If you are looking for 5V DC power supply for the digital circuit. But you have a 12V source, battery. I will show you, 12V to 5V converter step down regulator. In many ways for using, it depends on parts you have and other suitability.
Page Contents, Click
- 1 How to Choose a 5V converter
- 2 5V Zener diode Regulator—Lower 50mA
- 3 100mA 5V converter circuit
- 4 200mA, 5V Regulator
- 5 500mA, 5V regulator from 12V
- 6 How to convert 12VDC to 5VDC 1A
- 7 12V to 5V converter 1.5A output
- 8 12V battery to 5V 1.5A DC converter
- 9 Higher current transistor for 7805 Regulator
- 10 3A current output, 5V converter
- 11 Simple 5V overvoltage protection
- 12 5V 2A power supply using 78S05
- 13 Also 5V DC adapter
How to Choose a 5V converter
We should use a suitable circuit. How? The saving is the best. I use these guidelines.
- Save money—if I have it in my store it is so good. Also, save time to buy, not a long wait.
- Easy to build—simple and worked circuits are always good.
- Small size—some project has a Limited space.
Look at the load first!
Suppose that the load uses a current about 30mA. You should use the 5V converter at 60mA. It is enough for this case. When the current is small, it is small and easy to build. Also, save energy.
You should not use a large 1A current source circuit. It is like to ride an elephant to catch a grasshopper. Which it is wasteful and unnecessary.
For example circuits
- 3A current output—if you have the load that uses current more than 2A. For example, digital camera, GPS, Raspberry Pi, Arduino and more.
- Lower than 50mA—Small circuit for example, CMOS digital
- How to convert 12VDC to 5VDC 1A
- 12V to 5V 2A converter circuit
5V Zener diode Regulator—Lower 50mA
Some circuits draw the current from 20mA to 50mA (0.05A) only. You can a Zener diode voltage regulator circuit.
The Zener diode keeps a fixed voltage, 5V. It needs a resistor to limit the current to it and load.
How to calculate the device
To power it from a 12V source. You look at in the circuit again. There is three current.
- IZ = Maximum the Zener diode current
- IR = The Current through R1
- IL = Maximum the load current
IR is a constant at all time. Even IL will change from 0 mA to planned maximum value (50mA). IZ need to change to keep the voltage across the output is 5V.
First, use 5V Zener diode because we need 5V, VZ. Then, IR is about IL, 50mA.
R1 = (Vin – VZ)/ IR
= (12V – 5V)/ 50mA
= 140 ohms
or about 150 ohms.
PR is Power of R1.
PR = VR x IR
= 7V x 50mA
= 0.35 watts or use 0.5 watts.
But we forget, a power of the Zener diode, PZ
PZ = VZ x IZ
Note: IZ is about IR, 50mA.
PZ = 5V x 50mA
PZ = 0.25 watts
So we use 5V 0.5W Zener diode.
Also, C1 is a filter capacitor to smooth DC voltage.
100mA 5V converter circuit
In the digital circuits that have many parts. They may use the current more than 100mA but lower than 300mA.
We can use many circuits. In the previous circuit, it has a low current. If you want 100mA. You need use a low resistance (R1) and higher watts of the Zener.
It is a better idea. If you add a transistor in the circuit. It will increase a higher current more. But the output voltage is 4.4V only. Because of some voltage drop across BE of transistor Q1, 0.6V.
Since a transistor is good to increase the current. So, we can change R1 to be 1K as circuit below. To reduce a bias current Zener diode and the Base of Q1.
200mA, 5V Regulator
5V transistor series voltage regulator
If you use 2N2222 instead of BC548. It can use 200mA @ load. Because 2N2222 has current collector (Ic) about 0.8A in a datasheet. But in a real using it can use max 0.5A only.
500mA, 5V regulator from 12V
If you need to use with 300mA to 500mA load. You should change the transistor is BD139. It has Ic about 2A max in spec. But I can get it about 0.5A only. While it is working. It may be warm. So often perform better with a heat sink.
C1, C2 capacitors are used to reduce the ripple output. And C3 will reduce a spike voltage.
How to convert 12VDC to 5VDC 1A
Many friends want to convert 12VDC to 5VDC at 1A. It is the popular rate in the most circuits.
I have two choices to choose from. It depends on The suitability of your parts and time.
First, 5V 1A transistor Regulator. It is similar to the above circuits.
I use TIP41 power transistor. Because it can get 4A max in spec. But in real using it can give me about 2A max only. Also, its body is TO-220 so easy to use with any size heatsink.
Before I like this circuit. If I have all the components in my store. I will make it first.
But lately, I like to use this component, 7805 Regulator.
Second, 7805 Regulator popular.
This is so easy, faster than other. Because its body is the same as TIP41, without Zener diode and a bias resistor.
12V to 5V converter 1A using 7805
Also, it has low ripple output about 10mV, with the electrolytic capacitors(C1, C4) on the input and output. And both filter capacitor, C2, C3 to reduce the spike voltage.
Note: 7805 pinout
Since it is the linear regulator. So while it is working. There is the voltage across input and output of IC1 about 7V. When full load has a current of 1A. So the output power is about 7 watts. It is hot. We should mount it on enough heatsink.
12V to 5V converter 1.5A output
Sometimes we need an output current about 1.5A. We have 3 ways to do.
- Connecting 7805 in parallel
- 12V battery to 5V 1.5A DC converter
- Higher current transistor for 7805 Regulator
- 2A transistor Regulator
Connecting 7805 in parallel
If we connect 7805 in parallel. It makes higher current more. This is suitable for who carry up or do not have any power transistors. But it is not good for a long time. You can try it!
Both IC-7805 should be exactly the same.
12V battery to 5V 1.5A DC converter
If we need to use 12V to 5V voltage regulator. This is the 5V 1500mA DC regulator circuit. Which is a simple circuit using IC-7805, to the fixed regulator 5 volts and TIP41-NPN power transistor to increase current up to 2A.
I use the 7805 power supply with the 12V battery. To reduce the constant voltage of 5 volts.
I try to use the load is 4.7 ohms 5W resistors. As principles, it will use the current about 5V/4.7ohms = 1A. I measure the current of about 0.7A and the voltage drop is 4.9V, but it is still usable. As Figure 1
It requires a transistor to increase the output current up.
I use the TIP41 transistor. In principle, it can supply the current about 2A. Which it is enough to use.
In the circuit diagram.
Then, I test circuit with about the load, 2.4 ohms resistor. Next, measure current approximately 1.3A, and voltage drop is 4.9V. It can be used as we want.
I put a Diode-1N4007 voltage to offset the loss of the transistor between pin BE.
We insert LED1 to indicated power on of this circuit and a series resistor-R1 is used to limits current to a safe value.
C1, C3 is filtered capacitors to smooth dc current input and output sequence.
C2, C4 is noise filter spark current.
While Q1 is working it will very hot so we must mount it with a large heatsink.
Note: It has cons. If it short circuit. IC-7805 may be damaged.
Higher current transistor for 7805 Regulator
If you want the current more than 1A using 7805 in better than 2 circuits above.
It needs helping from a power PNP transistor, with circuit below.
The high current will flow through the power transistor Q1, TIP42. While 7805 gets lower current. Because R1 reduces this current down.
So 7805 keeps a fixed regulated voltage, 5V only. It works well without the heatsink.
While Q1 is working. It is so hot. We need to mount it with enough heatsink.
If you have the ready parts. You can use this circuit for a long time.
Then, if you want 3A current. It is easy just you use MJ2955 instead of TIP42.
Although this circuit can be used well. But it still has disadvantages.
When a short circuit, the power transistor may be damaged.
Look at below.
3A current output, 5V converter
This is 12v to 5v converter step down Regulator at 3A load.
The digital camera can take pictures and videos as well. But it has a disadvantage is not long the battery runs out. When using outdoor. We needed to often recharge the battery. It is quite a waste of time.
If buying additional spare batteries. It is expensive and still often to change as the same.
Its side has a slot for plugging DC 5V adapter, 2A current. If we modify 12V lead-acid battery to reduce a voltage down to 5 volts. It is so good idea.
Because this battery is cheaper and long time using. For example, 12V 10Ah battery you can take the camera for 5 hours.
How it works
We have many ways to do. But I will show you this circuit below. I like a linear circuit than a switching mode circuit.
There are many components in the circuit. As above this circuit can power up to 3A current with an increasing current of Q3-MJ2955. Also, it has many parts interesting.
When the load is overload or short circuit. Then, a voltage is across R2 about 0.6V. So, Q2 gets a bias voltage, it works. After that VBE of Q3 is low, Q3 works lower until stop.
While Q1 works to connect current through LED1. It indicates now overload.
Component list of 12v to 5v voltage regulator
IC1: LM7805, 5V dc regulator IC
Q1: BC558, 0.4A 40V transistor
Q2: BD140, 1.5A 30V PNP transistor
Q3: MJ2955 or TIP2955, 4A 50V PNP power transistor
C1: 4,700uF 25V, Electrolytic
LED1: LED any color as you like
R1: 330 ohms 0.25W
R2: 0.22 ohms 5W
R3: 470 ohms 0.5W
R4: 47 ohms 1W
R5: 18 ohms 1W
Heatsink, wires and more etc.
I have old GPS, usually, I use it in my car. We need to have DC to DC converter circuit that can reduce voltage 12V to 5V at current more than 2A.
Which circuit diagram can make this. I like it that need to buy some parts since I have also in my stores.
As Figure 2 I assemble them on universal board
Simple 5V overvoltage protection
In normally you can use the above circuit. Because it is simple and inexpensive.
You just add Fuse-F1 to protect overload more than 2A. Also, if the circuit powers high voltage more than 5.1V. It has too many currents through ZD1 and D1 as overcurrent. So the fuse will burn suddenly.
5V 2A power supply using 78S05
Another way, My friend wants a 5V 2A power supply circuit. In model to be simple, use little equipment, build easy. Then, I choose this circuit for him. Why is it? It uses pillar equipment, a positive voltage regulator 5V / 2A in TO220, 78S05. And few parts see in a circuit, is high-quality and low noise.
The circuit will work without the extra components, but for reverse polarity protection, a 1N5402 diode is provided at the input, extra smoothing being provided by C1-220uF 50V. The output stage includes C2-47uF 25V for extra filtering.
Also 5V DC adapter
- Microprocessor dc regulator supply 5V 3A by LM323K
- 5V 3A switching power supply by LM2576
- LM2673 -5V 3A Switching Voltage Regulator
- Top Linear power supply regulator 5V 5A with 7812 and LM723
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