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1

Consider the two transistors in the current mirror. With only 1.8R emitter resistors, any 1mV difference in Vbe between both transistors will cause a difference in current of 0.55mA. This transistor's hFe stays above 200 even at 100mA, so base current is not the main cause of your current mismatch. The two transistors are in the same package, but the ...


1

You are missing the base currents of Q1.1 and Q1.2: they all go through the LED string, so your mirror is unbalanced at low currents.


0

The transformer is a typical automotive coil: LP = 8mH RP = 15Ohms RATIO = 1:100 LS = LP * POW(RATIO,2) RS = 100kOhms I did not have time to put proper models in for suitable transistors and diodes, so pick a 27V zener diode in a large package. A push-pull driver rather than a single PNP will produce larger voltages as it's the rate of change of current ...


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You are operating the transistor as an emitter-follower in reverse mode (it's not a good circuit, you should just buy an S8050, they're good to have around so get a few NPN and PNP). The beta is quite low so the Pi will try to sink a lot of current (on= low, by the way) and with it high @3.3V, it may not fully turn the LED off. As an emitter-follower, it ...


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The RPi outputs are usually \$3.3\:\text{V}\$ and they probably can handle saturating your PNP as a switch. But the problem is the \$3.3\:\text{V}\$ limit (which assumes zero sourcing current.) The I/O pin can sink current when driven high, but that's not controllable as it is only through a protection diode -- which, in general, you should NOT rely on for ...


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Not sure how you are getting the 3.5mA, but I can comment that this is a pretty weird driving arrangement. You should put the PNP up top, with the emitter at 5V and the collector attached to the LED. Alternatively you could swap out the PNP for an NPN and drive it from the low side. *** To be clear I find this weird because this really isn't a switch ...


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