ChatterBank2 mins ago
Electronics/Maths question
I have an electrically assisted bicycle (mobility scooter in disguise) but the LED lights are not powered from the main battery. I find this quite strange and would like to remedy the situation.
I thought something like this would do the job of dropping 12v10Ah to 3v100mA
http://www.maplin.co....oogle&u=8067&t=module
But I want the 100mA version not the 1.5A CM and the only application diagram I can find is for the CM.
http://oi41.tinypic.com/5xpog.jpg
Question is threefold, Is this schematic relative to the 100mA regulator, how would I visually identify which pin was what and can anyone please suggest values for R1 and R2?
Thank you in advance for your time, I really would sincerely appreciate any help at all.
I thought something like this would do the job of dropping 12v10Ah to 3v100mA
http://www.maplin.co....oogle&u=8067&t=module
But I want the 100mA version not the 1.5A CM and the only application diagram I can find is for the CM.
http://oi41.tinypic.com/5xpog.jpg
Question is threefold, Is this schematic relative to the 100mA regulator, how would I visually identify which pin was what and can anyone please suggest values for R1 and R2?
Thank you in advance for your time, I really would sincerely appreciate any help at all.
Answers
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See this link here for the device data sheet:-
http://pdf1.alldatash...8839/NSC/LM317LZ.html
The device operates by developing 1.25V between the output pin and the reference (Adj) pin.
The Iadj is typically 100uA and can be ignored if sensible values for R1 and R2 are selected.
If R1 is given a value of 240 Ω – with 1.25V = 5.2mA. Therefore the value of R2 is such that 1.75V (3V-1.25v) passes 5.2mA = 336 Ω (or 330 Ω as a preferred value).
So to clarify, you will have a 240 Ω resistor between the output voltage pin and the Adj pin, and a 330 Ω resistor between the Adj pin and 0V.
http://pdf1.alldatash...8839/NSC/LM317LZ.html
The device operates by developing 1.25V between the output pin and the reference (Adj) pin.
The Iadj is typically 100uA and can be ignored if sensible values for R1 and R2 are selected.
If R1 is given a value of 240 Ω – with 1.25V = 5.2mA. Therefore the value of R2 is such that 1.75V (3V-1.25v) passes 5.2mA = 336 Ω (or 330 Ω as a preferred value).
So to clarify, you will have a 240 Ω resistor between the output voltage pin and the Adj pin, and a 330 Ω resistor between the Adj pin and 0V.
Before you rush out and build the circuit – in your proposed application the device will be dissipating just under 1 Watt. The device has a thermal rating of 180°C/W. Therefore the package will shutdown due to getting too hot, so you might need the 1.5A version, given the power dissipation of the device.
Is the traction battery only 12 volts? the ones that I have come across are normally between 24 and 36 volts.The reason your lights are powered by separate batteries is that when the main battery is completely flat (unlikely I know) you will still have lights for pedal cycling and you can carry spare batteries too.
OK Hymie, in light of your disclaimer I may have been barking up the wrong tree and if what you say is correct then I realise my idea is fundamentally flawed. Sounds ridiculous, a 3V100mA circuit drawing close to 1W. I think that I'll have to keep buying AAA batteries for the lights despite the fact I have a huge battery on board.
Looked at your answer again Hymie, if you meant 240K and 330K that would make more sense and reduce the current drain to slightly less than 100mA?
Thanks for your answer, the maths seems to work. I'll get these bits and pieces together and see what happens. I didn't notice the '3' version I want is the same order number as the LZ model you have linked to. Same device but different case style and surprisingly a different price.
Thank you.
Thanks for your answer, the maths seems to work. I'll get these bits and pieces together and see what happens. I didn't notice the '3' version I want is the same order number as the LZ model you have linked to. Same device but different case style and surprisingly a different price.
Thank you.
By increasing the resistance values by a factor of 1,000 – the Iadj current will affect the output voltage.
With my values, 5.2mA is effectively ‘wasted’ in acting as a control current. If you double the resistance values to 470 Ω and 660 Ω, the control current will be less than 3mA.
The figure of 100mA is the available current to supply the LED lights at 3V.
But you will still have the power dissipation issue.
I will take a look for a more suitable device to do what you want and advise later today.
With my values, 5.2mA is effectively ‘wasted’ in acting as a control current. If you double the resistance values to 470 Ω and 660 Ω, the control current will be less than 3mA.
The figure of 100mA is the available current to supply the LED lights at 3V.
But you will still have the power dissipation issue.
I will take a look for a more suitable device to do what you want and advise later today.
DO NOT TAP ONE STRING FROM THE BATTERY !!!!!!!!!!
This is a very quick way to ruin the battery because it will unbalance the cells resulting in those cell being over discharged and the other cells ruined by overcharge.
Even run off twelve volts (don't even think about it) the regulator is dissipating most of the power. The regulator is a terrible solution.
The best solution would be a 48 volt lamp run from the whole battery. These are avialable but I have not reseached if a suitable power rating is available. Google 48 volt lighting.
One do it yourself option is to place twelve cheap plastic white LEDs in series. Whilte LED operate at 3.3 volts so this will take about 40 volts to run. Use the spare voltage to operate a constant current source to supply the LEDs about 20 mA. This will be about 0.8 Watts.
You can increase the number of parallel strings to get more power if you want brighter lights. If you want to pursue this option post back and I will explain a couple of incredibly simple constant current sources.
You could even include a bypass to run the circuit on very flat batteries in emergency by bypassing a couple of LEDS.
If you want substantial wattage you could consider CREE LEDs. They make high powered glass encapsulated LEDs.
This is a very quick way to ruin the battery because it will unbalance the cells resulting in those cell being over discharged and the other cells ruined by overcharge.
Even run off twelve volts (don't even think about it) the regulator is dissipating most of the power. The regulator is a terrible solution.
The best solution would be a 48 volt lamp run from the whole battery. These are avialable but I have not reseached if a suitable power rating is available. Google 48 volt lighting.
One do it yourself option is to place twelve cheap plastic white LEDs in series. Whilte LED operate at 3.3 volts so this will take about 40 volts to run. Use the spare voltage to operate a constant current source to supply the LEDs about 20 mA. This will be about 0.8 Watts.
You can increase the number of parallel strings to get more power if you want brighter lights. If you want to pursue this option post back and I will explain a couple of incredibly simple constant current sources.
You could even include a bypass to run the circuit on very flat batteries in emergency by bypassing a couple of LEDS.
If you want substantial wattage you could consider CREE LEDs. They make high powered glass encapsulated LEDs.
Well, it's looking increasingly more likely that I'll opt for rechargeable AAA's. But as an after thought... would it be so glaringly obvious to take a separate feed from across the whole battery and drop the voltage / limit the current with a single 450Ω bog standard 1/2W metal film resistor?
That would give 3V100mA, or would it be ridiculously inefficient. Could it be that simple?
That would give 3V100mA, or would it be ridiculously inefficient. Could it be that simple?
Such a solution would only work on the proviso that the 3V LEDs were guaranteed to draw 100mA. If for some reason the LEDS only drew 50mA, then there would be a voltage of over 20V across the LEDs, resulting in their destruction.
But to solve this problem, you could add a 3V zenner diode across the LED load, which would guarantee a maximum of 3V across the LEDs.
I’m not sure how you calculated a ½ Watt resistor; 100mA through a 450 Ω resistor will dissipate 4.5 Watts.
But to solve this problem, you could add a 3V zenner diode across the LED load, which would guarantee a maximum of 3V across the LEDs.
I’m not sure how you calculated a ½ Watt resistor; 100mA through a 450 Ω resistor will dissipate 4.5 Watts.
It's far more complicated than I'd hoped and I was out of my depth from the start. I cant run the risk of ruining an expensive battery pack or burning stuff out. As there is apparently no easy answer I think the wisest move for me is to leave it well alone, rechargeable AAA's it is.
I appreciate your patience Hymie. Thank you once again.
I appreciate your patience Hymie. Thank you once again.
Hymie: The variation in voltage drop across the LEDs is why it is imperative to use a constant current source to drive them. A resistor is a poor solution to current limiting because it is entirely passive.
A very good constant current source is an LM317 regulator with a single resistor between adjust and output such that the constant outpuut current develops 1.5 volts between those terminals. It requires about 3V to run it.
A very good constant current source is an LM317 regulator with a single resistor between adjust and output such that the constant outpuut current develops 1.5 volts between those terminals. It requires about 3V to run it.
The LM317 has an input to output voltage limit of 35 volts. It is one of the benefits of the 317 over the 78xx regulators which are limited at 32 V input to ground.
Consequently the LM317 can even comfortably regulate a 48 volt rail from a 75 Volt input. The is no ground terminal on it so it only see the voltage across it.
But in the application with the LEDs the 317 is acting as a current limiter. The output is connected to the adjust pin via a suitable resistor. The supply to the LED string is taken at the connection of the resistor to the adjust pin.
The resistor is selected to produce 1.2 volts at the desired output current. So 12mA would require a 100 Ohm resistor. A fully charged 48V lead acid battery has a terminal voltage of nearly 50 volts going upto about 55V on a normal charge so lets design for 63 volts to be safe.
The twelve LEDs in series have a total voltage of near enough to 40 volts. So with the regulator's own loss (1.5V) we have at worst 25 volts across the regulator for a dissipation of one quater of a watt. It won't even need a heatsink.
The resistor will always have 1.2 Volts actoss it. At 12mA the disipation is only 144mW which would be more than comfortable with a 1/4W resistor.
I chose 12mA because plastic LEDs have a maximum current of 20mA. This will provide enough current for a 5W ouput lamp (.012A x 40V). With losses it will draw less then 6W.
Your original lamp had a power of 300mW or just 1/15 th of the LED. LEDs are about three to four times as efficient as incandescant bulbs so cleary out will have a very powerful light. You can reduce the power if you like to conserve energy by simple increasing the value of the sense resistor.
Be sure to mount the LED well apart as they do get hot when clumped together. The less power you choose to run them at the better too.
You could even have a dimmable setup with a variable current sense resistor. You should configure this resistor in a bridge such that when it goes open circuit it results in the regulating of a lower than normal current but not totally off. If you know about bridges then you should be able to work this out.
Capacitors should be placed on the regulator as specified. However the LED is a very steady load and they are not absolutely essential.
I will continure to keep an eye on the thread so do post back if you want more information.
Consequently the LM317 can even comfortably regulate a 48 volt rail from a 75 Volt input. The is no ground terminal on it so it only see the voltage across it.
But in the application with the LEDs the 317 is acting as a current limiter. The output is connected to the adjust pin via a suitable resistor. The supply to the LED string is taken at the connection of the resistor to the adjust pin.
The resistor is selected to produce 1.2 volts at the desired output current. So 12mA would require a 100 Ohm resistor. A fully charged 48V lead acid battery has a terminal voltage of nearly 50 volts going upto about 55V on a normal charge so lets design for 63 volts to be safe.
The twelve LEDs in series have a total voltage of near enough to 40 volts. So with the regulator's own loss (1.5V) we have at worst 25 volts across the regulator for a dissipation of one quater of a watt. It won't even need a heatsink.
The resistor will always have 1.2 Volts actoss it. At 12mA the disipation is only 144mW which would be more than comfortable with a 1/4W resistor.
I chose 12mA because plastic LEDs have a maximum current of 20mA. This will provide enough current for a 5W ouput lamp (.012A x 40V). With losses it will draw less then 6W.
Your original lamp had a power of 300mW or just 1/15 th of the LED. LEDs are about three to four times as efficient as incandescant bulbs so cleary out will have a very powerful light. You can reduce the power if you like to conserve energy by simple increasing the value of the sense resistor.
Be sure to mount the LED well apart as they do get hot when clumped together. The less power you choose to run them at the better too.
You could even have a dimmable setup with a variable current sense resistor. You should configure this resistor in a bridge such that when it goes open circuit it results in the regulating of a lower than normal current but not totally off. If you know about bridges then you should be able to work this out.
Capacitors should be placed on the regulator as specified. However the LED is a very steady load and they are not absolutely essential.
I will continure to keep an eye on the thread so do post back if you want more information.
Another altenative for the LEDs would be to put four 12v lamps in series. Some automotive lamps are rated 12/24 volt so you could get away with two but it would be a waste of power.
The 12 LEDs in series is a very good option. They will dim as the battery discharges extinguishing entirely at about 42 volts which is the maximum discharge point of a 48 volt lead acid battery.
If you need to have a more reliable light reduce the number of LEDs accordingly. You could even have a switch that just shorted out one or two LEDs for emergency situations. You could similarly add an extra LED for when the battery is fully charged if you liked.
The 12 LEDs in series is a very good option. They will dim as the battery discharges extinguishing entirely at about 42 volts which is the maximum discharge point of a 48 volt lead acid battery.
If you need to have a more reliable light reduce the number of LEDs accordingly. You could even have a switch that just shorted out one or two LEDs for emergency situations. You could similarly add an extra LED for when the battery is fully charged if you liked.
Thank you for your continued interest and ideas beso. I have decided to do nothing as this has proved to be more complicated than I would like.
I regret asking the question in so much as wasting your time, but I do realise this is too much for me to take on.
Thanks for pointing that out. Reality check, got it.
I regret asking the question in so much as wasting your time, but I do realise this is too much for me to take on.
Thanks for pointing that out. Reality check, got it.
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