Here is a 7805 Adjustable voltage regulator circuit. I used to recommend many projects that use IC-7805 or 78xx types. Today, it will modify from the old circuit, IC-78xx series has many numbers. We can choose the desired voltage. such as “12V” should use IC-7812, but sometimes we don’t have this number for any reason.
But we have too many IC-7805 numbers since it is applied often in digital circuits. And with the reason above, so it is cheaper than others and easy to look for in a store.
Therefore, I think to apply the IC-7805 to adjust various voltage level, such as 7.5V, 8.4V, and 10V, etc. As application and low budget.
Finally, it is this adjustable dc voltage regulator that can supply output of 5V to 22V at 1A
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Adjust 7805 voltage Regulator
With a habit of playing the circuit and set up suspected in various circuits.
In this case, too:
1. Firstly, Assemble components as the circuit in Figure 1, a negative lead, then use a digital multimeter measure voltage at various point.
Appear voltage at the output and negative pin are about 16.6V And also measure current.
Release the pin negative and measure voltage at various point
2. Then I take 1K-resistor joint between negative-ground pin as Figure 2 and next measure at the output of 9.5V. Next measure across R1 has a voltage of 4.5V. After that measure the output-negative lead of IC1 same 5-volts that a steady voltage at all time.
It means that we begin to see the way. Thus, replace R1 with a 5K-potentiometer. And then rotate it, can measure the output of 9.5 volts, cause voltage output during 5V to 14.8V, show that IF put the resistor drop across between the negative and ground lead it can change output voltage and measure voltage the output lead compare with negative lead of IC1 of 5V at all time.
A formula of 7805 variable Regulator
I always like to calculate the various devices. For ease of use. When see in Figure 2 will see that the current flow through R1-1K will cause a voltage drop across it around 4.5-volts. Following sequence, we came to the current flowing through it together.
The simplest and most obvious way is measured by an ammeter. Which I measure current about 4.6mA Thus try instead value below:
R1 = VR1/IR1
VR1 = 4.5V , IR1 = 4.6mA
Results : R1 = 4.5V / 4.6mA = 0.98K omhs
The value was close to R1. But we still inconclusive. I replaced R1 with the potentiometer again. Then adjust voltage since 6V,7.5V,8V,9V,10V and 12V etc. I found that all the voltage value have one current value flow the potentiometer same together is 4.6mA at all time.
Then we try to calculate the resistance again, We want to supply voltages up to 12V, we would have thought from the formula :
By: VR1 = Vout – Vic
Set to: Vout = 12V, Vic = 5V
Instead value :VR1 = 12V – 5V = 7V
Then replacement in (1) the formula :
R1 = 7V / 4.6mA = 1.5mA
Thus can find the resistance 1.5K replace on R1, the output appears voltage of 11.9V nearly well.
Conclusion is my the formula is R1 = (Vout – Vic) / IR1.
Real 7805 adjustable voltage regulator circuit
Next, we see a real use for me. It can adjust the supply voltage 5V to 22V and under 1A. Which is enough for normal use of us. As Figure 3 as the full circuit in the application.
The concept for the device in the circuit
This 7805 variable voltage regulator circuit is nothing special, Similar circuit ago, So it is recommended only priority is device specific:
1. T1 is a transformer to provide to the entire circuit, From the conditions of the circuit, we should select the transformer and 1A is sufficient and can use more than this, such as 2A. But it will be larger and more expensive.
Section voltage ratings required to calculate it.
Determined from (Vout/1.4) +3 V, in when Vout = 22V.
Try to replace the value (22V/1.4) +3 V = 18.7V. I chose 18V Size, may be used transformers 9V CT 9V (don’t use CT) or Use an 18V CT 18V (use only, 18V and CT lead, but 18V lead the rest not used).
The origin of the …
-1.4 is a constant number. from voltage between DC and AC is the relationship between the VDC = 1.4 x VAC.
-3V is the voltage value lost in the IC.
2. C5 is capacitor act as voltage filter. I calculated about 1,515 uF, but this value without sell. Thus use 2,200 uF 35V instead.
3. R1 is a resistor to reduce current to LED1. I calculated about 3.12 K and chose value 3.3k.
4. C7 is the capacitor act as noise filter from between of the variable resistor-VR1. I select the value of 0.01uF 100V.
5. D6 is Diode acts as Prevent reverse flow of electricity to the power circuit, may cause damaged circuit, as mentioned above. In this circuit, I chose the diode type 1N4007 diode.
Build 7805 adjustable voltage regulator
First order for the device to complete first. Details are as follows:
1. IC1__LM7805_5V dc fixed voltage regulator_or equivalent___= 1 pcs.
2. D1-D6____1N4007__1000V 1A Diodes ______= 6 pcs.
3. LED1__LED as you need________= 1 pcs.
4. C1-C4,C6,C7,C8___0.01uF 50V___Mylar or Ceramic capacitors______ = 6 pcs.
5. C5____2,200uF 35V___Electrolytic capacitor_________________= 1 pcs.
6. C8____100uF 35V____Electrolytic capacitor_________= 1 pcs.
7. R1____3.3K 0.25W +5%_______= 1 pcs.
8. VR1___5K Single Potentiometer type B______= 1 pcs.
9. F1___0.5A Fuse____________= 1 pcs.
10. F2___1 A Fuse_____________= 1 pcs.
11. T1___Transformer__18V 1A___= 1 pcs.
Look at 7805 pinout
Further, makes the PCB layout as Figure 4 or use the universal PCB board. Because of a few components and easy to builds. Then place components layout as Figure 5 and wiring to the parts successfully. Check error to make good again, before applying the circuit.
What I want to warn, to be careful about polarity electrolytic capacitor and diodes, should connect the voltage measurement lead, before the input voltage is fed, and rotate potentiometer-VR1 lowest before will have the voltage of 5 volts, then adjust voltage rises.
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