From: Jonathan Kirwan
Subject: Re: PIC controlled variable power supply
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NNTP-Posting-Date: Tue, 21 Jan 2003 10:30:43 GMT
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Date: Tue, 21 Jan 2003 10:30:43 GMT
On Sat, 18 Jan 2003 15:53:55 -0700, Jim Drew
>Yes, this is powered by a 9v battery. What I need to accompish is to
>activate a medical solenoid for a period (variable) between 4ms and
>20ms, and I need to be able to adjust the coil output voltage from 5v to
>about 8v to accomodate different solenoids, by just programming the
>PIC's EEPROM on the fly. Once the voltage is set, it won't really
>change, and any type of manual switch is out of the question. The coil
>draws nearly 1 amp of current the first 2ms, and then has a hold current
>of about 470ma there after. During the idle time (when the solenoid is
>not being used), the circuit must have a low power mode, where the
>entire piece of hardware draws no more than 5ma, which is easy enough to
>accomplish unless something has to be sitting drawing power as you have
Just some notes for your consideration, since some time has gone
by without more on it...
If your 9V battery is typical of alkaline (Zn/MnO2) batteries,
then their voltage will drop to some 8V after using up only 8%
or so of their stored energy. If, say, you program your unit to
supply 8V to an 8V solenoid coil and the battery voltage drops
below this (plus some minimum headroom), then you cannot depend
on linear regulation and will need a switcher involved. In this
case, a buck-boost I imagine. Otherwise, where do you imagine
you'll get the voltage?
If you really imagine that you don't need a switcher and can
survive on only, say, what's needed for a low-dropout linear
regulator as the source voltage for this variable regulator,
then you will have to figure on the system failing once the 9V
battery drops to some 8.2V or so, at best. Perhaps higher than
that. At that voltage, the 9V battery has only supplied some 5%
to 7% of it's total capability.
The spec I'm reading says about 600mAh for a 9V battery. Let's
assume 7%, that's .07*600mAh, or 42mAh. At say an average of
8.6V or so, let's call it about 1300 Joules, total. Let's
further assume you can deliver that energy at a 100% efficiency
to the solenoid coil -- nothing dissipated in your variable
regulation scheme, nothing tossed away for a micro, nothing used
at other times, nothing wasted anywhere -- all of it goes to the
coil, all the time, every time.
The coil requires some 1000mA for 2ms at 8V and some 470mA for
the rest of the time, which can be an additional 2ms to 18ms.
Let's assume the worst, 18ms. We'll round it up to 0.1 Joule
per relay actuation. Of these 'worst case' switches, you'll get
some 13,000 total pulses before needing to replace the battery.
Assuming 100% efficiency, of course, and 7% of the battery and
being very general about how the solendoid actually consumes
energy. Batteries might give you 5% or less, at times, too.
With a 1mA draw from your battery, you'll also be using over 32
Joules/hour when not activating the solenoid, giving a total of
40 hours or so, even if you don't activate it once. And each
activation takes 11 seconds off of that. I can see your desire
for low consumption during inactive times, if you want to
stretch past a day or two.
So some questions are... how long do you need the battery to
last, at minimum, and how many solenoid actuations must it
deliver in that time, at minimum?