How Catu Daya Linear works
Monday, May 24, 2010 by sms
Electronic devices should be powered by direct current supply DC (direct current) are stable to well. Battery or batteries are a source of DC power supply is best. But for applications that require larger power supply, the source of the batteries is not enough. A major source of power supply is the source of alternating AC (alternating current) from power plants. For that we need a power supply device that can convert AC into DC current. In this article presented the principles of the power supply circuit (power supply) linear ranging from the simplest rectifier circuit to the power supply was regulation.
Rectifier (rectifier)
Principle rectifier (rectifier) The simplest is shown in Figure-1 below. Transformer is needed to lower the AC voltage from the grid at the primary coil becomes smaller AC voltage on the secondary coil.
image 1: a simple rectifier circuit
In this circuit, diodes just continue to contribute to a positive voltage to the load RL. Which is called a half-wave rectifier (half wave). To obtain full-wave rectifier (full wave) is required transformer with center tap (CT) as in the picture-2.
image 2: full-wave rectifier circuit
Positive voltage phase of the first forwarded by D1, while the next phase which is passed through D2 to the load R1 with the transformer CT as a common ground .. Thus R1 burden gets a full wave voltage supply such as the picture above. For some applications such as for supply to small dc motors or dc incandescent lamp, the shape of this voltage is sufficient. Although the voltage ripple seen here from both series are still very large.
Image 3: half-wave rectifier circuit filter C Dengah
Figure 3 is a series of half-wave rectifier with filter capacitor C is parallel to the load R. Apparently with this filter discharge voltage waveform can be blended. Figure-4 shows the DC voltage output from circuit half-wave rectifier with capacitor filter. Bc line is approximately straight line with a certain slope, which in this state for the load current supplied by voltage capacitor R1. Bc line is not actually a straight line but exponential in accordance with the nature of the discharge capacitor.
image 4: filter capacitor waveform with
Bc curve slope depends on the large current I flowing to the load R. If the current I = 0 (no load) then the curve will form a horizontal line bc. But if the load the greater the flow, the slope of the curve will become sharper bc. The voltage that comes out will be shaped with a sawtooth ripple voltage of the magnitude is:
Vr = VM-VL ... ....... (1)
and dc voltage to the load is VDC = VM Vr / 2 ..... (2)
Good rectifier circuit is a circuit which has the smallest ripple voltage. VL is a voltage discharge or discharge the capacitor C, so it can be written:
VL = VM e -T/RC .......... (3)
If equation (3) disubsitusi to formula (1), it is obtained:
Vr = VM (1 - e -T/RC) ...... (4)
If T <
so if this disubsitusi to the formula (4) can be obtained by a simpler equation:
Vr = VM (T / RC) .... (6)
VM / R load is none other than I, so that with this visible relationship between the load current I and the value of capacitor C to the ripple voltage Vr. This calculation is effective to obtain the desired ripple tengangan value.
Vr = I T / C ... (7)
This formula is to say, if I load current was higher, the greater the voltage ripple. Conversely, if the capacitance C was higher, the voltage ripple will be smaller. For simplification is usually considered T = Tp, ie a period of one sine wave of the grid frequency of 50Hz or 60Hz. If the grid frequency of 50Hz, then T = Tp = 1 / f = 1 / 50 = 0.02 sec. This applies to half-wave rectifier. For full-wave rectifier, of course, the tuner frequency doubled, so that T = 1 / 2 Tp = 0.01 sec.
Full-wave rectifier with C filter can be created by adding a capacitor in the circuit of figure 2. Can also use a transformer without CT, but by assembling four diodes as shown below-5.
Image 5: full-wave rectifier circuit with filters C
For example, you designed the full-wave rectifier circuit of the power supply 220V/50Hz grid to supply the load of 0.5 A. What is the required capacitor value so that this circuit has a voltage ripple of no more than 0.75 Vpp. If formula (7) be inverted so obtained.
C = I.T / Vr = (0.5) (0.01) / 0.75 = 6600 UF.
For this size capacitor elco widely available type which has a maximum working voltage polarity and specific. Working voltage capacitors used must be greater than the power supply output voltage. Barangkalai now you understand why you create a series of audio hum, try to check the power supply rectifier circuit that you created, if the voltage ripple is quite disturbing. If not available in the market of such a large capacitor, can certainly memparalel two or three capacitors.
Regulator
The rectifier is good enough if its a small ripple voltage, but there are stability issues. If the voltage of electricity rise / fall, then the output voltage will also rise / fall. Like the rectifier circuit above, if the current higher voltage dc discharge was also decreased. For some applications this voltage changes quite annoying, so it requires an active component that can regulate the output voltage becomes stable.
The most simple regulator circuit shown in Figure 6. In these circuits, zener works on regional breakdown, resulting in an output voltage equal to or Fout Zener voltage = VZ. However, this circuit is only useful if the load current not exceeding 50mA.
6 pictures: Regulators Zener
The principle of such a power supply circuit is called a shunt regulator, one of his trademark is a regulatory component in parallel with the load. Another feature of the shunt regulator is susceptible to short-circuit. Notice when Fout connected short (short-circuit) then the current is fixed I = Vin/R1. In addition to a shunt regulator, there is also a series called by the regulator. The main principle of such series regulator circuit in Figure 7 below. In this circuit output voltage is:
Fout = VZ VBE ........... (8)
VBE is the base-emitter voltage of transistor Q1 of the magnitude of 0.2 - 0.7 volts depending on the type of transistor used. By ignoring the IB currents flowing at the base of the transistor, to determine the magnitude of R2 resistance that is needed is:
R2 = (Vin - VZ) / Iz .........( 9)
Iz is the minimum flow required by the zener diode to achieve the zener breakdown voltage. These large currents can be detected from a datasheet which amount to approximately 20 mA.
Image 7: regulator zener follower
If the supply current required is greater, of course the base of IB in the current calculation circuit above can not be ignored anymore. Where such a known, large currents will be proportional to IC flows dirumskan with IB or IC = BIB. For such purpose, the transistor Q1 is used can be replaced with Darlington tansistor which typically have large b values. With a Darlington transistor, the current small base that could produce a greater flow of IC.
Techniques that better regulation is to use the Op-Amp to drive the transistor Q, as in a series of eight images. Zener diodes are not here to give feedback directly to the transistor Q, but as a reference voltage for the op-amp IC1. Negative feedback on the op-amp pins are excerpts from the voltage regulator out, namely:
Vin (-) = (R2 / (R1 R2)) ....... Fout (10)
If the stress out Fout ascending, then the voltage Vin (-) will also be rising until the voltage is equal to the reference voltage VZ. And vice versa if the voltage decreases Fout out, for example because of supply current to the load increases, op-amp will keep the stability of the reference point by giving the current VZ IB to the transistor Q1. So that at any time maintaining the stability of op-amp:
Vin (-) = VZ ......... (11)
Figure 8: regulator with op-amp
By ignoring the VBE voltage of transistor Q1 and mensubsitusi formula (11) into formula (10) then obtained a mathematical relationship:
Fout = ((R1 R2) / R2) VZ ........... (12)
In this circuit the output voltage can be regulated by regulating a large R1 and R2.
Now it should no longer need to painstakingly search for the op-amps, transistors and other components to realize a series regulator as above. Because this kind of circuit has been packaged into a single fixed voltage regulator IC. Are now widely known as the 78XX series component of the voltage regulator remains positive and 79XX series which is a voltage regulator to remain negative. In fact, these components are usually already equipped with current limiting (current limiter) and also limiting the temperature (thermal shutdown). This component is only three pins and by adding some components alone can be a power supply circuit was good regulation.
Image 9: regulator with IC 78XX / 79XX
For example 7805 is a voltage regulator to get a 5 volt, 12 volt voltage regulator 7812 and beyond. While such 79XX series is the 7905 and 7912 which are respectively the negative voltage regulator 5 and 12 volts.
Apart from the fixed voltage regulators have a voltage regulator IC also can be arranged. The principle is the same as OP-amp regulator packaged in a single IC for regulators such as LM317 LM337 variable positive and negative variables for the regulator. The difference between resistors R1 and R2 are outside the IC, so that the output voltage can be adjusted via external resistors.
It's just important to know that with the IC regulator circuit can work, tengangan input must be greater than the output voltage regulatornya. Usually the difference in voltage Vin to Fout recommended in the datasheet component. Use heatshink (Aluminum cooling) is recommended if these components to be used to supply a large current. In the datasheet, these components can pass through the flow reaches a maximum of 1 A.
- End -
2 2 Data 0 Out Data 
