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newbie question

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captaincaveman

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hi,newbie to all this.I have a question about dropping voltage from 12v to 9v, i have a kind of strobing circuit with two bright l.e.d's(i'll post when i have scematic i can upload)its currently powered by a 9v pp3 battery and was wondering how to power it from a 12v car battery. Is there a formula avaliable without knowing current drawn to drop voltage from 12v to 9v? can a resistor be put in series with input from power supply or does it have to be done with a voltage divider?

can anyone help, it would be nice to have a formula for both if possible, to save me asking the question later:D

also another quick one. can anyone tell me a good program to draw scematics(with electronic symbols)thanks


Thanks again

Captaincaveman:)
 
I use Microsoft Paint program to make schematics. Straight lines are made with the Shift key down. I copy components from other schematics or datasheets then Paste them into the Paint schematic I am making.

Your circuit can probably work at 9V or 12V, but the value of the current-limiting resistors in series with the LEDs will need to have their value increased for 12V to be used.
 
audioguru said:
I use Microsoft Paint program to make schematics. Straight lines are made with the Shift key down. I copy components from other schematics or datasheets then Paste them into the Paint schematic I am making.

Your circuit can probably work at 9V or 12V, but the value of the current-limiting resistors in series with the LEDs will need to have their value increased for 12V to be used.


ok thanks but still dont know how to select right resistor value, also the above equations would be helpful for future projects as im trying to learn as much as i can as i go along:)
 
R = (Vsupply - Vdrop)/If

If = LED forward current, normally 10 to 20mA.
Vdrop = the LED's voltage drop, 1.9V for red & orange, 2.2V for yellow and yellowish green, 3.5V for some green and all blue LEDs. Refer to the manufacturer's datasheet.
 
Without knowing the current you really can't determine the proper value of the resisitor except by experimentation. If you have a DMM and with a current measuring scale you might run the 9 v battery thru that.

If your current consumption is less than an amp you might use an LM7809 three terminal regulator to get the voltage down from 12 volts to 9 volts.
 
thanks that helps, what about the equation for a potential divider(for future reference) i used to know all the basics years ago and have lost alot through not using it for 10 years;)

its the old saying use it or loose it:D
 
Sounds like what you need is a worksheet that describes the problem, with a simple sketch - and the formula. I'd be willing to bet that there's a website or two around with just this kind of thing - common, relatively simple problems with an example worked out. There are likely electronics workbooks with things like this. I'll bet I moved one aside the other day when I was at Borders Bookstore - a pocket reference of some kind.

If you are learning it's better to struggle a little but I'd say there are times when you just don't have time to work back thru and dust all the cobwebs off your memory.

As a practical matter - the formula that you see requires input and the current is one of those inputs. You might estimate that current with info provided in previous posts or measure it.
 
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