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Long projectiles suck. The pull force is not directly proportional to the length of the bullet. Doubling the length will not double the force. If you go extreme, double the length will not increase force measurable at all. But the time coil must be energized (and dissipate heat) depends on the time bullet passes the coil. The faster the better. This means that short projectiles mean high efficiency (but higher currents if the energy is the same). I use projectiles 1.25-1.5 times longer than their diameter. Long projectiles mean low speed, high pulse (m*v), low efficiency, mild currents. But long projectiles are not as bad as long coils, so you still can be on a good track, though.
Registered Member #4932
Joined: Thu May 17 2012, 01:42PM
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Posts: 59
Actually, I didn't think spear like projectiles would be better, I just thought they would be cool. they might be more aerodynamic and they have a greater inertia.
Edit: Hey Yan, shouldn't there be a small resistor between the control point and the - of the previous comparator and maybe decrease the resistance of the 100K resistor. maybe 5K and 20K? Also maybe a resistor from the first control point to ground? This way you know for sure that when the projectile passes the second fototransistor, the second stage actually switches on and you know for sure that the - of the first comparator is dominant when they both see the projectile. Or am I wrong? schematic
Edit 2: shouldn't there also be a small resistor between the fototransistor and the control point? Let's say 50 Ohms? This way there's always a small voltage in the circuit, even if there's light everywhere. you would have 0.2v at the - of every comparator and almost 0.05 volt at the + of every comparator. schematic
The inputs of comparators have couple pF capacitance and dozen pA leakage current. It is absolutely insignificant - so just connect that input directly to the control point (resistor in series will simply have no effect). If you want to replace 100K with 20K - it is OK, but 1M must also be replaced with 200K then. This gives 10% voltage hysteresis to ensure difference between inputs. Just build that thing "wires in air", connect LED with 10K resistor to the outputs of comparators and move the finger between gates to see if the thing works. 3-4 gates is enough to be sure. If any output LED will glow dim, this means output oscillates. We expect 1 LED glow solidly having 2 gates crossed at once. For that test, you do not need to build the output stage with MOSFET drivers, neither barrel and coils required - just an optogates glued with thermal glue to some board and wires all around. :)
Registered Member #4932
Joined: Thu May 17 2012, 01:42PM
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Hey Yan, I just tried out your idea (the second schematic) and it didn't work. The LED of the second stage just glowed continuously. I first thought I did something wrong, but I triple checked and everything is according to the second schematic. I can't try out my idea, because the guy at the electronics shop gave me 51K resistors instead of 51 ohm resistors. I'll try out my idea tomorrow and see the results.
Wait a sec. I understand why. Sorry. Try the following: insert 200 Ohm resistor between 1K and phototransistor (each stage). Then - of each comparator goes to the point between 1K and 200 Ohm resistors now. Will you try? P.S.: 200 Ohm could be replaced by a resistor in range [150-510] Ohm. P.P.S: did you measured voltages at phototransistor's collector when it lighted and dimmed?
Registered Member #4932
Joined: Thu May 17 2012, 01:42PM
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Posts: 59
well, I don't have 200Ohm resistors, but I'll get some tomorrow together with some 51Ohm resistors. (I think 51 Ohms is enough isn't it?). Also, I still think there should be a resistor between the control point and the - of the comparator. (Wouldn't there else be no voltage going to the + of the comparator, because it's actually a voltage divider and if you have a 20K or 100k path and a 0Ohm path, wouldn't there be no voltage at the + of the comparator?) So I would think you should put a resistor that's about four times as small as the 100K or 20K resistor between the control point and the - of the comparator. So you would have 1V across the + of that comparator and 4V across the - of the previous comparator if the fototransistor doesn't see light and you would have between 0.05V and 0.2V across the + and between 0.2V and 0.7V across the - of the previous comparator when the fototransistor does see light. This was my idea for the fourth schematic I made.
You know what, I'll try again this saturday. I can make a better test setup this friday and then I'll try everything out saturday. I'll hook up my bro's oscilloscope to two outputs and I'll hook up my multimeter to another output. this way I can test it with 3 stages so I'll have a good idea of what works. For the fototransistors and leds I use BPW96B and OPE5685. I'll get 2 (multiplex?) boards and drill a few 5mm holes (every 5cm?). I'll let them face eachother and slide a board in between and watch if something happens. If I can find a way it works, I'll tell you. If I can't find a way it works, I'll tell you too of course. (This is the schematic I'm going to test next.)
OK, I've been masturbated enough to draw a final schematic for you. It is simple, fast, and will surely work (because it features my favourite long-used FET drivers). Order a dozen of MCP14E3 (or MCP14E9 if E3 are not available anymore) and here is the simple solution with just 1 resistor per stage:
Those drivers are referred as "Double Inverting MOSFET driver with Enable". Single chip has two equal drivers in it. Each driver has inverting input and "enable" input. If voltage on Enable pin is low, the output of the driver is forced to be low no matter what input wants. If Enable is high, then output is determined by the input - if input goes low, output switches to high; if input goes high, output switches to low. So here is what we are doing: we connect Enable to the control point of next gate. This way output is allowed to be high ONLY if next gate is lighted. If it is lighted, then output value depends on the input: if control point of THIS gate (connected to In) has high potential (when ph is lighted), then output is low. If it falls close to ground (gate dimmed) output swings high turning on the IGBT. Once the gate got lighted again (or next gate got dimmed) the driver closes the IGBT.
The main difference between MCP14E3 and MCP14E9 is that the old E3 is 2 times more powerful (4A peak output current) - it gives faster switching time for your IGBT. The Enable function is not very common among FET drivers, but I found it extremely useful, though. Inverted In and Enable is equal to logical AND element with one inverted input. Mention that those drivers are certified for capacitive loads - that means no resistor required between output of the driver and IGBT/MOSFET gate. It doesn't mean that they will not burn if you SC their output with Vcc or Gnd, though - it just assumes that heat dissipation limitations are not exceeded. It will drive your IGBT directly just fine, but if IGBT will burn melting the gate - driver will go in smoke too.
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