# Create a DC Power Supply

57,318

49

5

Published

## Introduction: Create a DC Power Supply

How to make a DC power supply using the AC wall voltage!

## Step 1: Build a Basic Full-wave Bridge Rectifier

1. A transformer with a turn ratio of about 6. Connect your transformer to a power strip with an on and off switch for safety.

The voltage from the wall is generally about 120 volts in the United States. This voltage it too large for our ¼ watt resistor to handle, so we will use a transformer to decrease the amplitude of the AC voltage in. We pick a transformer that has an appropriate turn ratio. For transformers, the turn ratio Nprimary /Nsecondary = Vprimary/Vsecondary.  Vprimary = 120V and we are shooting for Vsecondary to be about 20V, so our transformer should have a turn ratio of about 6.

2. Four 1N4007 diodes or a bridge rectifier.

Our circuit diagram shows 4 diodes. However, these can also be replaced by a bridge rectifier like the one we used in our circuit in the lab.
How to hook up your bridge rectifier: Connect the legs with the squiggles to the outputs of the transformer, the minus side to ground, and the plus side to the rest of the circuit.

3. A 1/4 Watt 1kΩ resistor.

Important: When building your circuit, make sure that you have the polarities of the diodes correct. ALWAYS TURN YOUR POWER STRIP OFF BEFORE MODIFYING YOUR CIRCUIT.

Your graph of the output voltage should appear to be the absolute value of a sine wave with an amplitude of about 20 volts. This circuit eliminates the negative portions of the voltage. However, the voltage retains a lot of variation. We can add other components to make the voltage less variable.

## Step 2: Add a Low-pass Filter

Add a capacitor to your circuit to reduce some variation in the output voltage.

1. Take the previous circuit and add a 100µF capacitor in parallel with Rload.

The capacitor creates a low-pass filter which filters out the lower voltages.

Important: Make sure you get the polarity of your capacitor correct before you turn on your power strip. The negative side should be connected to ground.

Lower voltages are filtered out in the output voltage. However, this circuit retains variation in V_out.

Fun Fact! The ripple in the voltage can be calculated if you have the average voltage.
∆V = ∆t*I / C = (∆t*Vavg) / (R*C)

You can further improve this DC power supply by adding a zener diode or a 7805 Terminal Regulator.

## Step 3: Add a Zener Diode

1. Add a zener diode and a resistor to your circuit to create a 5V power source and stabilize the voltage even more.

Add a 5.1 V zener diode (we used a 1N4733A) in series with a resistor, R_zener (we used a 68 Ω and 680 Ω resistor in series). These two components should be added in parallel with the 100µF capacitor.

You should pick an RZener that accounts for the worst-case scenario, which is when the current splits evenly between the load and the zener diode. For a load current of 10 mA, the worse-case scenario is 20 mA through the zener diode.
RZener = V*I = 15V*0.02A = 750
Ω at most.
RZener should be a 1/4 W resistor of not more that 750 Ω. We used 68 Ω and 680 Ω resistors in series to obtain an equivalent resistance of 748 Ω (remember that resistances simply add in series).

Important: Make sure that you have the polarity of the zener diode correct before turning on your power strip. The current flows opposite the direction that the zener diode points.

The resultant voltage out should have an average of about 5V and a small ripple.

2. An even better alternative to the circuit with the zener diode is using a 7805 Terminal Regulator in place of RZener and the zener diode.

Remove your RZener and zener diode. Hook up your 7805 terminal regulator in parallel with your capacitor.

How to hook up your 7805:
From left to right, the 3 terminals are your INPUT, COMMON, and OUTPUT.
INPUT should be hooked to the capacitor. COMMON should hook to the ground. And OUTPUT should be your V_out that is measured by your oscilloscope.

You should observe a flat signal of about 5V with no ripples and only some noise.
Congratulations! You've created your 5V DC power supply from the AC wall voltage!

## Recommendations

• ### Large Motors Class

8,821 Enrolled

• ### Clocks Contest

We have a be nice policy.

## Questions

Hi, I'm quite familiar with schematics, used them for years since NAVY days years back. What I would like is to build a 1.2 vdc @ 600 milliamps power supply for LED-Christmas Lights outdoors. I'm quite familiar LED's are fineky on voltage. Since the voltage above is for 1-set of 30 led stars 3" diameter & I have 3 sets of. Pwr supply would power all 3 sets, should go to at least 2.5amps. Would pushing voltage to 1.3 or1.5 be ok? So, how would I change the 5vdc pwr. supply on your site? Thank you

what are the calculations for 120v ac to 9v dc (0.5amp) power supply.

Wow! I remember using that exact breadboard back in highschool!

Memorieeeeeees.... pressed between the pages of my mind...

Usually I see an output filter capacitor used with fixed 3 terminal voltage regulator ICs. If you include one of those it may increase the stability of your current. I also usually use a much larger input filter capacitor too. I mean just look at the size of the capacitor I used in this power supply:

https://www.instructables.com/files/orig/F25/WU6P/GSUSNEZL/F25WU6PGSUSNEZL.jpg

OK perhaps that is a bit of an extreme example. Even in this power supply I used larger capacitors though. I also run off both sides of the bridge network too to have a dual positive negative supply:

https://www.instructables.com/id/Dual-POS-NEG-Power-Supply/

On the negative side of the bridge positive gets connected to the ground because ground is higher potential.

You don't necessarily need the charger or hub. The amount of current applied to the load (whatever you are charging) is determined by the load itself. An iPhone, for instance, has a charging circuit inside that limits current to 1A. An iPad limits its charging current to 2A. The Amp rating for a given voltage on the PSU is the maximum that the PSU can supply, not a constant rating.

Remember that Current is proportional to Voltage and inversely proportional to Reisistance, so

I = E / R

Since your E is a constant (regulated) 5 volts, the Resistance, or your load, will be the one that affects the amount of current.