Re: impulse generator circuit design


brandon.joseph.moore@xxxxxxxxx wrote:
What voltage? What capacitance? What load? (Is it resistive,
inductive, what? What sort of peak current are you looking for? How
long do you expect the output pulse to be? How much energy per pulse?)

The load is going to be mainly inductive (an electromagnet actually).
Not sure about my voltage source yet. Probably be in the 5-10V range.
Figured I'd adjust the capcacitance to get a frequency-pulse power
combination I could live with.

Basically, I'm just looking for a general circuit design that I can
play with to get what I want after I get some idea of what I'll be
using for voltage source and load.


Are you thinking to essentially just have the circuit connect the
capacitor to the load and let it discharge, then disconnect?

Yeah, I think so.

How did
you have in mind charging the capacitor? (That is, would you just put
a resistor between the voltage source/battery and the capacitor, or do
you want to do something more complicated? -- A resistor will dissipate
half the energy provided by the voltage source, if the capacitor is
allowed to discharge to zero volts.)

Well, if I wind up using a battery with a small maximum current, I
might just connect it straight to the capacitor. If there's a better
way to do that, I'd be open to suggestion.

OK, I've looked over the other suggestions, and there seem to be some
good ideas. I'd avoid a PIC: the 555 is a lot better at handling a
range of voltages gracefully, especially ones above 5V. And, it's easy
to use. It can generate narrow pulses easily. I'd recommend the CMOS
version (with due care about zapping the CMOS inputs). You didn't say
what peak current you want, but at such low voltage, and running from a
battery, certainly a power MOSFET can be found to do what you want.

Because you want to have fairly fast recharging of the capacitor and
you're planning a small (?) battery for power, I'd recommend you
consider a couple different things: first, do like one poster
suggested and disconnect the recharge circuit when you're conducting
current in the load. Second, provide some much larger capacitor across
the battery, so you don't draw relatively high current pulses from the
battery. A capacitor likely will be much better at delivering the AC
part of the recharge current.

And finally, if you want the best battery life, by all means use a
resonant recharge circuit. If you connect the battery "directly" to
the capacitor, you will lose the same power in the battery's internal
resistance and the circuit resistance as you would if you put in an
external series resistor. The net resistance changes the time it takes
to recharge, but not the energy lost in heating the resistance. A
typical resonant charging circuit would be: battery, with large energy
storage capacitor, C2, in parallel. Negative side to common/ground.
Positive side to a P-channel power mosfet source. Pull the gate of the
power mosfet low to turn it on when the load is disconnected. Power
mosfet drain to the anode of a diode (Schottky nice, to keep the
voltage drop low). Cathode of the diode to an inductor, L1. Other end
of the inductor to the capacitor you're recharging, C1. Pick the
inductance of L1 such that the resonant frequency of C1 and L1 has a
period LESS THAN two time the length of time you have to recharge C1.
For example, if C1=1000uF and you want a 200Hz rep rate with a
discharge time of 1 millisecond, that leaves 4 milliseconds to
recharge. You need to pick L1 to give a resonant frequency with C1 of
at least 125Hz, which is 8 milliseconds for a full cycle. The recharge
takes place in half a cycle. So L1 should be less than 1.6
millihenries. Since a 1000uF capacitor will have a pretty loose
tolerance, I'd suggest 1.0 millihenries. NOTE: the capacitor will
charge to a voltage such that the average of the final voltage and the
initial voltage (just before you started the charging) is equal to the
battery voltage less a diode drop -- not accounting for a little
additional drop in resistances. So if you have a 9V battery, C1 could
charge to nearly 18V with this circuit. The peak current in the
inductor (important that the inductor can handle this current without
saturating) will be such that Ipeak^2*L1=0.25*Vpeak^2*C1. (That
overstates the current, if you don't discharge C1 to zero each cycle,
but it's a safe limit.) In this case, note that it's a pretty high
current, about 9 amps! Why? Because 1000uF charged to 18 volts
represents (18^2)*0.001/2 joules = .162 joules, and at 200 discharges
per second, that's over 32 watts. From a 9V battery, that's an average
of 3.6 amps. -- In other words, be sure to set you expectations
correctly with regard to how much power it will take to run your little
experiment, before you start. And...C2>>C1; I'd say C2 >= 10*C1 Be
sure to size the parts right for the job.

Cheers,
Tom

.

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