Re: Heathkit IP-18 Battery Charging and Fire

Oct 23, 2010 6 Replies
F


I don't see the need for that whole tap on the


> transformer. It's whole purpose seems to be to bring the anode of the
> 1N191 negative for full power. Wouldn't tying it to ground work just
> as well? Maybe its isolating that transistor from the output current.

Tying the 1N191 anode to ground would appear to cut off the 2N3416 it is attached to, which would provide maximum voltage output from the supply. But then you could not adjust output voltage.


The more I study the circuit, the more I think of Rube Goldberg! It might be possible to equal the performance by using one 723 voltage regulator IC with the 2N3055 pass transistor. In that case, the 40 Volt winding might be useful for independently powering the 723. If interested, you might find a suitable circuit on a 723 data sheet.


A fuse would be nice in the event of catastrophic failure. (A fuse is a $1 device that protects a $.10 transistor!)


Fred

Vote
Twitter Facebook LinkedIn
W

=A0

t
a

Yes, you're right. I meant tying the pot between ground and the input voltage, but that wouldn't work. That bottom transistor is the feedback loop. It's like a puzzle to see how few transistors you can use. Just five transistors but look at all the technology. A JFET, three kinds of NPNs, silicon rectifiers, a zener and a germanium diode.

Vote
Twitter Facebook LinkedIn
F

Yes. Another approach is the LM317. Just one power IC and a pot. Its only fault is that output can't be adjusted below about 1.2 volts. I wonder if a negative power supply would help?

Fred

Vote
Twitter Facebook LinkedIn
E

Yes. You can drag adj pin below ground & get the range from 0 to +15 (or whatever, withn ratings).

Ed

Vote
Twitter Facebook LinkedIn
J

actually, its not the adjust pin that is doing it, its the fact that you take the supplies common (-) and drag that down a bit via a couple of diodes for example to the outside world..

This also takes away from the top but most of the time this isn't an issue..

Vote
Twitter Facebook LinkedIn
E

Wrong. The 317 Vout pin will always be at least Vref (about 1.2V) above the adj pin. To get a range starting at 0, you must drag adj at least Vref below 0.

Perhaps you'll post a schematic of what you have in mind.

Ed

Vote
Twitter Facebook LinkedIn
W

. =A0

n

The cap on the half-wave rectified circuit popped. I think it was just old. Funny, the IP-2728 schematic is the same as the IP-18 with 2

1N4002 diodes removed and a fuse added.
Vote
Twitter Facebook LinkedIn

Join the Discussion

Have something to add? Share your thoughts — no account required.

Didn't find your answer?

Ask the community — no account required