Re: 18 volt charger


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
The 2n3055 conducts 130 mA. Divide that by the gain of the transistor,
and you have the base current. Ry has to allow at least that much
current to flow into the base. Size Ry to allow twice that much. To
do the arithmetic you need to know the voltage across Ry. It is supply
voltage minus two diode drops (the base-emitter junctions of the
transistors). So if the base of the 2n3055 needs say 13 mA, and the
voltage across Ry is 22 or 23 volts, Ry is in the 15 or 20 kilohm
ballpark.

.

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