Power Source, transformer, rectifier, filter, regulator
Many circuits needed for operation, a power direct current (DC) , but is usually found power alternating current (AC) .
In order to get current, the AC input must follow a process of conversion as shown in the diagram.
The following chart is the operation of a source, using a block diagram.
also shows the expected waveforms at baseline (AC Input), end (DC output) and between each of them.
- The input signal, which is the primary of transformer , is a sine wave whose amplitude depends on where you live (110/220 Volt AC or other).
- The transformer secondary delivery to your signal with an amplitude less than the input signal and it must have a value that is in keeping with the voltage (voltage) final DC power is desired.
For example if you want a final voltage of 12 volts direct current, the transformer secondary must have an AC voltage of no less than 9 volts, this value being very tight (remember that the secondary peak is : Vp = 1.41 x Vrms = 1.41 x 9 = 12.69 V).
If one takes into account the voltage drops at different stages (blocks) of the power supply may no longer be able to obtain the desired 12 volts.
In this case, choose a transformer with a secondary voltage of 12 volts AC. This AC voltage to obtain a voltage Peak: Vp = 1.41 x 12 = 16.92 volts.
- The rectifier converts the signal prior to a pulsating DC waveform, and in the case of the diagram, using a rectifier 1 / 2 wave (eliminates the negative wave .)
- The filter, consisting of one or more condensers (capacitors) , smooths and flattens the previous wave by eliminating the component of alternating current (AC) gave the rectifier. The capacitors are charged to the maximum value of voltage delivered by the rectifier and slowly discharge when the pulsed signal disappears.
See the diagram above and process of discharging a capacitor
- The controller receives the signal from the filter and delivers a constant voltage regardless of variations in load or supply voltage .
- transformers are used to lower or raise AC voltages.
- Rectifiers are made up of diodes and used the process of transforming an AC signal to DC, allowing the passage or not semicycles AC wave.
- Filters can be of various types and are used to remove unwanted components of CA .- The regulators are a group of elements or an electronic element
Regulator zener diode voltage
Features, design example
The Zener can be used to regulate a voltage source . This semiconductor is manufactured in a wide variety of tensions and powers
These range from less than 2 volts to several hundred volts, and power will be dissipated from 0.25 watts (W) to 50 watts (W) or more.
power dissipated by zener diode is simply multiplies the voltage for which it was manufactured by current flowing through it.
x Vz Pz = Iz
This means that the maximum current that can pass through a zener diode is:
Iz = Pz / Vz.
Where: - Iz = current through the Zener-Pz = power zener diode (manufacturer) - Vz = Zener diode voltage (manufacturer)
Example: The current maximum zener diode 10 Volt and 50 Watts (W) may hold is: Iz = Pz / Vz = 50/10 = 5 amps
Calculation limiting resistor Rs. (See diagram of the regulator with zener diode)
The calculation of the resistance Rs is determined by the load current request (which we to connect to this source.)
This resistance (resistor) can be calculated using the following formula:
Rs = [Venmin - Vz] / 1.1 x ILMAX
where: - Come (min) is the minimum input voltage. (Remember that a voltage is regulated and may vary) - IL (max) is the value of the maximum load current request.
Once earned Rs, we obtain the maximum power zener diode, using the following formula:
PD = [[Venmin - Vz] / Rs - ILmin] x Vz
Example of a Design:
A 15 volt source to feed a load with 9 Volt, which consumes a current that varies between 200 and 350 mA. (Milliamps). Choosing a 9.1-volt zener diode as there is no 9.
- Calculation of Rs: Rs = (15 - 9.1) / (1.1 x 0.35) = 15 ohms (ohms)
- Calculation of power zener diode, PD = [(15 - 9.1) / 15] x 9.1 = 3.58 watts or watts.
Since there is no zener diode of 3.58 Watts, and choose one of 5 watts which is the closest
- Rs Power: An additional calculation is the power of resistance Rs. This is done with the formula: P = I2 x R.
Current data are: I (max) = 350 milliamps = 0.35 amp = Rs 15 ohms (ohms) using the formula, PRS = 0,352 x 15 = 1.84 Watts (Watt) This means that when you buy this resistance (resistor) must be 2 Watts or more
