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L7805-based 5V (Fixed-Voltage) Power Supply Kit

This ECROS Technology power supply kit uses the ST Microelectronics L7805CV voltage regulator.  This device has the following features:

  • very low cost
  • fixed output voltage of 5.0V (4.75 to 5.25 V over temperature, load and input voltage)
  • dropout voltage of 2.0V typical (2.5V maximum) at 1.0A output current
  • over 1A output current with internal protection against overcurrent
  • internal thermal shutdown to protect against overheating
  • operating input voltage of up to 35V (subject to power dissipation)
  • 0.33F input and 0.1F output bypass capacitors required for stability
  • regulation down to 5mA load current (see Technical Specifications, below)

Price and Ordering:

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Mini-PSU Parts Kit contents:

U2 L7805CV Regulator †
J1 2.1mm Power Jack
D1 1N4006 (or equivalent) Diode
C1 0.33F Ceramic Capacitor
C3 0.1F Ceramic Capacitor
C4 470F 25V Electrolytic Cap.
- Heatsink, Screw and Nut

Additional Parts in Flexi-PSU Kit:

R3 470 ohm Resistor
LED1 Red LED, T1 (3mm).
SW1 Switch (0.5A @ 50V) ‡
- Terminal Block, 2-way
- Larger Heatsink

† The regulator must be inserted as U2, not U1.

‡ When switching more than 0.5A, an external switch is recommended.

(Use the menu on the left hand side to go back to the Mini-PSU introduction.)

Technical Specifications

Input Voltage - For the regulator to operate properly, the input voltage to the PSU must always be the output voltage plus the regulator "dropout" voltage plus the voltage dropped across D1, if used.  For example, with this 5V regulator and D1 installed as shown, the input voltage must be at least 8.3V (5 + 2.5 + 0.8).  This must allow for supply voltage "ripple".  Continuing the example, if, at full load, ripple causes the input voltage to drop 0.5V below its average value during each half-cycle, then the average input voltage should be 8.8V.  This can be reduced to 8.0V if D1 is omitted, in which case care must be taken not to connect the input the wrong way round.

Power Dissipation - Heat dissipated in the regulator, in watts, will be the product of the load current in amps and the voltage dropped across the regulator.  Again, using the above example, suppose you apply an average input voltage of 9V.  The regulator will drop 3.2V (9 - 5 - 0.8).  At 0.5 amps, the heat dissipated will be 1.6W (3.2 x 0.5).  ECROS Technology recommends that you limit power dissipation in the regulator to 3W with the compact heatsink supplied in the kit.  This is based on personal safety concerns.  At higher power, the circuit will continue to function but the heatsink will become very hot.  If you touch it, you may not pull away fast enough to avoid some lasting discomfort.

Minimum Load Current - Inexpensive regulators, such as the 78xx series, do not function properly if the load current falls below some small minimum.  In some applications, this doesn't matter, but if you will use the PSU to supply circuits that take very little current, it is a good idea to add a small load to the PSU itself to guarantee proper regulation.  Fortunately, the current to light a small LED is sufficient and the LED acts as a power-on indicator.  You can add a power LED to a fixed-voltage Mini-PSU using the locations of components intended for adjustable PSU circuits.  Put the LED in the location marked for C2 (anode to the positive side).  Put the current setting resistor in the location marked for R2.  For a 5V output supply, a suitable value is 470 ohms.  This will draw 6 - 7 mA, which guarantees the minimum load current.  Lower values will give a brighter LED, but don't exceed the maximum rated current of your LED, which is typically 20 mA, which in this case means not using a resistor smaller than 180 ohms.

(Use the menu on the left hand side to go back to the Mini-PSU introduction.)