetic Flux
Load
2 : 1
Calculating Secondary Current
PS = PP
Electronic Fundamentals I Page 6- The Transformer
We know that power is voltage times current then:
And
A quick look at this last formula should tell you that the current ratio is the inverse of the voltage ratio. This means that:
This simply says that if the voltage on the secondary increases, then the current in the secondary decreases.
Eg:Hydro ® Bump up voltage to 500KV to 1MV
For a step-down Transformer I > I (Seconday V I ) For a step-up Transformer I < I (Seconday V I )
S P S P
Using our simple transformer in Fig. 4 as an example:
We found before that the secondary voltage was 12 Vac when the primary voltage was 24 Vac.
Calculate the secondary current Is if the primary current, Ip is 1A ac. Show your work.
Figure 4
V (^) P
N (^) P
V (^) S
N (^) S
Magnetic Flux
Load
2 : 1
V (^) S I S =V (^) PI (^) P
V (^) S I (^) S V (^) P
I (^) P = =
N (^) S N (^) P
Electronic Fundamentals I Page 6-
Transformer Ratings
While some manufacturers list their transformers by their turns ratios, others list them by secondary voltage ratings. In the lab, the transformer we use is listed as a 12.6 V transformer. This means that the secondary output will be 12.6 Vac at 120 Vac input.
It should be noted that many small transformers are designed to deliver their rated voltage at a rated current. When the transformer is operated below the rated current, it is normal for the secondary voltage to be above the rated voltage. This is because some of the output voltage is dropped across the secondary winding resistance.
Primary (Input)
Secondary (Output) Primary & Secondary 180 Oout of phase
Primary (Input)
Secondary (Output) Primary & Secondary in phase Figure 5
The Transformer
Types of Rectifier Circuits There are three basic types of rectifier circuits:
The most commonly used rectifier is the bridge rectifier followed by the full wave rectifier. The half wave rectifier sees limited use in today’s circuits.
**_1) The Half Wave Rectifier
- The Full Wave Rectifier
- The Bridge Rectifier_**
Page 6-8 Electronic Fundamentals I
The Half Wave Rectifier The easiest rectifier to understand is the half wave rectifier. It is simply a diode and a load as shown in Figure 6.
It is used to either the or to
eliminate negative swing of the ac input waveform eliminate the positive swing of the input waveform.
The Half Wave Rectifier
D 1 = on
V 0.7V
D 1
+VS (pk)
IF
V (V - 0.7V)
L S (pk)
VP (pk)
Positive Half Cycle
Basic Operation
(using the practical diode).
During the positive half cycle of the input, diode D is forward biased and provides a path for the
current through the load. This allows a voltage V to develop
across the load resistor (R ). This voltage is approximately equal
to V - 0.7 V
During the negative half cycle, the diode D is reverse biased and no current flows through the load resistor (R ) Now the voltage V is approximately 0 and the voltage across the diode V is the peak secondary voltage V
1 L L S (pk)
S (pk)
1 L L D .
R (^) L
V D1 = -VS (pk)
-VS (pk)
-VP (pk)
V = 0L
Negative Half Cycle
Figure 6
Figure 7
Page 6-10 Electronic Fundamentals I
The is found as:
this uses the standard 0.7 V for V
The is found as:
Note that the formula above assumes that the input to the transformer is given as a peak value. More often than not, source voltages are given as an rms (root mean squared) value. When this is the case, use this formula to convert it to a peak value.
F
Once the peak load voltage is determined, we can find the peak load current
Example 3-2, 3-3 and 3-4 work with peak load voltages and currents
peak load voltage
peak secondary voltage of the transformer
The Half Wave Rectifier
V pk V^ rms 0.
V S(pk) (^) = V^ P (pk)
N (^) S N (^) P
V L(pk) =V S(pk) V F
V L(pk) =V S(pk) 0.7 V
or
Calculating Load Voltages and Currents
where (^) = the ratio of transformer turns N (^) S N (^) P V (^) P(pk) (^) = the peak transformer primary voltage
I L(pk) (^) = V^ L(pk) R L
Electronic Fundamentals I Page 6-
The - V is the reading you would get if you used a dc voltmeter to read the pulsating dc output across the load of our rectifier. The meter averages out the dc pulses and displays this average. The formulas to calculate V are:
Either equation will find V. This is also called the dc equivalent voltage for our half wave rectifier.
Average Load Voltage (^) ave
ave
ave Keep in mind that this formula finds V (^) avefor our half wave rectifier only.
p
V ave (^) = V (^) L(pk)
Average Load Voltage & Current
or V ave (^) = 0.318V (^) L (pk)
The Half Wave Rectifier
Example 3-5 finds V (^) ave for the half wave rectifier.
Secondary Voltage
Equivalent DC Voltage Vave
Output Voltage
+V (^) (pk)
0V
Vave = 0.318 V (^) (pk) 0V
Just as we can convert a peak voltage to average voltage, we can also convert a peak current to an average current. The value of the average load current is the value that would be measured by a dc ammeter. This value is called the equivalent dc current.
Average Load Current
Figure 10
Electronic Fundamentals I Page 6-
Given an input of 120Vrms, and V
of 0.7V, calculate the following: Vp(pp), Vp(pk), Vs(pp), Vs(pk), Is(rms), Is(avg), V (rms)
Sketch the input and output wave form. Hint: Remember to show your units.
F
L.
Input Wave Form Output Wave Form
Electronic Fundamentals I
The reasoning for the PIV equation is that when the diode is reverse biased, there is no voltage across the load. Therefore, all of the secondary voltage (V ) appears across the diode. The PIV is important because determines the minimum allowable value of V for any diode used in the circuit. Remember that in ordinary circumstances.
Spk
RRM a replacement diode should have a V of at least 1.2 times the PIV
RRM
The Full Wave Rectifier
D 1
D 2
R L
Example of a 12.6 VAC Centre-Tapped Transformer
12.6 V
6.3 V
6.3 V
Primary Secondary
For example The transformer you will be using in the lab has a 12.6 Vac centre tapped secondary winding. This means that the voltage output from the centre to each outer terminal in one half of the total voltage or 6.3 Vac. See Figure 12.
Basic Circuit Operation
The transformer has a secondary winding. This secondary winding has a lead attached to the centre of the winding. The voltage from the centre tap to either end terminal on this winding is equal to one half of the total voltage measured end-to-end.
centre - tapped
The Full Wave Rectifier
Peak Inverse Voltage (cont)
Figure 11
The full wave rectifier consists of two diodes and a resister as shown in Figure 11
Figure 12
Page 6-
Electronic Fundamentals I
Using the practical diode model, the peak load voltage for the full wave rectifier is found as:
The full wave rectifier produces twice as many output pulses as the half wave rectifier. This is the same as saying that the full wave rectifier has twice the output frequency of a half wave rectifier.
For this reason, the average load voltage is found as
Page 6-
2
V L(pk) V^ S(pk) - 0.7 V
Calculating Load Voltage And Currents
= p V ave 2V^ L(pk)^ or V^ ave^ =0.637 V^ L(pk)
+VL(pk)
0V
V ave = 0.637 VL(pk) 0V
Peak Load Voltage
Average DC Voltage V (^) avefor a Full Wave Rectifier
Average DC Voltage V (^) ave
Figure 16 below illustrates the average dc voltage for a full wave rectifier.
The Full Wave Rectifier
Figure 16
Example 3-9 calculates the load voltage for a full wave rectifier. Example 3-10 determines the peak load current and the average dc voltage for a full wave rectifier.
Electronic Fundamentals I Page 6-
Negative Full - Wave Rectifiers
If we reverse the directions of both diodes in a positive full wave rectifier, we get a negative output across the load. Figure 15 shows this. Note both diodes point toward the transformer and that the output waveform is negative.
Negative Full Wave Rectifiers
Secondary Voltage
Output (^) 0V
0V
D 1
D 2
R (^) L
The Full Wave Rectifier
Peak Inverse Voltage
Figure 15
When one of the diodes in a full-wave rectifier is reverse biased, the voltage across that diode will be approximately equal to This
point is illustrated in Figure 16. The 24 V across the primary
develops peak voltages of +12 V and – 12 V across the secondary (when measured from end to centre tap). Note that V equals the
difference between these two voltages: 24 V.
V 2
pk
S(pk)
Electronic Fundamentals I Page 6-
From the information given in figure 3.18, calculate: Vs(pk), Vs(p-p), VL(pk), VL(avg), IL(avg) & PIV. Sketch the output waveform.
Page 6-20 Electronic Fundamentals I The Full Wave Bridge Rectifier
The Full Wave Bridge Rectifier
Basic Circuit Operation
The bridge rectifier (Figure 17) is the most commonly used rectifier circuit for the following reasons:
No centre - tapped transformer is required.
The bridge rectifier produces almost double the output voltage as compared to a full wave center-tapped transformer rectifier using the same secondary voltage.
!
!
During the positive half cycle (Figure 17) , both D and D are forward biased. At the same time, both D and D are reverse biased. Note the direction of current flow through the load. On the negative half cycle (Figure 18) D and D are forward biased and D and D are reverse biased. Again note that direction of current through the load is in the same direction although the secondary winding polarity has reversed..
3 1 2 4
2 4 1 3
D (^1)
D (^2) D (^3)
D (^4)
Positive Half Cycle
+V S (pk)
Figure 17
