Low power consumption

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The formula for eletric power is P = I^2 * R, where P is power, I is current, and R is resistance.

Given that the current is constant, power will increase/decrease linearly with an increase/decrease in resistance, respectivly.

Since U = R * I, U beeing the voltage, P = (U^2)/R, U beeing the voltage potential, when U is constante, P varies inversly with R.

This is only for DC, AC is a whole other story.

What is your load anyway? Or is this your homework?
 
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Resistors need to be calculated to meet specific requirements in the circuit they are used - you can't just randomly increase them to reduce power consumption.

The circuit you posted seems to have no point at all, what is it supposed to do?.
 
The LED is drawing about 13.6mA. Increase the value of the 200 ohm resistor then the current will be decreased but the brightness of the LED will be dimmer.
The 10 ohm resistor is doing nothing.

The 9V battery is wasting about 7V because the LED needs only about 2V if it is red. A blue or white LED needs about 3.5V. Two AA alkaline cells will power a red LED for about 5 times as long as a 9V alkaline battery at the same brightness. The cells inside a 9V battery are the tiny and skinny AAAA size.
 
firstly, why does everyone keep saying "is this my homework?".
I may come across foolish, but im just starting off in this game.

The circuit does have a point,
im going to replace the light in my car glove compartment. Im going to mount the switch where the glove compartment door closes, so that when its pressed, no light will be on.
When the switch is open the light will be on.

My circuit inverts the output of the switch and turns on a super bright led.

I will take the 10ohm resistor out.

I thought that the 9v battery would power my light for longer?

thanks
 
I have included another diagram. When i made this circuit the resistor got hot when the switch is closed. i.e the 360ohm resistor is directly connected to the + and - rails.

The second transistor allows me to control the brightness of the led without a resistor getting hot.

Is this incorrect?


thanks
 

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you could put the led in series with the transistor instead of in shunt and put a potentiometer to the base then you could adjust the brightness
 
If i put the led in series with the transistor then surely it would be on when the button was pushed? This is not my aim.

thanks
 
bitem2k said:
firstly, why does everyone keep saying "is this my homework?".
I may come across foolish, but im just starting off in this game.
Click to expand...

Sorry, no offence meant. It just initially sounded like a homework question...
 

You could put in a Zener diode instead of the resitor to control the brightness. They won't get hot that easily.

Edit: For arguments sake, lets say the break-down voltage for the zener diode is 5.6V, then on a 9V rail that leaves 3.4V for your LED. Zeners come with many different break-down voltages and current ratings, just select one that comes the closest to what you want the LED to have. The zener diode is put the "wrong" way around as compared by a "normal" diode.
 
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Sorry to cause confusion (just trying to disapate a little of my own).
just a little 2 letter slip there. sorry.
instead of replace try place.
 
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