I want to make a battery powered light, powered by a PIC.
I have a rechargable 6.6 volt battery, and a 5 watt halogen bulb.
The battery is 2.3 amp hours.
Am I correct that to estimate the length of burn time this way:
5 watts/6.6 volts = current = around 757 milliamps
2.3 amp hours / 757 miliamps = a little over 3 hours.
Now, if I use a PIC and pulse the bulb at about 50% to get a dimmer output, is it safe to guess that I would get about 6 hours light out of this battery?
Also, at that power rating, can anyone suggest a transistor to use that won't get too hot?
A BJT switch, at those currents, is probably as good as anything. And fairly cheap. To keep the power dissipated in the BJT down, of course, you need to keep the collector as close to the emitter as possible -- the saturated case. That means planning on 1/10th to
1/20th of the 3/4A needed for the bulb as base current drive. Something on the order of 50mA, or so. A lot more than what you should plan on a PIC pin providing. This suggests a two-BJT solution to the problem.
With those currents, the Vce will likely be more like 0.4V, I think. Power in the larger BJT will then be 50mA*Vbe+0.4V*750mA. The Vbe will be something on the order of 0.9V, I suspect. So about
45mW+300mW or likely under 400mW. That doesn't sound too bad.
Since your PIC supply voltage may be different (you don't say) than the 6.6V, it's output pin probably won't go to 6.6V. So that probably means that you need an NPN for direct connection with the PIC. It will invert your signal sense and also provide the current drive for the PNP BJT that is your main power BJT. This means the PNP BJT will be the big one. They dissipate a little more power than an NPN would in the same type of application, I think, and are somewhat harder to find in larger-than-small-signal sizes (at least, in my junk box.) But I think that's probably the way you need to go because the PIC is likely running on lower supply. The circuit might look like:
Before I go through the design, I'll let you respond.
Your calculations are correct. However, the bulb should be rated for that voltage.
If it's rated for less than 6.6 V, it will draw a higher current than you calculated, and the bulb life will be shortened. It will also run brighter than what it is designed for.
If it's rated for more than 6.6 V, it will draw a lower current, have a longer life, but will be dimmer than it's designed operating point.
You're right...My bulb is rated for 6 volts. (I do have some 12 volt ones that I was going to use for testing since I don't need much light output for testing) I wasn't sure how to handle the difference. I was hoping I could just run it at 6.6 volts, realizing that the bulb life would be shortened. Otherwise, I was not sure if I could use a voltage divider to bring it down to 6 volts.
I have no schooling in this, only in programming PICs, and have learned a few things on my own...so I guess I would be in the newbie catagory, so I may be way off on these guesses. But, it is a fun project, so I figure I'll learn on the way.
The problem is, when you know as little as I do about this, sometimes you don't know where to start. I'm sure I'll fry some bulbs and PIC's along the way...lol!
Ok, that makes a lot of sense to use 2 BJT's...I never would have thought of that. I probabally would have gone with a mechanical relay after a lot of failed attempts...which would have wasted a lot of $ and battery power. Not to mentioned fried chips. lol!
Your way makes much more sense...but I have never used BJT's, so I'll look up some data sheets. Have any numbers/packages in mind that I can google for a datasheet?
I haven't picked out a PIC yet, but have some old 16F84's that I was going to use. I have run those in the past at 5.0 volts...so was planning on the same here. So, the pin would be at 5 volts.
Great! Thanks! The ON/OFF is the PIC control line, right? I'm not used to reading these ascii schematics, and I'll need to switch to a fixed font to get a good look at it, but I think I am following.
Using a PNP at the PIC pin means you'll have a "0" at the PIC output means the lamp is on. If you want a "0" at the PIC output pin to mean OFF, try this:
I don't think your time rating is correct. The 2.3 Ah rating is likely based on the 20 hour rate - that is, by discharging at 2300/20 or 115 mA. Your actual discharge time at 757 mA is likely to be lower than
3 hours. The basic concept - PWM the bulb - will give you a longer discharge time than if you run the bulb at full current.
Another ugly little fact: a cold bulb will draw
5 - 10 times the current of a hot bulb. That means that when you PWM it to 50 %, each time the bulb is turned on it will draw more than 757 mA. It won't draw near what it would if it was completly cold, but it will draw more than 757 because it is not 100% hot.
Setting aside the time calculations, you don't need a PIC for this. A 555 and a 2N3055 power transistor would work. The 555 can provide up to 200 mA to drive the transistor - limit that to roughly 100 mA with a 68 ohm resistor from the output pin of the 555 to the base of the 2N3055. But if you decide to go with the PIC, then the two transistor solution you've been given is needed, but you should use the 2N3055 instead of the TIP41.
You questioned running a 6 volt bulb at 6.6 volts. Might be a good idea to verify that. The bulb should work fine, possibly with some sacrifice of life if it really is being over driven.
Ok. I will be using the new Higher powered Lithium Ion batteries I read about in MIT Technology review. They are supposed to be significantly better than regular Lithium Ion Batteries.
I just got their developer's set of batteries and they look great! Can't wait to test them.
Their specs say "Nominal capacity: 2.3Ah" I guess that is what you mean about the 115ma rating.
They have a bunch of charts on their specs page that I don't understand totally.
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WOW! That IS ugly!!!!!!! Know of a better way to dim them?
Thanks! I appreciate the info and may use it on another project. The reason I am using a PIC is that I am wiring about 12 of them together via an I2C interface and am going to use one IR remote control to turn on/off and change modes: Ultra bright, dim, party mode, sequence, etc...
My girlfriend will be doing some funky glasswork around each one.
Ok. Do you think a voltage divider will work for this? Or, will it be a lot of wasted battery power?
Glad to help, Gerald. Andy's ASCII circuit is a beerware program written by a gentleman in Denmark (Beerware: if you like the program and happen to meet him in Denmark, buy him a beer if you like the program). It's free of spyware and other gunk, and is really easy to use. If you have any experience with any CAD program, it's totally intuitive. And you do draw the schematic -- it doesn't convert. But for simple stuff, it's very quick as long as you're not fighting your mouse.
Go to the website, download it, and enjoy. I definitely owe him a beer if I'm ever in Denmark.
By the way, please bottom post -- it's good form, and helps keep things in context with a lot of newsreaders. Since you're a newbie, please read Google Groups Help Topic "What's good 'netiquette' when posting to Usenet?"
That's not too far off. With a 10% higher voltage, the current will be less than 10% higher than it's design value, or somewhere between 750 and 830 mA. You're battery should still last more than 2.7 hours, so you're not in bad shape.
Bulb life is another matter, but they are cheap and if you experiment you'll find this out. There's a chance the battery's resistance will provide enough voltage drop; have you measured the battery voltage while it is producing 750-800 mA?
Myself, I'd love to start learning about PIC's. Have promised myself I will finish 2 or 3 other projects before tackling them though.
Good luck!
Mark
p.s. A voltage divider will waste energy. If you do need to drop the battery voltage, a series resistance of around 0.8 ohms will drop the
6.6V to 6V on the bulb (for a current of 750-800 mA). As I said, first measure the battery while it is producing current before bothering with the extra resistor.
The discharge curve they show indicates almost the same Ah capacity whether discharged under a light load or a heavy load, so using the computation you did in your initial post would be valid.
Not in terms of energy consumption - I think PWM is best.
Ok, good on the PIC. You'll need the 2 transistor circuit posted earlier.
You can use a resistor in series with the bulb. .5 ohms would drop about .38 volts, and Vce across the TIP41 will be about .1, so about 6.12 would be delivered to the bulb. Alternatively, you could use a 1N540x diode in series with the bulb. It will drop about .7 volts, meaning ~5.8 volts would be delivered to the bulb.
The wasted power in either case, using your figure of 757 mA, is Vbatt-Vbulb * .757 where Vbulb is the voltage across the bulb.
I wouldn't worry about bulb life, until and unless it becomes a factor. I don't know how to compute the effect of running them at 6.6 versus the effect of running them in PWM. Bulbs burn out when run exactly to specs anyway. A 10% higher voltage may reduce the overall life, but it won't cause instant failure or anything like it.
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Great idea!
I\'m going to change all of the 100 watt lamps in my house to 60
watts, since not only will they be brighter, It\'ll cost me less to
run them.
Thanks ever so much!!!!
Really, John. Its just so sad that the lamp that comes on over Eeyore's head is a 12 V grain of wheat bulb, and he only as a single AAAA cell to power it.
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Service to my country? Been there, Done that, and I\'ve got my DD214 to
prove it.
Member of DAV #85.
Michael A. Terrell
Central Florida
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