Re: 18 volt charger
- From: "Chris" <cfoley1064@xxxxxxxxx>
- Date: 19 Nov 2005 23:54:45 -0800
Nelson Johnsrud wrote:
> The charger for my 18 volt rechargeable drill met with an untimely and
> violent demise this week, and I am trying to build a charger circuit for
> my battery pack. I am starting with this basic circuit (attached -- I
> hope attachments are ok on the group). The cells in the 18 volt Ni Cad
> pack are rated at 1300 mah. If I use the 10 percent capacity rule for
> charging, I will need to build a charger which outputs 130 ma, constant
> current. To do this, I will need to make a few changes to the basic
> circuit.
>
> [I am somewhat low on the experience ladder, so feel free to jump in and
> correct my thought process if some alarm bells go off.]
>
> To start with, I will need a higher voltage supply. Lacking a suitable
> transformer, I dug in the closet and found a 24 volt, 2.5 amp regulated
> and filtered supply that was a take-out from a machine control. It's
> overkill, but it's what I have. That will take care of the supply side
> of the circuit, as long as I control the current.
>
> I do not have the specified BC547 or the TIP41. Since this is a general
> purpose NPN transistor driving an NPN power transistor, I plan to use a
> 2N3904 to drive the base of a TIP3055 from our friends at Radio Shack.
> How am I doing so far?
>
> Now how do I calculate the resistance values for Rx and Ry? The
> information I got with this circuit diagram says to calculate Rx as 0.7
> divided by the charge rate. If I want 130 ma, I use 0.7/.13=5.38 ohms.
> Correct? That resistor will also need to be a beefy one to carry the
> wattage. How do I calculate Ry to give the proper voltage output?
>
> I will also want to add an LED and limiting resistor ahead of this
> circuit for a power-on indicator. That's easily done. Maybe I can put
> it in series with Ry?
>
> I would also like to add another LED in the charging output end to
> indicate that the batteries are drawing current. Either the LED stays
> on as long as the battery is drawing current, or it lights up when the
> pack is full, whichever is easier. I'm not quite sure how to do this.
> It would seem that I would have to shunt off the output somehow. Any
> suggestions?
>
> Thank you for reading this long post. Feel free to tell me where I've
> gone wrong. I will appreciate any help someone is willing to offer. I
> am trying to use what is on hand, or can be had very cheap. If I have
> to invest too much in this project, I will simply buy a new drill/charger.
>
> Nels
Hi, Nels. The answer for your question is:
Rx = 5 ohms
Ry = 1500 ohms
Here's how we got there. Looking at your circuit redrawn with
component values added (view in fixed font or M$ Notepad):
24Vo-----o---------------o----.
| Vb+ |
| ---
.-. -
| |Ry Vb- |
| |1.5K .-----o----'
'-' |<--.13A
| |
| |/
1.4V o-------| Q1
| |>
\| |
Q2 |-------o 0.7V
<| |
| ___ |
0Vo-----o--|___|--'
Rx <--.13A
5 ohm
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
several things are immediately apparent. First, if the circuit
actually does work as a quick&dirty current source, 130mA should be
flowing through the collector of Q1 (TIP41). As a first cut
approximation, you can assume that about 130mA will also be flowing
through the collector. Since the transistor Q2 (BC547) is going to
work to keep its base-emitter voltage at no more than 0.7V, you can
size Rx so that about 0.7V will be impressed across Rx when the 130mA
is flowing through it. Thus, Rx = 5 ohms. (By the way, .13A * 0.7V =
..09W. A 1/4 watt resistor should work fine for Rx.)
Now let's look at Q2, and Ry. You want to set things up so Q2 will
always be on at least a little, no matter how low the beta of Q1. So,
let's make a worst case assumption that the beta of power transistor Q1
can never be worse than 10. That means that Q1 will never need more
than 13mA of base current to drive 130mA of collector current. That
means that if Ry is less than 1700 ohms, there will always be enough
current for the base of Q1, and Q2 will shunt away the rest. (If Q1
gets too much base current, collector current and the voltage across Rx
will try to rise, making Q2 turn on more, which will pull down the
voltage at the base of Q1.) A 1.5K (or 1.6K, if you've got one) 1/2
watt resistor will do the trick. You can also use two 10 ohm resistors
in parallel for Rx, and two 3.3K, 1/4W resistors in parallel for Ry).
Your transistor replacements should work OK. I wouldn't worry too much
about using a TIP3055 in place of the TIP 41. The 2N3904 should also
be a good replacement for Q2.
By the way, if your regulated 24VDC power supply is a linear with a
tweaker adjustment, try to bump it up a little to 26V or more. The
voltage across your 18V battery is going to rise as it's charging, and
if the battery voltage gets much above 21 or 22V, this circuit may not
be able to keep the full charge current going, especially if your set
regulated voltage is a little on the low side. And also, if you want
your charger to last more than a few minutes, use a heat sink for Q1
that can dissipate several watts, along with a thin coating of thermal
grease.
In order to keep the concept of a quick&dirty charger, you might just
want to just put a 2700 ohm resistor in series with your LED across the
power supply terminals. Anything in series with the main charging
circuit will cause a voltage drop which will keep your circuit from
working properly. Also, putting the LED in series with Ry won't work
to show charging, because if you remove the load from the circuit, the
current from Ry and the LED will just flow through the base-emitter
junction of Q1 and the 5 ohm resistor. It will be on whether charging
is occurring or not. But, if you happen to have an LM358 single supply
dual op amp around, you can use one half of the IC to drive the current
sink transistor, and the other half as a comparator sensing the voltage
across the current limiting resistor, and driving the LED directly.
Post again if this is of interest (which I guess means, if you've got
an LM358 or something like it in your junkbox). Be sure to also
mention whether your power supply is a linear open frame, and whether
it has a tweaker pot.
Good luck
Chris
(By the way, most of the time, attachments are frowned upon in the
newsgroups. If you want to post a picture, and ASCII art doesn't do it
for you, post in binaries to keep those "silly perfectionists" happy.
Also, some newsreaders just don't allow pictures or attachments. By
using ASCII art or posting in binaries, you're not cutting out the
"silly perfectionist" part of your audience. After all, they might
have something to say, too -- possibly even a great answer approaching
perfection. After all, they _are_ perfectionists. ;-) )
.
- Follow-Ups:
- Re: 18 volt charger
- From: Nelson Johnsrud
- Re: 18 volt charger
- Prev by Date: Re: Interface 4x4 matrix keypad to AT91EB55 board
- Next by Date: Re: Any good magazines for Electronics?
- Previous by thread: Re: 18 volt charger
- Next by thread: Re: 18 volt charger
- Index(es):
Relevant Pages
|
