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I've removed a voltage regulator from my circuit and replaced it with a series of BJTs. The regulator was just switching voltage. But I need the circuit (which happens to power/wake the micro) to be on by default. So I've come up with this scheme. I've modeled it and it seems to work well for both turning on and turning off. The load will be limited to 100mA which is also the limit of the BJTs I believe. In truth it will be around 10mA at most. PWRIn will be from 4.5V to 21V. PWR1_Dis will be most likely 3.3V GPIO.

Have I violated any rules with my design? Should I have any protections that I have over looked? Do I have any extra components? Any general advice for this?

Thanks!

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    \$\begingroup\$ So far I only see that you could add something like 10k-47k resistor between base and emitter on Q6B to ensure its turn off. Maybe even on Q6A if the input signal doesn't go low or low enough (below 0.5V). Other than that, I might be missing something in your description. \$\endgroup\$ Mar 31, 2020 at 23:59
  • \$\begingroup\$ @EdinFifić really. If it was a Mosfet I'd do that. I thought bjt cared about current not voltage so with no current path Q6B should just be off? I can't put one on Q6A because pwrin can go 20V and I don't want that on the gpio. Although it's actually within injection current limits. \$\endgroup\$ Apr 1, 2020 at 3:15
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    \$\begingroup\$ BJT cares both about the current AND the voltage, so if you have anything above 0.5V between its base and emitter, it will likely conduct. All transistors have leakage currents, normally for small BJTs that's around 50nA but it could be significantly more. That's why you place a high value resistor (10k-1M) between the base and the emitter of the next transistor in chain, or between gate and source if it's a MOSFET. And you're right, a MOSFET is more sensitive to these small current leakages as it needs almost no current to turn on, but BJTs are still affected to an extent. \$\endgroup\$ Apr 1, 2020 at 7:33
  • \$\begingroup\$ @CarlGilbert Is this circuit to handle the initially high impedance (input pin) behavior that usually is the case during the early part of a power-up process of an MCU? \$\endgroup\$
    – jonk
    Apr 1, 2020 at 9:04
  • \$\begingroup\$ @jonk no it's just a power source. \$\endgroup\$ Apr 1, 2020 at 12:45

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