| Electronics > Beginners |
| Comments about this simple circuit that is teach in a university...... |
| << < (4/4) |
| rstofer:
--- Quote from: Belrmar on June 12, 2018, 11:39:19 pm ---for this kind of circuit i tend to prefer yo use a MOSFet, as it is less likely to overload the pin when correctly used, and i would add a resistor to the capacitor to prevent LC oscilation --- End quote --- And, specifically, a 'logic level' MOSFET so that we don't need to deal with 10V Vgs. Still, it pays to read the datasheet because there is a minimum Vgs which will turn the device fully on. 'Logic level' is often more marketing than engineering. A 5V output pin might get the device turned on, a 3.3V output pin probably won't. It's the same story, Vsd needs to be small, the device needs to be fully turned on. |
| westfw:
--- Quote --- Base current is 10 mA and the input to the diode will be around 1.4V (2 diode drops). So, for a 3V Vdd on a 16F877A, we need a Voh of 1.4V @ 10 mA. --- End quote --- Gain of the transistor is irrelevant to loading of the pin. Nothing LIMITS the current through the diode or base/emitter junction. Two diode junctions between the output pin and ground (heh. About equivalent to directly connecting a red LED to the pin...) |
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