Are we losing the plot here. How much current is the main circuit still pulling ? Surely mA for battery monitoring is absolutely peanuts against the main circuit draw. Any battery monitoring circuit is only good until its supply fails. This takes me back to having a separate supply for the monitor to isolate the battery but even that will fail if its left on its own without some drunkard turning it off.
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Couldn't agree more, I use a monitor circuit to remind this drunkard when to plug in and recharge
The only way to be absolutely sure that the supply is removed when the battery is exhausted is to use a relay to disconnect the battery at the point of exhaustion. BUT, that would imply that the relay has to be energised all the time so that it de-energises when the supplies fail. Using a PIC keeps coming to mind, then you could have an audible alarm. Crikey you could get the PIC to phone the local Police to get some sheriff to come and warn you that your battery is about to fail.
Use a MOSFET. It can switch slowly, so even a large die P channel will work and be an almost perfect switch. There are single chip solutions for battery management that can do this for you.
That's actually not a terrible idea. A low end PIC will have a voltage reference and a pair of comparators built in. The sleep function can be used to wake the PIC every minute or so, check the battery voltage, and go back to sleep. Average power drain will be micro-amperes.
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Thank you all for such enlightening inputs! I've perhaps mistakenly understood that a low current led consumes less power, hence less drain on the batteries than a high current LED? Using a resistor for a voltage drop of any size dissipates that energy in the form of heat & a drain on the battery. That means that we’re just wasting that energy on heat instead of getting more light efficiently out of the LED circuit, yet, I am only concerned about the drain on the batteries & not the brightness!
Lets compare a 25mA & a 5mA rated LEDs
5mA LED, 15Vs x 5mA = 75mW (ignore Vf)
25mA LED, 15Vs x 25mA = 375mW (ignore Vf)
Let us also assume that both LEDs have a Vf =2v
Now, connecting 1k & then a 2.2k respectively in series to the 25mA LED
Using this formula.... (Vs-Vf)/R=I........we'll get the led will draw
1. 13mA .......VxI= 15 x 13mA= 195mW..........50% efficient?
2. 5.9mA.......VxI= 15 x 5.9mA= 75mW..........25% efficient?
Is it then correct to say that the rest is wasted across the R as heat while the resistor + the LED are still drawing 25mA?
Now, if we connect a 2.7k R to a 5mA rated led..then
It'll draw 4.8mA which will translate to
15 x 4.8=72mW.....nearly 100% efficient but still very low power consumption!
Does this mean, the higher your power supply voltage, the more power you are wasting while the higher the voltage drop of your LED, the less power you are wasting?
Please don't laugh if my math is worse than that of a 10 year old's! I wasn't any good at it! (No need to confirm this .....just show me how to do this correctly)
Finally, so what is the advantage of using a low current LED over a high current LED apart from the brightness in this particular instance
K&D: I did consider using a small LCD thingy,but I am so desperately restricted by space,besides, even the ones on e-bay costs a few bob!
As for the circuit, it has 5 x NE5532..8 x 5 =40mA?
1 x OPA2132 = 4mA ?
Red/green bicolor LED assuming 2 x20mA = 40mA?
No specs available! so just guessing! If only I can find a 10mA bicolor LED? wishful thinking perhaps?
Btw,#11 b.monitor works nicely...albeit for the high current LED!
Guys, Thanks for some clever solutions, but my circuitboard is just 90mm x70mm completely populated & the enclosure is just 110 x 75 mm! I have no other choice but to keep to this size It's an onboard guitar pre amp! I hope you'll understand.
Lets compare a 25mA & a 5mA rated LEDs
5mA LED, 15Vs x 5mA = 75mW (ignore Vf)
25mA LED, 15Vs x 25mA = 375mW (ignore Vf)
Let us also assume that both LEDs have a Vf =2v
Now, connecting 1k & then a 2.2k respectively in series to the 25mA LED
Using this formula.... (Vs-Vf)/R=I........we'll get the led will draw
1. 13mA .......VxI= 15 x 13mA= 195mW..........50% efficient?
2. 5.9mA.......VxI= 15 x 5.9mA= 75mW..........25% efficient?
Is it then correct to say that the rest is wasted across the R as heat while the resistor + the LED are still drawing 25mA?
Now, if we connect a 2.7k R to a 5mA rated led..then
It'll draw 4.8mA which will translate to
15 x 4.8=72mW.....nearly 100% efficient but still very low power consumption!
Does this mean, the higher your power supply voltage, the more power you are wasting while the higher the voltage drop of your LED, the less power you are wasting?
Please don't laugh if my math is worse than that of a 10 year old's! I wasn't any good at it! (No need to confirm this .....just show me how to do this correctly)
Finally, so what is the advantage of using a low current LED over a high current LED apart from the brightness in this particular instance
K&D: I did consider using a small LCD thingy,but I am so desperately restricted by space,besides, even the ones on e-bay costs a few bob!
As for the circuit, it has 5 x NE5532..8 x 5 =40mA?
1 x OPA2132 = 4mA ?
Red/green bicolor LED assuming 2 x20mA = 40mA?
No specs available! so just guessing! If only I can find a 10mA bicolor LED? wishful thinking perhaps?
Btw,#11 b.monitor works nicely...albeit for the high current LED!
Guys, Thanks for some clever solutions, but my circuitboard is just 90mm x70mm completely populated & the enclosure is just 110 x 75 mm! I have no other choice but to keep to this size It's an onboard guitar pre amp! I hope you'll understand.
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I detect some confusion. Hopefully I can help clear some things up.
Supplying more current is done often... that's how many chips and circuits function. Energy is stored, usually in an inductor, and then dumped through the LED. That's how a single 1.5V cell can power a white LED with its ~3V Vf. The duty cycle is low enough that the junction has time to cool before the next burst, and the average current through the LED can be surprisingly low.
Supplying less current is done just as often.
The LED will draw what the circuit is designed to provide it. See above. If you want the LED to draw 10mA, design for it. The difference in brightness between 20mA and 10mA through the same LED will be virtually insignificant. That ball is in your court!
Beginning here, let's further define what these ratings mean. The 25mA and 5mA is the maximum continuous current recommended for that device. It can be supplied less current (if the resulting decrease in brightness is acceptable) or it can be supplied more (as long as the junction does not overheat).
Supplying more current is done often... that's how many chips and circuits function. Energy is stored, usually in an inductor, and then dumped through the LED. That's how a single 1.5V cell can power a white LED with its ~3V Vf. The duty cycle is low enough that the junction has time to cool before the next burst, and the average current through the LED can be surprisingly low.
Supplying less current is done just as often.
And let us also assume that Vs = 17V.
Efficiency is the ratio of output to input, expressed as a percentage. That isn't what is being calculated above.
Nor here.
Yes indeed. Any voltage over and above the Vf of the LED has to be dropped somewhere else in the circuit. That voltage drop, along with current flow, is energy wasted.
I hope it's making more sense now.
Since the voltage in the circuit is a set quantity, the lower the current then the lower the power wasted. P=V*I
Remember that each of those op amp current specs are for a specified voltage. Datasheets are your friends.
The LED will draw what the circuit is designed to provide it. See above. If you want the LED to draw 10mA, design for it. The difference in brightness between 20mA and 10mA through the same LED will be virtually insignificant. That ball is in your court!
That really seems to be your biggest stumbling block, and it's self-inflicted. Design for the lowest LED current that offers suitable brightness, and use virtually any bi-color LED available to you.
Thank you indeed Sofaspud for your patience & taking the time to explain this in detail. The best explanation so far! I think I'm now almost completely cured of my LED phobia!
You have indicated the mistake in my "efficiency" calculations, so could stretch my luck & ask you if you'd mind just showing me how to do this? I need to learn this once & for all.
Thanks again for the great help.
You have indicated the mistake in my "efficiency" calculations, so could stretch my luck & ask you if you'd mind just showing me how to do this? I need to learn this once & for all.
Thanks again for the great help.
It was just a single LED out of my "optos" drawer. You would have to look at the specs for the LED's. What you want is a high optical output (millicandella or mcd). A quick look at a couple of suppliers didn't turn up any really high brightness ones (not like the single LED's). If you couldn't find one then does it have to be a bicolour or could two singles be used, red and green ?
I'm not sure what you're asking for, but if you rephrase your question I'll help if I can.
Efficiency is just output/input (times 100 to make it a percentage), for example an amplifier's power supply is delivering 20 volts at 5 amps (100 watts input) but the output is just 50 watts. That would be 50% efficiency.
Your calculations were percentages of the capability or capacity of the LED circuit. For example, with an amplifier that is capable of 100 watts peak output. Running it at 50 watts output doesn't equate to 50% efficiency.
Quote: "Since the voltage in the circuit is a set quantity, the lower the current then the lower the power wasted. P=V*I" This was exactly what I was seeking an answer for!
I see, phew what a relief, I thought the math itself being wrong! Anyway, thanks to you, I now hve a far better understanding of LEDs. Have loads of different type of leds to experiment with......phew, what a learning curve!!
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