| Electronics > Beginners |
| High-side vs. low-side NPN transistor |
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| iMo:
@Zero999: his base is not 3.3V and the emitter resistor is not 1k.. |
| rstofer:
--- Quote from: imo on January 12, 2019, 05:22:58 pm ---@Zero999: his base is not 3.3V and the emitter resistor is not 1k.. --- End quote --- That doesn't matter. His base is 3.1 and the emitter is about 0.5V less which is as expected. Within reason, the value of the load resistance doesn't change things very much. There will always be a one diode junction drop between the base and emitter because there actually is a diode in there. You simply don't use NPN transistors for high side switching unless you can get the base one diode junction drop higher than the required output voltage and this is usually 0.2V less than Vcc (the Vbe(sat) voltage) In other words, the base needs to be about 0.3V higher than Vcc to get the maximum output voltage. This is often very difficult to do. Short answer, don't use NPN for high side switching. That's why they invented the PNP variant. |
| iMo:
It does not matter from EE perspective, sure. It does matter when the OP is showing a specific schematics and asking on specific voltage at specific node :) |
| rstofer:
--- Quote from: imo on January 12, 2019, 05:40:02 pm ---It does not matter from EE perspective, sure. It does matter when the OP is showing a specific schematics and asking on specific voltage at specific node :) --- End quote --- The OP shows an output voltage and no load resistor, hence infinite output resistance and zero output current. Therefore, Vbe is going to be as low as it will ever be - about 0.5V for the 2N3055. This value will increase as the load increases as shown in Figure 8 of the datasheet I linked above. At 10A, Vbe(sat) could be as high as 1.8V but the OP circuit isn't delivering 10A, it's delivering zero current. The reason the voltage has to rise is that the transistor is going to require a LOT of base current - probably on the order of 1A, and to push 1A through the base-emitter junction is going to require higher base voltage. |
| permal:
Lots of people chiming in now. I appreciate the attention. :-+ @Spec: With a 1k resitor what I get is this: V across resistor (so to ground): 2.50V Vb = 3.07V --- Quote from: spec on January 12, 2019, 04:39:36 pm ---It would be much better to use MOSFETs for the switching rather than BPJTs like the ZXTN25050FHTA, Are you OK with MOSFETs? --- End quote --- I don't know. I went with "regular" transistors because I though I understood them (which I've now learned that I didn't fully). All I want is to be able to turn on/off 5/12V (250-500 mA) on eight different outputs. I've already overcome lots of other obstacles in this project so MOSFETs would just be one more thing to learn. What's the benefit, lower gate(?) voltages? @imo / @Zero999: I see your discussion, but opt not to comment on it at this time as I'm trying to understans what you're saying. It has become apparent to me that my understanding of transistors have been slightly off. |
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