  PICKIT 3 External Dual Power Supply (3v3-5V) AuthorGiorgosÂ LazaridisAugust 28, 2013

PAGE 1 of 2 - PICKIT 3 External Dual Power Supply (3v3-5V) Circuit  First of all, why... Why use such a circuit since the PICkit 3 has a built-in power supply software controlled. There are basically two reasons to build this circuit: First of all, it may happen - The internal PSU may be destroyed. Actually, i fried the PICkit3 PSU myself and i know for sure that it may happen...

There is also another very important reason: To increase the current supply capability. According to the PICkit 3 Programmer/Debugger Users Guide:

 When using the USB connection, PICkit 3 can be powered from the PC but it can only provide a limited amount of current, up to 30 mA, at VDD from 1.8-5V to a small target board.

Consider this: A single LED may consume as much as 20mA...

The Circuit
So simple: Based on the LM317, its just a constant voltage power supply. To select between 3.3V and 5V, a miniature slide switch is used (click to enlarge): First, the input capacitor. I've left the voltage range open on purpose. Typically, a 60V or higher voltage rating is suitable for any input voltage rating. But I'd rather suggest you select the proper capacitor as per your input supply voltage. If you plan to use a 12Vdc PSU for input, then a 16V input capacitor is enough, you don't need to select higher rating to keep the capacitor size down to minimum.

The input voltage must be DC from a power supply. Make sure the polarity is correct. The LM317 can regulate anything between 1.5 and 40Vdc. But since we need to have 5V output, the input voltage must be at least 6.5V (there is typical 1.25 to 1.5 voltage drop across the adjustment resistors).

The power dissipation on the LM317 depends on two factors: The voltage difference (Vout - Vin) and the current:

 P = Vdiff x I

Here is an example. Say we set the output voltage to 5V. The input voltage is 24V. The circuit draws 100mA (0.1A). The power dissipation is:

 P = (24 - 5) x 0.1 => P = 1.9 Watt

It may not sound much, but if the circuit runs for a long time, the temperature on the LM will increase significantly. I strongly suggest to use proper heatsink for power above 1.5 Watts. To keep the power dissipation low, select a lower input voltage. If for example the input voltage is 9V instead of 24V:

 P = (9 - 5) x 0.1 => P = 0.5 Watt

As you can see, there is absolutely no reason to us higher supply voltage than 9 Volts. After all, the output voltage will never exceed the 5V.

The switch SW1 is used to select between 3.3 and 5V output. Here is how it works. Say that the switch is OPEN (as in the schematic). The two 150 Ohms resistors (R3,R4) are connected in series with the 360 Ohms resistor. The total series resistor is then 150+150+360=660 Ohms. The output voltage is then:

 Vout = 1.25 x (1 + (R2/R1)) = 1.25 x (1 + (660/220)) => Vout = 5 Volts

Now say that the switch is CLOSED. In that case, the two 150 Ohms resistors are bypassed by the switch and are not connected in series with the 360 Ohms resistor. The output voltage is:

 Vout = 1.25 x (1 + (R2/R1)) = 1.25 x (1 + (360/220)) => Vout = 3.3V Volts

One may ask "Why use two 150 Ohms resistors and not one 300 Ohms?". Well, i just found it easier to connect them on the pre-perforated PCB. You can definitely use one 300 Ohms resistor instead...

Keep on reading to see how I soldered the parts together on the pre perforated PCB (click here) Continue reading. Click here to go to the next page >>>.

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