Hello guys. If you want to learn electronics and want to make electronics project then first thing for doing the experiment of electronics you need is the power supply. You will make different projects so different projects require different voltage power supply. So you need a variable power supply for this purpose lets build a variable power supply. In this tutorial, I will guide you on how to make variable power supply.
So let’s start….
Components Detail,
1.Transformer,
We require 1 transformer whose rating is 220/18V ac and the curretn rating is 2A.
2.Bridge rectifiers,
For converting ac voltage to dc voltage we require rectifier IN4001 quantity.
Vp = Vrms /.707 => Vp =18 / .707 = 25V
3. Filtering capacitor,
we will use C1 that is filtering capacitor 3300uf and 35V rating.
4.Voltage regulatore
We require a voltage regulator LM317. It is a variable positive voltage regulator capable of supplying more than 1.5A with voltage rating 1.25 V to 37V.It also comes with internal current limiting, thermal shutdown, and safe area compensation. All this makes it a very good candidate for a regulator if we need a moderately accurate supply with medium power output. For more details, you may refer to its data sheet. As you can see it has three pins,
• INPUT – This is where we give the unregulated input
• OUTPUT – This is where we will get the regulated output
• ADJUST – The variable resistor connected to this pin, controls the output voltage
Now using the datasheet we can conclude the value of the resistor using this formula.
Vout = 1.25 *( 1+R2/R1) + I adjust * R2
Vout = output voltage
R1 and R2 = resistor value
Iadjust = current through adjust pin of the regulator
The important point to remember,
1…Adjust current must be 50 to 100uA
2…the value of R1 should be fairly small less than 500 ohm
C2 and C4 are also filtering capacitors. These are enough components for power supply to work but to improve efficiency and work performance increase we need some more component.
Diode D1 is used to protect the voltage regulator by the excess current if a voltage source or a battery is connected at the output terminal of power supply. Diode D2 also used for this purpose.
It is also clear from the data sheet that the worst case scenario dropout voltage for the LM317 is almost 2.3 V. So, to be in the safe side it is advisable to select a transformer with at least 4V greater than the required output voltage ( 2.3V of 317 + 1.4V of bridge rectifier)
For more detail about LM317 please see its datasheet.
So let’s start….
Components Detail,
1.Transformer,
We require 1 transformer whose rating is 220/18V ac and the curretn rating is 2A.
2.Bridge rectifiers,
For converting ac voltage to dc voltage we require rectifier IN4001 quantity.
Vp = Vrms /.707 => Vp =18 / .707 = 25V
3. Filtering capacitor,
we will use C1 that is filtering capacitor 3300uf and 35V rating.
4.Voltage regulatore
We require a voltage regulator LM317. It is a variable positive voltage regulator capable of supplying more than 1.5A with voltage rating 1.25 V to 37V.It also comes with internal current limiting, thermal shutdown, and safe area compensation. All this makes it a very good candidate for a regulator if we need a moderately accurate supply with medium power output. For more details, you may refer to its data sheet. As you can see it has three pins,
• INPUT – This is where we give the unregulated input
• OUTPUT – This is where we will get the regulated output
• ADJUST – The variable resistor connected to this pin, controls the output voltage
Now using the datasheet we can conclude the value of the resistor using this formula.
Vout = 1.25 *( 1+R2/R1) + I adjust * R2
Vout = output voltage
R1 and R2 = resistor value
Iadjust = current through adjust pin of the regulator
The important point to remember,
1…Adjust current must be 50 to 100uA
2…the value of R1 should be fairly small less than 500 ohm
C2 and C4 are also filtering capacitors. These are enough components for power supply to work but to improve efficiency and work performance increase we need some more component.
Diode D1 is used to protect the voltage regulator by the excess current if a voltage source or a battery is connected at the output terminal of power supply. Diode D2 also used for this purpose.
It is also clear from the data sheet that the worst case scenario dropout voltage for the LM317 is almost 2.3 V. So, to be in the safe side it is advisable to select a transformer with at least 4V greater than the required output voltage ( 2.3V of 317 + 1.4V of bridge rectifier)
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