## More Volts from the LM7805 Regulator

I began development of an S-meter/Power meter interface for the MMR40 today.
The plan calls for a 10 volt supply, using a 10 volt zener diode for regulation.  Since I didn’t have a 10-volt zener on hand, I decided to use an LM7805 regulator chip, but expand the voltage output to 10 volts from its normal 5 volts.
To do that I used the method set out in an article by Sam Ulbing, N4UAU in the January, 1999 QST, page 45.

The basic circuit looks like this:

In the article, Sam gives the general guidance of setting the current through R1 at about 3 times the normal quiescent current for the 7805 regulator.Â  He goes on to give the formula for the Output voltage from the circuit:

• Vout = 5 volts + (5v/R1 + Iq)*R2
• Vout = 5 + (5/470 + .003)*390
• Vout = 5 + (0.0106 + 0.003)*390
• Vout = 5 + (0.0136)*390
• Vout = 5 + 5.304
• Vout = 10.304 volts

The device he was using had a quiescent rating of about 8 ma, while the 78L05 (smaller, 100ma device) that I am using has a quiescent rating of about 3 ma.

So I created a spreadsheet with the pertinent formulas and breadboarded a circuit.  I found that my voltages were slightly less, so I fudged the quiescent current value from 0.003 to 0.0026 to make the expected match the actual a little better. They were probably close enough already.

I suspect that for the range of values that are reasonable (say, 5 – 13 volts output) with an input at least 2 volts higher, the table should provide a good starting point for whatever voltage is needed.  Just stick with the ones in which the IR1/Iq ratio is at least 3 to 1, and then pick ones with lower IR1 values for better efficiency.

Here is the test setup, with a 470 ohm/390 ohm divider, and a 220 ohm load (about 46ma):

And here is the result: