# Changing the Output Voltage of a Cheap Power Supply

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This instructable show how to change parts inside a small power supply to chnage the output voltage to suite your needs.
For DIY project I needed a stabilized voltage of exactly 7V dc and about 100 mA. Looking around my parts collection I found a small dc power supply from an old mobile phone that was unused.
The power supply had written 5,2V and 150mA written on it. That looked fine only the voltage needed to be pushed up a little bit until it was 7V.

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## Step 1: Reverse Engineering

BE CAREFUL! THE PARTS MAY STILL CONTAIN HIGH VOLTAGES IF TEARING APART SHORT AFTER USAGE!

It was easy to tear the power supply a part. It only had one screw that kept the case together.
After opening the case a small circuit board fell out ... containing just a few parts.
It is a simple switching power supply. The stabilization of the output voltage is done using a TL431. This is a shunt regulator with a refence voltage and an input pin to adjust the output voltage. The data sheet of this device can be found on the internet. I located the resistors that are responsible to set the output voltage. They are named R10 and R14 on the pcb. I took the values of them and put them in the calculation formula that is written in the data sheet.
Vo=Vref*(1+R10/R14) . Using R10=5.1kOhm and R14=4.7kOhm the Result is exactly 5.2V as it is written on the power supply.

## Step 2: Calculating New Parts and Modifying the Device

I wanted to keep the sum of R10 and R14 about the same as it was in the original circuit. That is round about 10kOhm. To get a higher output value I needed to modify the resistors according to the data sheet. I also needed to replace the protecting zener diode.
For the protective zener I choosed a 10V type because I found it in my parts collection. This voltage protects the output capacitor.
Calculating the new resistor values I started with R10 using the formula of the TL431 data sheet and kept the 10kOhm in mind. The calculated resistor would be 6.5kOhm. That is not a resistor value that is common. I selected a near value of 6.8kOhm.
Now I calculated the value of R14 using the choosen value for R10. The calculation leads to a value of 3.777kOhm for R14. I choosed a value of 3.3kOhm and added a 500Ohm trimmer potentiometer.
Because of the tolerance of the circuits it seems to be a good idea to insert a trimmer to adjust the output voltage.
After removing the original parts from the soldering side of the pcb I added the new parts on the components side because I did not use smd parts.

## Step 3: Results

The voltage meter shows exactly 7V (ok .. it's 7.02V). That's what I wanted :-)
Now I can use the power supply for my beetle bot project ... coming soon ...

## 35 Discussions

Sorry to say but I tried the same with my Philips 9V/1.56A adapter, but the output though was constant 5V, not able to charge my smartphone with 3000mAH battery, where as an old samsung adapter with 5V/0.7A was able to charge at 700mA, sometimes as as high 1019mA. I checked that using "ampere app" from. As per my understanding the displayed current would is -400mA used by the display. Hence the charge current might be as high as 1400mA. But 7805 was showing 800mA in the app, but the charge percent never increased even after waiting for 10-15minutes.

Don't do that. The LM7805 is a linear regulator, and given the common laptop power supply provides about 19 V, the efficiency would be 5V/19V * 100 which is about 26.3%. That means 73.7% of the power is lost into heat, so unless you have a liquid nitrogen cooling setup, I wouldn't do that. If the case is regulating the output you can use a switching step-down DC-DC converter for about \$3 on eBay.

The regulated modules usually have a potentiometer which you can use to adjust the output voltage. The output amperage limit is set by the input power and the efficiency. For example; for the case you put above, and assuming an efficiency of 80%: 19V * 2A * 0.8 = 5V * xA thus theoretically the maximum output current would be 6.08A. You should take a look into the specs of the module, as switching regulators also have current limits.

The simplest way to do this and get full power aka watts/amps out of it, is to use a PMW ( Pulse Width Modulator ) controller, these are very inexpensive and some can control from 0 to 100 volts so if you put in 19v the maximum that can come out in its regulation is a bit short of 19v if you need a pointer, just contact me

Is there a way to reduce the voltage, I have Philips adapter with 9.02V/1.56A.

I want to use it as high power mobile charger. The circuit does have the secondary side of the transformer a TL431 to regulate voltage through opto-coupler. Same as R10 and R14, here in this circuit some other numbering, the value of R10 is 11K and R14 is 4K, providing a voltage of 9V. Now to bring down the voltage to 5.0V I tried to change R10 and R14 as per the formula, but the voltage never drops below 6.7V. Any idea why is it so.

Hi. I got here following TL431. Your article was quite interesting. I'm looking for technical ideas to cheat our dishwasher's temperature sensing. It's highest available temperature is 50°C (and in reality it's even lower, around 46°C), which happened to be not enough, grease accumulates in the corners, bad washing performance. I believe that the temperature is controlled by a TL431. The question is, how could I raise the temp from 50 to 65°C? One technical detail, it might help (this is written on the panel): WQP8-9239H S-S(9239H). Every help appreciated. Thanks in advance.

In my case my PS output was 5.25v and I wanted to reduce it to no more than 5.05v. You mentioned the sum of R1 plus R2 being 10Kohm and I'm a bit worried about it because R1 was 5,1K and R2 was 4,7K originally and I simply put a 56K resistor in paralell with R1 to achieve 4,67K and get the 5.05v. The problem is that 4,67k + 4,7k ~ 9,3k and the lower resistor is R1 instead of R2. Does it make any difference? The datasheet sheds no light upon it.