That circuit will take about 6 to 10 mA roughly. That means that base currents on the order of 10 to 30 microamps or so will partially turn on the transistor, doesn't it? Would we get these kinds of emitter-base currents if there is only 400 mV across this junction?
That circuit will take about 6 to 10 mA roughly. That means that base currents on the order of 10 to 30 microamps or so will partially turn on the transistor, doesn't it? Would we get these kinds of emitter-base currents if there is only 400 mV across this junction?
That circuit will take about 6 to 10 mA roughly. That means that base currents on the order of 10 to 30 microamps or so will partially turn on the transistor, doesn't it? Would we get these kinds of emitter-base currents if there is only 400 mV across this junction?
According to a simulation of a 2N3906, 400mV Vbe will yield Ib=257.8pA. 700mV gives 27uA.
This is just a sim, but still, 400mV ain't gonna turn a real one on.
Yeah, how come? Good question. Even if those zeners had leakage currents on the order of 25 uA or so, pretty much all of that would flow through the 1K ohm resistor and not through the e-b junction so that doesn't explain the turn-on of the transistor.
Did you make R5=10K, R4=1K, and change the program so I/O P4 is a high impedance input during the pulser-off periods...as I had suggested back on Jan 27th? I noticed the schematic you posted was my earlier version that indeed had those problems. If my circuit didn't work I would like to know about so I can correct it.
Did you make R5=10K, R4=1K, and change the program so I/O P4 is a high impedance input during the pulser-off periods...as I had suggested back on Jan 27th? I noticed the schematic you posted was my earlier version that indeed had those problems. If my circuit didn't work I would like to know about so I can correct it.
Ken, i went back and looked at all your Jan 27 posts, and I couldn't find any schematics posted.
I don't think high-impedance is gonna help. The PIC has a diode from each I/O pin to vcc, so the current will flow through that diode, and the voltage will get clamped to +5.7V. It's not like a true open-drain output.
I was emailing Dan off-forum. I had supplied my circuit, with an untested output section, in another thread. He came back with the stated problems. I recommended the changes in the attached schematic, and to the program. The last I heard from him it was working. ??? You are right about the diode...forgot about that...thanks! I guess the next step would have been a NPN inverter for isolation between the P4 and R4...and changing the program accordingly. Or, no R6, and a 2N7000 instead of a 2N3904...or a duplicate of the right side's P-MOS and 2N7000.
Ken,
I didnt forget about you,Ive been having email problems(computer freezes when I try to send) I figured youd drop in on this eventually.
Everything worked fine till I increased the batt voltage.The pulser output was dropping too low before the 2nd cycle had finished.I didnt realize until Pearl started developing a cough again.
Somebody already asked this, but I don't think you answered.
Do you really have to switch the high side? Since the zapper was originally battery powered, I don't see why you can't switch the low side. It will save you a transistor, and a resistor or two.
The original schematic is switched on the high side,but I dont think it really matters as long as The output turns off completely.
If the low side is switched, wouldnt the high side stay energized all the way to the output? If so, then I would rather switch the + side. I dont know if that would affect the treatment.
The original schematic is switched on the high side,but I dont think it really matters as long as The output turns off completely.
If the low side is switched, wouldnt the high side stay energized all the way to the output? If so, then I would rather switch the + side. I dont know if that would affect the treatment.