Is there a way to reduce the current of a circuit?
For example, I have this circuit:
and I want to reduce the total current so it can last longer in a 12V battery.
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There are a few ways:
You are wasting a considerable amount of power in your dropping resistors. One way to reduce the total power consumption would be to put all of the LEDs into one series string along with a low-value current sense resistor, and then use a boost converter to drive the entire string with a constant current. It would be easy to make such a converter adjustable, which would give you the option of dimming the LEDs for additional power savings.
The voltage input of the LEDs can be pulsed so as to reduce the effective total power consumption. So long as the frequency isn't too low, any flicker shouldn't be noticed by human observers.
The circuit below uses an op-amp and single voltage supply to produce a 12V oscillating output. I assume the LEDs operate at roughly 2V.
There's really no free lunch, if you need the brightness you have you will have to live with approximately the current you have given the existing LEDs.
Reducing the current draw by 10%, 20% or even 25% might be possible with your present LEDs, but much more is unlikely without reducing the brightness.
If the existing LEDs are low quality, you might be able to substitute more expensive LEDs that can operate with similar brightness at significantly less current.
If you're looking for a 5:1 change, it probably isn't going to be possible without significantly reducing the brightness and changing the LEDs, and then only if they used particularly crummy LEDs to begin with or you're willing to live with a lot less brightness.
Lets start by analyzing one "leg" of the circuit ( we can multiply the results by 7 for the total circuit ). Assume the forward voltage of a LED = 0.4V and current of 12 mA. Voltage of 3 LEDs = 1.2V, leaves 10.8V across the resistor, which makes the value of the resistor = 10.8 /12 mA = 900 ohms.
The power across the 3 LEDs is EI =1.2V x 12mA = 14.4mW. The power across the resistor = 10.8V x 12mA = 129.6 mW.
From this analysis, one can see that the power "wasted" in the current limiting resistor is about 9 times the power used by the LEDs. If you were to use 3V instead of 12V, a much smaller resistor would be required ( about 150 ohms ), and the power "wasted" would now only be 1.8V x 12mA = 21.6mW, which is only 1.5 times the power used by the LEDs.
The total power used by the first circuit (leg) = 12V X 12mA = 144mW. The power used by the second circuit would be 3V X 12mA = 36mW.
What this shows is that if you can "tap" to a lower voltage source on some "other section" of the "project," your battery would last about 4 times longer.
Typically, on designs like this, there is usually a source of 5V. Connecting your circuit to this source would save you power, although not as much as a lower voltage. If there is no other source of lower voltage, then the options mentioned above, should be used.
If you needed to keep the LED circuit intact, you could use a 555 timer configured as an astable oscillator with variable duty cycle - this could switch on and off a MOSFET. The MOSFET can be wired in series with the array of LEDs and you could dim the LEDs by controlling the duty cycle. Theoretically no power loss in the MOSFET and typically power loss will be less than 5%.
A 555 doesn't have full duty cycle control but there are other chips that would work. I used the 555 because virtually everyone has heard of it.