Tutorial 14 B - Smoothing and Voltage Regulators
The picture shows a circuit board with a rectifier bridge, a smoothing capacitor, and a voltage regulator:
Note that one of the AC terminals is missing...
Here is a circuit diagram of a smoothed power supply for a logic tutor kit:
Notice the 2200 mF capacitor. The smoothing capacitor acts as a kind of electrical spring. The springs on a car smooth out the bumps in the road. The capacitor acts as a kind of electrical spring, smoothing out the bumps in the supply. The picture of the stylised CRO screen below shows the idea.
The output shown by the CRO is not far off a smoothed DC. Electronics engineers may refer to the wave form as a ripple. The lower the current taken, the smoother the ripple becomes. Here is the ripple when there is a low current.
When the current demand increases, the ripple becomes larger:
This might well introduce a hum.
Voltage in smoothed circuits
If we measure a smoothed output, we can be surprised to find that the voltage is somewhat higher than we expect. This is because the voltage given out is the voltage across the capacitor. The capacitor stores the peak voltage, while the expected voltage is in fact the RMS voltage.
The RMS voltage (Vrms) is the "DC equivalent" of an alternating voltage. The maximum voltage is called the peak voltage, which is given the Physics code V0. These quantities are linked by the equation:
The output of a rectifier is measured as 13.4 V. A capacitor is then placed across the terminals to smooth the output. Calculate the output voltage after this has been done.
V0 = Vrms × Ö2 = 13.4 V × Ö2 = 19.0 V (3 s.f.)
For most cases, this is not much of a problem, but if we have voltage sensitive components in our circuit, we need to be a bit careful. For example if we have capacitors with a working voltage of 16 V, we could have problems. If we have LEDs where the current must not exceed a certain value, we may need to change the value of the current limiting resistors.
The RMS voltage of the secondary of a transformer is 12 V. It is connected to a rectifier bridge.
(a) Calculate the rms voltage of the rectified AC.
(b) The supply is now smoothed by a capacitor. Calculate the voltage that would be read by a high resistance voltmeter.
When the current is very low, we get (almost) a perfect DC from this arrangement. However, if the current demand is increased, the supply is less smooth, and that can lead to problems. So this is where a voltage regulator helps.
In some circuits it can be important that the voltage does not exceed a certain value. Supply voltages can change their voltage depending on what current is drawn. Also when a big current is drawn, it is useful for the voltage to be constant. This is particularly important in computers. If the voltage drops too far, the computer may well crash.
Voltage regulators are placed at the output of the power supply to make sure that the voltage remains constant. You can buy voltage regulators for a few pence. They look quite like transistors, but do not use them as such.
The concept of the voltage regulator is shown in the diagram below:
The voltage regulator acts by providing a variable voltage within the power supply to the load. If we think of voltage as electrical pressure, we can extend that idea to think of a voltage regulator as being a pressure relief valve that opens when the pressure gets too high.
The key component in a voltage regulator is a Zener diode. This diode is used in a reverse-biased configuration, in which it is designed to break down at a particular voltage. This break down does not harm the diode in any way.
The simplest regulator is a Zener diode arranged like this:
The most common value for the breakdown voltage for a Zener diode is about 5.6 V. There is a limit to the current for a Zener diode of about 20 mA. The maximum power they can dissipate before being destroyed is between about 500 mW and 2.0 W. While in theory it is possible to make higher powered Zener diodes, this would be undesirable because of the waste. In general, the higher the breakdown voltage, the lower the current they can take.
Some common physics codes:
VZ – voltage across the Zener diode;
IZ – current through the Zener diode;
IL – current through the load.
A 5.0 V regulated power supply is required to be produced from a 10 V power source. The maximum power that the Zener diode can handle is 0.5 W. Work out:
(a) The maximum current that can flow through the Zener diode;
(b) The minimum value of the series resistor R;
(c) The load current if a load resistor of 2 kW is put onto the circuit;
(d) The Zener current at this load.
(a) I = P/V = 0.5 ÷ 5 = 0.10 A = 100 mA
(b) VR = 10 – 5 = 5 V Þ R = 5 ÷ 0.1 = 50 W;
(c) IL = 5 ÷ 2000 = 2.5 × 10-3 A = 2.5 mA;
(d) IZ = 100 – 2.5 = 97.5 mA
The higher the resistance of the load, the greater will be the current through the Zener diode.
What would happen if the series resistor was not there?
What do you think is the limitation of the Zener diode as a voltage regulator?
The answer to question 2 shows that the Zener diode is rather limited as a voltage regulator. A more sophisticated circuit has a transistor:
For this circuit the base voltage needs to be about 0.6 V above the Zener voltage. So if the Zener voltage were 5.6 V, the base voltage would need to be:
VB = VZ + 0.6 V = 5.6 V + 0.6 V = 6.2 V
With this set up, we can choose a Zener diode that has a small current requirement, such as 20 mA. Let’s suppose that our unregulated voltage is still 10 V. The base current for the transistor is 10 mA.
Voltage across the resistor is 10 V – 6.2 V = 3.8 V
Current through the resistor = 20 mA + 10 mA = 30 mA
Therefore the resistor would have a value:
R = 3.8 V ÷ 30 × 10-3 A = 127 W
The emitter-collector current would be:
ICE = hFE × IB
So if the gain is 50, ICE = 500 mA.
The emitter current will be 500 mA + 10 mA = 510 mA
Explain why the output current is 510 mA rather than 500 mA
Calculate the load resistance for a regulated 5.6 V output and a current of 510 mA
The power rating of the Zener diode is at its maximum when the load resistance is infinity, i.e. an open circuit. If there is zero current going through the transistor, the base current will be zero as well. So all the current goes through the Zener diode.
Show that the power dissipated by the Zener diode, if its current is 20 mA and its voltage is 5.6 V, is about 0.11 W
For this application, you would want a 200 mW Zener diode.
A more sophisticated circuit compares the output voltage with the reference voltage, and corrects the output voltage if there is any deviance. This helps to eliminate ripples and spikes from the supply. The concept is shown below in the block diagram:
As a circuit it looks like this:
This how it works:
TR2 acts as a voltage comparator, comparing the output from the potentiometer with the reference voltage from the Zener diode.
TR1 provides the current to the load.
If the voltage across the load falls, the transistor TR2 becomes less conductive, so the voltage at the base of TR1 rises.
This makes TR1 more conductive and increases the current to the load. The converse is true when the voltage across the load rises.
The system above uses negative feedback, which is a very common way of keeping electronic systems stable.
Voltage comparison can be carried out even more effectively using an integrated circuit called an operational amplifier.
AC to DC power supplies without transformers
Capacitor Power Supply
It is possible to make a power supply without a transformer using a circuit like this:
It is sometimes called a capacitor power supply.
The circuit works because the capacitor C1 has a reactance (or effective resistance). Remember that the resistance of the capacitor is infinite. The reactance is given by:
There is a large voltage drop across the capacitor, which must have a working voltage of 400 V. This means that the reactance is quite high.
Explain why a capacitor with a working voltage of 250 V would not be sufficient for this circuit.
Calculate the reactance of a capacitor of value 2.2 mF
The resistor R1 in parallel with C1 is there to discharge the capacitor when the supply is unplugged.
The rectifier bridge converts the low voltage AC to a rectified AC. Remember that the output of the rectifier bridge is has a peak voltage that is 2 × 0.7 V = 1.4 V lower than the peak voltage.
Capacitor C2 provides the smoothing. Remember that the voltage across the capacitor is the peak voltage.
R2 is the current limiting resistor for the zener diode.
ZD is the zener diode that clamps the voltage to 5.6 V.
R3 is the current limiting resistor for the LED.
Resistor Power Supplies
It is possible to drop the mains voltage across a resistor, but this is less efficient, as heat is lost across the resistor.
The circuit above needs to provide a current of 50 mA and the input voltage to the rectifier bridge is 10 VRMS .
(a) Calculate the value of the resistor needed.
(b) Calculate the power loss over the resistor.
Your answer to Question 14 should indicate that quite a lot of energy is lost across the resistor, which will get hot. Resistor power supplies are cheap to make, but are only useful where the output current is very low.
There are advantages to power supplies with no transformers:
They are lighter to carry (transformers are heavy lumps of metal);
They are cheaper (transformers are expensive);
They are smaller.
The disadvantages are:
These supplies can only provide a limited current;
Should the capacitor or the resistor fail, the full mains voltage will be at the output, which is highly dangerous.
Capacitor or resistor power supplies are most likely to be found in mobile phone chargers. Cheap versions of these are most likely to fail and can cause serious accidents.