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JC's Power Save Circuit

A

Anonymous

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******SEE THE ATTACHED PICTURE (I hope)*********
JC, I think that this is what you were talking about in your message. I have to give it a try.
By the way, posting a circuit means that you have a year to apply for a patent in the <IMG SRC="/forums/images/flag.jpg" BORDER=0 ALT="USA">. You do however lose rights to file foreign patents.
Others can copy what you have posted (unless you apply for a patent in the next year) They however cannot patent your work.
I like the idea a lot. I am sure that there must be a few other methods that will also work.
All the best, Dave Emery. * * *
 
Hi Dave,
More importantly, what program do you use to draw your schematics?
Thanks,
Charles
 
Hi Dave
You got it exactly. Many thanks. Part of the trick will be to get the value of C1 right, with the inductance of the coil and the transmit pulse width (depends on what you are looking for). Probably just experiment till the current rises up fast and flattens off. This is what you have to do after all the calculations anyway.
Again thanks, and for the patent info, I did not know how this works. Just trying to be funny about the patent stuff, seems to be part of the current topic. Too much to do to write up a patent, if it really is useful maybe someone will build it and everyone can benefit.
I haven't had time to build it and try it, I got even more exciting things to work on, at the moment. Still concerned with the transistor staying off while its emitter is flying up in voltage.
JC
 
Hi Charles,
I use an el cheapo program called "Design Works Lite". You can try it free for 30 days. If you like it, you get the software key for $39.95.
This is the best deal I have ever found for a schematic program.
My big trick is to use another cheap program called "Print Screen Deluxe" to capture the circuit from "Print screen Deluxe"
Print Screen Deluxe allows you to conver the image to gif or jpeg or about any other existing format that you can imagine.
I have provided the link to Design Works lite below. Just in case it does not work the URL is:
http://www.dwlite.com/index.html?2 All the best, Dave Emery. * * *
 
Whoops, one more thing, probably depending on the resistance of the coil, there will need to be a current limiting resistor in series with the coil or the source (bottom terminal) of the mosfet to keep the current at reasonable levels to start out with. Later on drop the resistor down and get the current up, still with fast current rise time.
If I get a chance I may give this a try as well.
JC
 
See below for the URL for "Print Screen Deluxe 4.0 that I use with the schematic program. This is also a great value and it is very useful.
I would really hate to be without this program. I use it many times a week for all sorts of things.
The program will open all sorts of files that you migh normaly have a problem with. It also allows you to do screen captures and let's you store or convert files from one format to another.
I will put the URL here in the message just in case of problems.
http://www.americansys.com/psd.htm?From=Main
Have fun!!! Dave Emery. * * *
 
Sorry, Chop off one of the extra http:// off of the URL. Dave Emery. * * *
 
Hi JC,
Just looking at the schematic, there may be a small problem with the drive pulse to Q2. After the voltage flyback pulse, the emitter of Q2 is going to be sitting at a +100 or more volts. To turn Q2 on, the drive voltage, with respect to ground, will have to exceed this.
Eric.
 
Of course being a PNP, the base need to be a bit negative w.r.t. the emitter.
Eric.
 
Should be able to pull the base down w.r.t. the emitter and turn it on. I guess I need to give this a try and see if it really does work. The only PNPs I know I have are 2907a and 3906 of which I have a hundred each, but they are lower voltage VCE. So I will have to do this at less than 100 volts. There are tricks to series xsistors and get higher voltage, or order some.
Aren't enough hours in the day, days in the week, or weeks in a year, to get all I want to do done.
JC
 
Thanks Dave,
Funny, its probably the capture program that would help me more <IMG SRC="/forums/images/smile.gif" BORDER=0 ALT=":)">
Thanks Dave, you're always helpful,
Charles
 
No need for current limiting resistor. Just don't turn the transmit transistor on so long that current rises to unwanted levels. Given the inductance and the drive voltage, that's a fairly straightforward calculation-- 1 ampere = 1 volt second divided by one Henry,.... and you just scale everything from that relationship.
Of course, in your "jump-start" system, you have to solve for two different volt-time intervals and then sum the currents.
--Dave J.
 
Yea, I know, but still need resistor just like without jump-start to finally limit current to some value if one is going three or more time constants of the coil/resistance combination.
JC
 
Some of us cannot see the jpg because the URL we get is bad. It seems to be too long and is getting cut off. Could you also put the URL in the text?
Robert
 
JC:
Seems to me that running the time-constant out kinda defeats the advantage of the jumpstart system.
If the idea is to lengthen the transmit pulse in order to find high-conductivity and/or iron targets, it would seem to make more sense to raise the inductance and Q of the coil in order to stretch out the pulse and have the whole thing be more efficient. If, however it is necessary to retain sensitivity to the tiny stuff, and an IB loop isn't being used, and there's plenty of battery horsepower available, then the series resistor might make sense, and the "jump-start" system would have so little advantage that it could probably be dispensed with.
--Dave J.
 
Following what others like Eric have been explaining, is that the eddy currents induced into the object during the first couple of time contants of the coil current (when the rate of change of current in the coil is such that it is still inducing eddy currents into the object) then for the next say three time constants the current is more constant and since it is no longer changing it is not inducing current into the object. The eddy currents in the object induced during turn on can now die down and depending on the objects of interest depends on how long this takes, and how long you need to keep the transmitter on.
The only advantage to the jump-start is that the coil can get to constant current quicker and the process of the eddy currents in the target dropping can begin quicker. This means that for the same constant current time during the transmit period the transmit pulse width can be shorter. This means that you can get more pulses in a second (my goal), get more sensitivity (more samples per second){higher frequency filters better in the integrator), since you are getting the same effect on each now shorter pulse.
Or you can shorten the transmit pulse and keep the rep rate the same and save that power and have the same sensitivity. And you get a little bit (not alot for long pulse times) of energy back from the storage cap. For small objects where the transmit pulse width can be as short as 50 us, you will get a much better return than if the pulse width needs to be 250 us (for larger high conductivity objects).
Or you can not use it at all. Your choice. I'm done.
JC
 
It's possible to scale everything so that during the main part of the "transmit pulse", current is relatively constant without having to throw in extra resistance. Explaining:
Say the coil is 1 millihenry, and assume no resistive losses anywhere. Its response is 1 ampere per volt-millisecond. If we apply 100 volts for 10 microseconds, we'll have 1 ampere.
Looking at it in terms of current, since the current ramped from zero to 1 ampere, the energy drawn from the 100 volt supply was 5 ampere-microseconds.
If at this point we short out the coil (i.e., apply zero volts), the current will continue at 1 ampere indefinitely.
If we apply 10 volts, the current will continue to ramp up at the rate of 10 milliamperes per microsecond. If we keep the thing turned on for 40 microseconds more, the current will ramp up to 1.4 amperes.
On flyback, if we clamp it to 100 volts, the duration of flyback will be 14 microseconds. Over those 14 microseconds, 1.4 amps will ramp down to zero. In terms of current, the energy put into the 100 volt supply during flyback was 9.8 ampere-microseconds. We came out ahead 4.8 ampere-microseconds compared to what we used up for jumpstart.
Notice in this example that we dumped nearly 100% more energy out of flyback, than we originally used to get the thing jump-started. Also note that of the 1.4 amperes delta current that happened during the total 50 microseconds transmit on-time, 71% of the change happened in the first 20% of the transmit duration.
It isn't a perpetual motion machine. The current through the transmitter switch starts at 1.0 amperes and ramps up to 1.4 amperes over 40 microseconds. From the standpoint of current, the energy drawn from the 10 volt supply is 1.2 amperes x 40 microseconds, or 48 ampere-microseconds. If you convert it to joules, it's equal to the energy difference between the flyback and the jumpstart, there being a 10:1 difference in the voltages.
Of course no system has zero loss. What I'm trying to illustrate using a simplified example is that the objective of keeping current relatively constant over most of the transmit period can be achieved without deliberately introducing losses, and with careful design, the efficiency could be limited primarily by the Q of the coil.
It is advantageous to have a large ratio between the high voltage and low voltage supplies. If everything is scaled so at the end of the jumpstart pulse, the voltage being dropped across the resistance of the coil is equal to the low voltage applied during the main transmit pulse, then the net voltage dropped across the reactance will be zero and the current will remain constant. If the coil circuit resistance is 3 ohms and the sustaining current is 2 amperes (just as an example), the system would be scaled to apply 6 volts during the main transmit pulse. (In most cases you'd apply a little overvoltage to make up for losses elsewhere in the system, rather than inject the extra energy elsewhere.)
--Dave J.
 
implement this then we can get rid of the current limiting resistor. And save a bunch of power.
JC
 
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