Circuit for 19volts in, 9 volts out....

[MichaelW]

Ok, I have a laptop supply that provides 19v and rated at 3.95 amps. I want to use this to power circuit that takes 9 volts. I was thinking that I could use a voltage divider with two resistors to get the output voltage I want, or I could use a voltage regulator, such as an LM7809C.

Something about the voltage divider approach seems wrong... wouldn't the ratio depend upon the current draw by the load circuit?  If so, then I'd have to know exactly how much the circuit is going to draw going in to set it  up correctly.

Using a voltage regulator, it would do what is necessary to keep the output at 9v, dissipating additional current as heat? I'm new in general and even newer when it comes to voltage regulators. haha

[Zero999]

Yes, you're right, a divider wouldn't work because the voltage would change with the load current.

There are two classes of regulator: linear and switched mode. A linear regulator just converts the excess voltage to heat. A switching regulator, is more efficient so if the output voltage is lower than the input, the output current will also be higher.

[AG6QR]

Don't use a voltage divider.  At least, certainly not for any kind of load that will vary its current draw.

Remember the basic voltage divider equation depends on the same amount of current going through the top resistor as the bottom.  So if any current comes in or out of the junction between the resistors, then the voltage won't be what the voltage divider equation said it would be.

A linear regulator like a 7809 will work, but it will dissipate excess energy as heat (a voltage divider would also dissipate heat).  The voltage regulator acts like a "smart resistor" that adjusts itself to provide just the right amount of resistance to drop the input voltage to 9V, regardless of the current output, provided you stay within the limits specified on the data sheet.

If your load will draw more than a few milliamps, you'll probably need a heat sink on that regulator.  If it will draw more than an amp or so, you're going to need to look into alternate strategies, either a switching regulator or a pass transistor with a massive heat sink.

A buck switching regulator will be more efficient, and therefore won't put out so much heat for large loads.  But the 7809 is cheaper and simpler, and would work well for a small load.

[MichaelW]

Thanks for the info guys.

I was looking for a fun thing to build, and this seemed like it would be both fun and useful:

E00CH31 555 Signal Generator Kit

I had most of the stuff already, just had to order a couple of items, which aren't here yet. According to the info it only draws 10mA so that shouldn't heat up an LM7809 much at all, I'd guess. I'll probably run it off of a 9v battery really, but I thought it would be kind of interesting to try to use the laptop power adaptor to power it.

[rolycat]

If you need more current, one option might be to use one of the LM2596-based DC-DC converter modules available from various vendors on Ebay for a couple of dollars. These are very simple to use - supply in one end and a 10-turn trimpot to set the output voltage.

Not recommended for critical applications, since the ICs are very likely unlicensed copies, but they mostly seem to work OK and can supply an amp or so with much less heat dissipation than a linear regulator.

[MichaelW]

Wow, those are pretty neat rolycat, thanks!

[deth502]

the 9v battery would be best. no matter how good your supply is, there is always going to be some noise coming through.

[MichaelW]

the 9v battery would be best. no matter how good your supply is, there is always going to be some noise coming through.

Good point, but that makes it more interesting to use the AC adapter. I'm basically making this for two reasons, the fun of building something, and something I an use to get experience with my oscilloscope. It might be interesting to examine that noisy signal compared to it being run on a battery. :-)

[AG6QR]

That 555 circuit can probably work on about anything between 5V and 15V.  It doesn't need to have precisely 9V.  But 19V will exceed the rating of a typical 555, so don't try running it directly off your 19V adapter. 

A linear regulator will be fine for such a low current.  A 7809 will drop about 10V, which, at 10mA gives 100mW, or a tenth of a watt of heat that it will have to dissipate.  That's not a lot. 

[G7PSK]

I built one of those Velman 555 signal generators about ten years ago, it is still running on the original battery. You could just run it with a resistance in series aim for 12 volts at no load and the drop on full load should still allow it to work.