Re: Liquid level indicator



MarkMc wrote:
> Hello again Chris
>
> >
> > 10K resistor R6 should be pullup, not pulldown. Its function is to
> > ensure turnoff when no current is being pulled from the emitter. By
> > that logic, it should go to +12V. In fact, if you leave it as a
> > pulldown, the LED transistor just won't turn off at all.
> Woops. Yes, I should have known that. I managed to go through school
> and college electronics and *never* managed to use a PNP transistor, so
> this is my first application! Although electronics was always a
> secondary subject for me, with Computer Science and Software
> Engineering being my primary focus.
>
> > You might have more luck finding a 470 ohm resistor (standard value)
> > for R4 than 480 ohms (a typo, I'm sure).
> Oops, yes, I meant 470.
>
> > When you calculate power
> > dissipation for the resistor, you have to subtract the voltage across
> > the LED and the saturated transistor. Let's say that there's 1.8V
> > across the LED, and 0.2V across the saturated ON transistor. That
> > leaves 10V across the resistor, which means about 21.3mA and .2128
> > watts. Actually, I'd think you can go with a 560 ohm resistor, and you
> > can save some power as well as not getting too close to the 1/4 watt
> > limit, especially if the LM7812 output voltage is a little on the high
> > side.
> Is 1.8v drop pretty standard for an LED then? I didn't know what
> values to use for the LED and c-e junction of the transistor, so I
> erred on the side of caution and just used +12v. Should've used at
> least 0.6 for the LED though, so that was a bit dumb of me.
>
> I only have a set of E3 resistors, so I don't have 560. Perhaps I
> should look to get a set of E12 to give me a few more choices?
>
> > As far as power requirements for the transformer, think expansion. You
> > want to have enough power so that, if you want to add another relay or
> > something else, you don't have to redo the power supply....
> A friend of mine is in to d-i-y Valve hi-fi, and I nicked this cct from
> him and simplified it for my own use, I guess it lost something in
> translation. He had other types of cap in parallel with C2 IIRC.
>
> I assumed that I'd just want as big a cap as possible for C1. I know
> PSU's are a big big topic, so I'll trust what you suggest and go with
> that. I do want to prepare for the future - the heater side of things
> for example
>
> >
> > KU50E Clip On TO220 Heat Sink 759 in Stock £0.39
> > N80AZ 10W 1x15V PCB Transformer 8 in Stock £5.99
> thanks for the suggestions.
>
> >
> > I'm looking forward to hearing how things are going -- feel free to
> > post back, or email. Good luck, and
> I'll definitely post back on progress. I'll probably put this together
> on a bread board first, and test each section in my spare time, so it
> might be a while!
>
> Thanks once again Chris. I'll be sure to give you credit on my web
> site!
>
> Cheers,
> Mark

Hi, Mark. LEDs vary quite a bit as to their forward voltage drop.
Red/yellow/green vary all over the place from 1.5 to over 2.5V. White
LEDs typically have a Vf of around 3V. Infrared LEDs typically have
about 1.7V. The key is to try to work around variability. If you've
got the basic collection of resistors, try two 1K 1/4W in parallel.
That will do the job nicely (net 500 ohm 1/2 W).

Your friend's circuit is perfect for this application -- a small linear
7812-based power supply. I was just kibbitzing to give a hand with
component selection. If you're using a 12.6V transformer, you need as
big a cap as possible, to minimize ripple. That's because once you
subtract two diode drops, the peak voltage at the cap will be around
16.2V at nominal line voltage. If you have any drops in line voltage,
you'll get close to the (Vout + 2.5V) = 14.5V dropout voltage of the
7812 regulator.

The rule of thumb is that an 8300uF cap will give you one volt of
peak-to-peak ripple with a current of 1 amp. That means 10000uF for
1/3 amp will give you roughly 1/4 volt of ripple, which would do just
fine.

If you've got a 15VAC transformer, you'll have 19.6V peak after the two
diode drops, which means you can afford about 5V of ripple. A 2200uF
cap will give you a little over 1-1/4V of ripple, which is very safe.
The extra capacitance is basically just extra expense. Actually, I was
looking for and couldn't find a 1000uF 25WV electrolytic on the Maplin
website (I don't have a catalog).

All of this kind of assumes you really want to do it this way. After
you put a PC mount transformer, bridge rectifier, caps, and linear
regulator on your PC board, you've still brought line voltage out into
the box. I'd definitely recommend purchasing a wall wart with a linear
12VDC regulator built in. That will do the job, keep line voltage out
of the control box (always a good thing), and simplify things by quite
a bit. You'd just put a 100uF cap on the board where the power supply
is connected. Here's one from your friends at Maplin:

JC92A 12V 500mA Mains Adpt 5 in Stock £14.99

If you're etching a board, it's good that you're testing everything
first. Remember that you should use a bench power supply instead of
the wall wart regulator until you're sure everything's OK. The one
specified says it has thermal fuse protection, which helps prevent
fires, but is a little unforgiving if you have an accidental short
circuit on the breadboard.

Feel free to post again or email if you need a hand.

Good luck
Chris

.

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