12 VDC
 

Who's a good tech?



I need a circuit drawn Using a 12 V DC 8 AH batteries, zener's and
components to get 9 Volts DC for a 1amp load.

And 7.6 Volts DC for a 2 amp load.



Would some one draw one?



Richard E. Lenker

12 VDC
 

On Mar 19, 5:14*pm, "Rich" wrote:
Who's a good tech?

I need a circuit drawn Using a 12 V DC 8 AH batteries, zener's and
components to get 9 Volts DC for a 1amp load.

And 7.6 Volts DC for a 2 amp load.

Would some one draw one?

Richard E. Lenker

You will need a MJE3055T transistor(Q1), a 1N758A(Z1),and a 100 ohm
resistor(R1) for the 9 volt out.
the 12v input goes to the collector of Q1, the output 9v comes from
the emitter of Q1. Tie R1 between the collector and base of Q1 and tie
the cathode end of Z1 to the base of Q1 and the anode end of Z1 to
ground. You should also add a capacitor(like .01 or .1uf) from the
emitter of Q1 to ground. You can use the same circuit for the 7.6V but
change Z1 to a 1N756A . R1 stays the same. You can get all of these
parts from www.mouser.com There is no minimum from them.
See if this helps
Don
W5DYV

12 VDC
 

Don wrote:
On Mar 19, 5:14 pm, Rich wrote:
Who's a good tech?

I need a circuit drawn Using a 12 V DC 8 AH batteries, zener's and
components to get 9 Volts DC for a 1amp load.

And 7.6 Volts DC for a 2 amp load.

Would some one draw one?

Richard E. Lenker

You will need a MJE3055T transistor(Q1), a 1N758A(Z1),and a 100 ohm
resistor(R1) for the 9 volt out.
the 12v input goes to the collector of Q1, the output 9v comes from
the emitter of Q1. Tie R1 between the collector and base of Q1 and tie
the cathode end of Z1 to the base of Q1 and the anode end of Z1 to
ground. You should also add a capacitor(like .01 or .1uf) from the
emitter of Q1 to ground. You can use the same circuit for the 7.6V but
change Z1 to a 1N756A . R1 stays the same. You can get all of these
parts from www.mouser.com There is no minimum from them.
See if this helps
Don
W5DYV

For battery operation, efficiency is usually important. Neither a linear
regulator doesn't satisfy that. With this, the demand on the battery is
still 3A (12V * 3A = 36W).

A switching regulator would be more involved but give much better
efficiency. National and Linear Technology both make fine units, and have
datasheets that will lead a person thru the design process. National's
LM257x or LM267x series would work fine for up to 5A load. With a switcher,
the demand on the battery would be (assuming a nominal 90% efficiency)...

(7.6V * 2A) / 0.9 = 16.89W
(9V * 1A) / 0.9 = 10.00W
Total: 26.89W

Bryan WA7PRC