Re: Switching +/-12V from 6/0V



George wrote:


As an amateur at all this, I'm afraid my knowledge is too
limited to understand all of these different versions
everybody is talking about. In this case, I don't
understand what keeps all the current from the 555 from
flowing through the base of the PNP to GND in your design (with
no resistor there),...

As long as the transistor remains in the active region, the base current Ib will cause Hfe x Ib of emitter to collector current Iec=Hfe x Ib. Since you know that the emitter terminal is at Veb~0.7V with respect to GND, and the input is at 6V, you must have 6-0.7=5.3V drop across the 1K emitter resistor. So the emitter resistor is carrying 5.3V/1KR=5.3mA and this is the sum of Ib and Iec, or 5.3mA=Iec+Ib=Iec+Iec/Hfe=Iec*(1+1/Hfe). With a typical Hfe of say 100, this makes Iec=5.3mA/1.01=5.25mA, leaving 5.3-5.25=0.05mA for Ib. So you see all the current does not shunt through the base to GND, because the base current causes the transistor to shunt the majority of current entering the emitter to the collector. The PNP collector is clamped to a Vbe above -10V by the base of the NPN, so this keeps the PNP junction well into reverse bias keeping the transistor in the active region. Your circuit on the other does not


The way I've done it seems to work. I've tested it for a
while, and it trips the shutter reliably. As for the
current requirement, I guess the maximum draw would be at
the beginning of the discharge when the voltage is at its
maximum of about 22V, with the DC resistance of the coil
being 100 ohms. So maybe it shouldn't work, but it does.



At a DC resistance of 100 ohms, the solenoid draws 20V/100R=200mA, so that as a minimum you would want 1/10 that into the base or 20mA. Use the calculation I just showed you to adjust the drive.

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