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I've created a low battery indicator circuit which lights an LED when the voltage drops below 3.75V. in this circuit, the 62ohm resistor is consuming about 240mW of power. why is this? is there any way to reduce it?

enter image description here

Here is the circuit when the battery voltage is below 3.75V:

enter image description here

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  • \$\begingroup\$ BTW. Your circuit is not a good design for a low battery indicator. Using this will ensure that you will have a depleted battery in short order. You want a detector circuit to draw low microamps of current up until you turn on the LED. Use of a microwatt comparator is far better than zeners and BJTs for this. \$\endgroup\$ Dec 23 '15 at 11:57
  • \$\begingroup\$ can you please explain how to use the comparator for this purpose? \$\endgroup\$
    – krishna
    Dec 23 '15 at 12:09
  • \$\begingroup\$ I will consider drawing up a different circuit and placing it in another answer. \$\endgroup\$ Dec 23 '15 at 12:13
  • \$\begingroup\$ Look at your schematic. Your resistor has 3.9 volts across it. Power equals V squared / R. Do the math \$\endgroup\$ Dec 23 '15 at 15:33
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As indicated in comments I have prepared a low battery detector circuit that uses a comparator to sense the low battery state. This circuit will place less than 0.2mA of load on the battery until the battery voltage falls below about 3.75V. At that point the LED turns on and the load on the battery becomes primarily the LED current which can be adjusted by changing the value of R5.

The 10meg feedback resistor around the comparator adds hysteresis so that the detection point does not cause flickering of the LED.

enter image description here

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In your circuit there should be no need for there to be a resistor as small as 62 ohms as the collector load of the left most NPN. Reduce the power and current at the same time by making this resistor much larger. A starting point might be 3.3K ohms.

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  • \$\begingroup\$ if i make the resistance larger, the LED will glow well before the voltage drops to 3.75V. thats why i chose this resistance value while simulating. \$\endgroup\$
    – krishna
    Dec 23 '15 at 12:06
  • \$\begingroup\$ And that is another good reason that in comments above I said that this circuit is not a good design for a low battery indicator. \$\endgroup\$ Dec 23 '15 at 12:09
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Just get rid of the 62 ohms and move the 10k like this: -

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Now, when the 1st transistor is on it will take a few hundred uA through the 10k and when that transistor is off that current will be allowed into the base of the final transistor and turn it on.

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  • \$\begingroup\$ i tried the simulation for this and the LED does not turn on at 3.7V. it only turns on after 2.95V. is there some other modification you did? \$\endgroup\$
    – krishna
    Dec 24 '15 at 6:45
  • \$\begingroup\$ With your zener at 2.4 volts there will need to be about 3V on the battery to activate the left transistor. That is how this should operate so, if you want it to operate at 3.7 volts maybe use a 3V zener. Try it and see. \$\endgroup\$
    – Andy aka
    Dec 24 '15 at 8:32
  • \$\begingroup\$ what will be the A/A rating for the transistor? i'm using BC547 transistors. when i change the A/A rating, the LED lights at different voltages \$\endgroup\$
    – krishna
    Dec 24 '15 at 8:40
  • \$\begingroup\$ I don't know what an A/A rating is. \$\endgroup\$
    – Andy aka
    Dec 24 '15 at 8:55
  • \$\begingroup\$ in this simulator we can change the parameter values. A/A it given as the beta value. is it necessary to change this value? \$\endgroup\$
    – krishna
    Dec 24 '15 at 9:03

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