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| Linear regulator boost with NPN transistor |
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| sureshot:
I was looking at this simple circuit, what are the negatives of using an NPN transistor to boost current of a linear voltage regulator ? I think there would be a forward volts drop. Would a diode in the ground path resolve that problem, i don't know the name for it but it increases the output by 0.7 Volts per diode. And finally what are the implications of there being no current limiting base resistor. Thanks for reading and any advise appreciated. |
| T3sl4co1l:
Yah, you've hit all the main points. Dropout voltage (minimum input = output + droppout) is ~0.7V higher. Regulation is worse by the "softness" of the diode curve. It's not inaccurate by a diode drop, because of the diode compensating it. But the diode's current is ~constant (whatever the regulator GND current is), while the transistor's Vbe depends on current draw. This, plus transistor resistance, gives maybe 100mV of variation. Worse than a 7812 -- but not bad for most purposes. No current limit, or rather it's hFE limited, which is a bad idea all the same. If you need that capability, expect to add more components to implement it. Note that it's not as simple as pulling down the base with a current sense resistor and transistor, because the 7812's current is what has to be sunk. Nor can you quite pull it off with a LM317, which has 1.25V output, more than the Vbe drop so you can't quite completely turn it off by pulling ADJ down to GND (with a few transistors wrapping around to do that). Tim |
| Zero999:
Yes, the base emitter voltage increases the drop-out voltage and worsens the regulation. No base resistor is required, because the base current is self-limiting. When the transistor turns on, the voltage at the emitter rises, reducing the potential difference between the base-emitter junction, therefore the base current, until an equilibrium is reached. This is an example of negative feedback. Look up emitter follower. An NPN transistor can be used, with a PNP transistor, as per the schematic I posted in the other thread. https://www.eevblog.com/forum/beginners/voltage-multiplier-120443/msg1648847/#msg1648847 |
| sureshot:
Thank you for the help, i think i I've got most of what you've explained there. I did think about the LM317 in that configuration. Just a question about this circuit using the 7812 voltage regulator. What would be the implications on the transistor at say 3 Amps, would it thermal runaway ? And where would you see the voltage drop under that load ? With a zener diode and fuse for protection, would it be capable of supplying a cb radio with an RX current of 250mA, and a TX current of 1.2 Amps ? I know the 7812 is almost capable of that. But thought it might be suitable for that application. I do have some thyristors if a crowbar circuit would be better than just the zener and fuse protection. Thanks again for your help. |
| sureshot:
--- Quote from: Hero999 on July 09, 2018, 10:43:42 am ---Yes, the base emitter voltage increases the drop-out voltage and worsens the regulation. No base resistor is required, because the base current is self-limiting. When the transistor turns on, the voltage at the emitter rises, reducing the potential difference between the base-emitter junction, therefore the base current, until an equilibrium is reached. This is an example of negative feedback. Look up emitter follower. An NPN transistor can be used, with a PNP transistor, as per the schematic I posted in the other thread. https://www.eevblog.com/forum/beginners/voltage-multiplier-120443/msg1648847/#msg1648847 --- End quote --- Thank you for the schematic, i could put that circuit together easily enough. I just like the idea of a circuit that needs little parts count, works reasonably efficient and gets the intened item powered up. I was just curious of the behaviour of the transistor with no base resistor. And the overall performance of the circuit up to 3 Amps maximum. Thanks again for the help with this. |
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