Need help checking my power calculations

G

Gerald

Jan 1, 1970
0
Hi,

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?

Thanks!
Gerald

J

Jonathan Kirwan

Jan 1, 1970
0
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:
: 6.6V 6.6V
: | |
: | |
: \ |
: / R3 |
: \ |
: / |
: | |
: | |
: | |<e Q2
: +-------| PNP
: | |\c
: | |
: | ',
: | |
: ___ |/c Q1 |
: ON/OFF ----------| NPN |
: |>e |
: | \
: | / LAMP
: | \
: | /
: \ |
: / R2 |
: \ |
: / gnd
: |
: |
: |
: gnd

Before I go through the design, I'll let you respond.

Jon

R

redbelly

Jan 1, 1970
0
Gerald said:
Hi,

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?

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.

Regards,

Mark

G

Gerald

Jan 1, 1970
0

Your calculations are correct. However, the bulb should be rated for
that voltage.

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.

you don't know where to start. I'm sure I'll fry some bulbs and PIC's
along the way...lol!

G

Gerald

Jan 1, 1970
0
Thanks for the detailed reply Jon!

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.

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? 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. 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. 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. Jon 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. G Gerald Jan 1, 1970 0 A BJT switch, at those currents, is probably as good as anything. Is a BJT different than a regular transistor? Like a 3906, etc? If I remember correctly, it means Binary Junction Transistor? I'l look it up in my books when I get back into the workshop. C Chris Jan 1, 1970 0 Gerald said: Hi, 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? Thanks! Gerald Hi, Gerald. Here's a possible two transistor solution (view in fixed font or M$ Notepad):

|
| +5V
| |
| .-. 6.6V
| 3.3K| | |
| | | +5V \.-./
| '-' | ( X )
| PIC ___ | |< /'-'\
| o-|___|-o-| 2N3906 |
| __ 3.3K |\ |
|OFF/ON | ___ |/
| o-|___|-|TIP41A,B,C
| | 270 |>
| .-. |
| 3.3K| | |
| | | ===
| '-' GND
| |
| ===
| GND
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

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:

| +5V
| |
| .-.
| 3.3K| |
| | | +6.6V
| '-' |
| | ___ |<
| o-|___|--| TIP42A,B,or C
| | 270 |\
|PIC ___ |/ |
| o-|___|-o-|2N3904 |
|OFF/ 3.3K | |> .-.
|__ .-. | ( X )
|ON 3.3K| | === '-'
| | | GND |
| '-' |
| | ===
| === GND
| GND
|
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Good luck
Chris

E

ehsjr

Jan 1, 1970
0
Gerald said:
Hi,

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?

Thanks!
Gerald

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.

Ed

J

jasen

Jan 1, 1970
0
Is a BJT different than a regular transistor? Like a 3906, etc?
If I remember correctly, it means Binary Junction Transistor?

It is a regular transistor, Bipolar Junction Transistor,

It think it was a term invented to distinguish BJTs from IGBTs
MOSFETs and JFETs

Bye.
Jasen

G

Gerald

Jan 1, 1970
0
Thanks Chris!

I can use either of them and just change the state in software.

I am new to this newsgroup, so I really like that you can send an
ASCII schematic. Do you draw them in that program: AACircuit, or does
it convert a schematic in another program?

G

Gerald

Jan 1, 1970
0
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.

Ok. I will be using the new Higher powered Lithium Ion batteries I
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.

http://www.a123systems.com/html/products/cells.html
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.

WOW! That IS ugly!!!!!!! Know of a better way to dim them?
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.

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.
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. Do you think a voltage divider will work for this? Or, will it
be a lot of wasted battery power?

Thanks for the feedback!
Gerald

C

Chris

Jan 1, 1970
0
Gerald said:
Thanks Chris!

I can use either of them and just change the state in software.

I am new to this newsgroup, so I really like that you can send an
ASCII schematic. Do you draw them in that program: AACircuit, or does
it convert a schematic in another program?

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
Usenet?"

Good luck
Chris

R

redbelly

Jan 1, 1970
0
Gerald said:

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.

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

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?
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.

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.
you don't know where to start. I'm sure I'll fry some bulbs and PIC's
along the way...lol!

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.

E

ehsjr

Jan 1, 1970
0
Gerald said:
Ok. I will be using the new Higher powered Lithium Ion batteries I
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.

http://www.a123systems.com/html/products/cells.html

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.

WOW! That IS ugly!!!!!!! Know of a better way to dim them?

Not in terms of energy consumption - I think PWM is
best.
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...

Ok, good on the PIC. You'll need the 2 transistor circuit
posted earlier.
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?

You can use a resistor in series with the bulb.
..5 ohms would drop about .38 volts, and Vce across
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.

Ed

E

Eeyore

Jan 1, 1970
0
Gerald said:
Hi,

I want to make a battery powered light, powered by a PIC.

Good Lord ! PICs are a free energy source too ?

Graham

E

Eeyore

Jan 1, 1970
0
Gerald said:
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?

Use a 2.5W bulb. The battery will last as long and it'll be a heck of a lot
brighter !

Graham

R

redbelly

Jan 1, 1970
0
ehsjr said:
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.

Ed, it looks like Gerald's batteries are still at least 2.2 Ah at 1 A.

http://www.a123systems.com/html/products/ANR26650M1specs.pdf

Regards,

Mark

J

John Fields

Jan 1, 1970
0
Use a 2.5W bulb. The battery will last as long and it'll be a heck of a lot
brighter !

---
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!!!!

M

Michael A. Terrell

Jan 1, 1970
0
John said:
---
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.

--
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

E

Eeyore

Jan 1, 1970
0
John said:
---
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!!!!

Idiot.

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