Simple circuit to generate 12VDC from a 9v battery?

If you can tolerate a quarter volt of p-p ripple:

I think that one is pretty efficient. You can tweak R1 and R2 to get closer to 12V, but figure about 500k-1Meg total resistance for the pair and that you need some 1.23V at the FB pin.

Jon

Hello everyone, I need to generate 12vdc from a 9v square battery, cheaply and simply, supplying about 20mA. I have looked at DC-DC convertors, but haven't come across anything suitable.

I found this circuit that does something similar with a 555 timer but it can only supply about 12mA. Any ideas how to increase the current capability a little?

Any other simple circuit ideas welcome Thanks Marius

Run the square wave from the 555 thru a custom wound transformer to get any voltage you want on the secondary?

That circuit creates a negative voltage that is smaller than the supply. You need a boosting converter with a positive output. You need something arranged more like this:

This should supply enough current, just put a 1K resistor on the output and a 12V Zener to ground (after the resistor) to clamp the voltage at

12V.

supply.

output

Oops, I just re-read your post and you need 20mA out. Lower the 1K resistor I suggested to about 220 ohms and use a 1W Zener.

^^^^ Ah, well 9V squared is 81V, plenty to use a resistor to drop down to 12V. ;o)

Tim

-- Deep Fryer: a very philosophical monk. Website:

And if you still don't get enough current out, reduce the value of R2. Output current from a doubler circuit like that is proportional to frequency.

Another approach is to use a switching converter IC, such as the MC34063A. Take a look at the drawing at

Brian

--- Try this: (View in Courier)

. . +9V +9V +9V . | | |A . [1000R] +---+---+ [1N4148] 78L12 . | |_ Vcc | | +-------+ . +------O|D OUT|--[10µF+]-+-[1N4148>]-+--|IN OUT|-->+12V . | | | | | GND | . [10kR] +--|TH 555| +| +---+---+ . | | |__ _| [10µF] | . +----+-O|TR R|O--+9V | GND . | | GND | GND . | +---+---+ . | | . [10nF] GND . | . GND .

-- John Fields Professional Circuit Designer

--
Or, better yet,:

.
.    +9V         +9V              +9V
.     |           |                |A
.  [1000R]    +---+---+        [1N4148]             78L12
.     |       |_ Vcc  |            |              +-------+
.     +------O|D   OUT|--[0.10µF+]-+-[1N4148>]-+--|IN  OUT|-->+12V
.     |       |       |                        |  |  GND  |
.  [10kR]  +--|TH  555|                       +|  +---+---+ 
.     |    |  |__    _|                     [0.1µF]   | 
.     +----+-O|TR    R|O--+9V                  |     GND
.          |  |  GND  |                       GND
.          |  +---+---+
.          |      |
.      [100pF]   GND
.          |
.         GND
.

Det var smuk og solrig dag da royalmp2001 skrev news: snipped-for-privacy@i39g2000cwa.googlegroups.com i sci.electronics.basics:

How about two 1.5V batteries in series with your 9V battery. That is simple, and you don't throw away a lot of energy in conversions.

--
Rune D. Jørgensen

--
Indeed.

In the same vein, why not just get rid of the 9V battery and run it
from a 12V battery?

This thread reminded me of a funny error from a freshly graduated electronics engineer that we hired some years ago. He needed something like

7.5V from a 5V power supply, so he simply wired a resistor voltage divider in "reverse", ie he connected 5V to the output... Luckily I spot the error before he made a pcb for it.

--DF

"royalmp2001" wrote in news: snipped-for-privacy@i40g2000cwc.googlegroups.com:

simply,

5volts.

Try this one at

roma