# L7805CV 5V Voltage Regulator 1.5A

I seem to be having some problems with hooking up the L7805CV Regulator. When I provide it 12v @2A I seem to be getting an output value of 10.xxV???

What could I be doing incorrectly? Does it matter that the power supply is 2A and the Regulator is 1.5A?

(TOP VIEW)
--------|====== Output
|       |====== Ground
--------|====== Input


I have my photo below hooked up looking at it from the back.

What would I be doing incorrectly?

• Not related to your question but on the picture: Do not bend the leads of components near the body/casing as they will be easily broken due to fatigue (minimum bending radius). Leave a 5mm length of lead unbent from the casing. It is highly recommended to test components using a breadboard as you will not need to bend any lead. – shimofuri Aug 10 '12 at 4:03

To me it looks like you've connected the power backwards.

If the green clips are input power and the unconnected lead is where you're measuring the voltage (and I'm assuming this, since there's no additional information in the question), then you've connected input voltage to the output of the regulator.

Next, how is your power supply set up? Normally the current rating is the maximum current that can be supplied and the voltage rating is the voltage that should be supplied, so if you have 12 V 2 A power supply, that means that it can supply up to 2 A when running at 12 V. It does not mean that it is actually supplying 2 A all the time.

The regulator is rated for currents up to 1.5 A, but again that doesn't mean that it's providing 1.5 A of current. It only means that if it's cooled correctly, it can tolerate output current as large as 1.5 A, so everything should work fine on that side.

• You are correct sir.... Thanks for pointing that out. Now it outputs 5v! :) – StealthRT Aug 10 '12 at 2:40

Your photo only shows two clip leads of the same color with no indication what you have done with the other ends or about the third connection.

But generally speaking, linear regulators require capacitors placed near them for proper operation. Try placing .1 uF and 10 uF capacitors in parallel across the output, and another .1 uF (with suitable voltage rating) on the input. Also look at the data sheet as there is typically a recommended or test circuit there.

Actually, on page four of your data sheet it's showing .33uF on the input. Build one of those circuits and try again.

Besides connecting it correctly and ensuring 0.1uF capacitors on the input and output make sure that you have a heat sink installed if you draw more than 50 mA. Please provide a better picture or a diagram if you still have issues.

• A heatsink for 350 mW??? – stevenvh Aug 10 '12 at 18:15
• Actually it is 5V x 50mA = 250mW. – Gregory Ion Aug 11 '12 at 6:57
• No, it's 50mA times the voltage difference between input and output voltage = 50 mA x (12V - 5V) = 350 mA. The 250 mA is what the load will dissipate. 350 mA is what a tiny SOT-23 can dissipate, a TO-220 will do much more. – stevenvh Aug 11 '12 at 7:00
• Before I wrote the comment I built the circuit to test what I just posted. At 50mA the TO220 is warm but at 100mA is very hot. You are right about total dissipation. – Gregory Ion Aug 11 '12 at 16:13
• A TO-220 has a thermal resistance, case to ambient of 65 K/W, so at 700 mW ,and 20 °C ambient it should only be about 65 °C. That's only just too hot to touch. It shouldn't be very hot, unless "too hot to touch" is your definition of "very hot" :-). Did you supply 12 V to it? – stevenvh Aug 11 '12 at 17:47