BJT switch

Thread Starter

colinb

Joined Jun 15, 2011
351
I need a circuit, simple, small, and cheap, that can switch on and off a 7.2 V power rail, controlled from a logic level (3.3V) digital signal. It needs to be very low leakage when the rail is off since it's in a low-power battery based system. The power rail is enabled for a small portion of the time (less than 1% duty) but may draw 300 mA or more during the several seconds of active operation.

The first design was a two-resistor version (base resistor on the lower transistor and a resistor between lower collector and upper transistor's base).

Later, I tried to reduce parts count by replacing the base resistor and collector resistor with a single emitter resistor. By my calculation, this creates a constant current sink through the top transistor' E-B junction when the logic signal enables the rail.

I have a few questions:

(1) Does this circuit look ok to you? Any possible problems?

(2) Is there any reason to prefer either the two-resistor circuit or the single-resistor circuit in this application?

(3) Is there a name for these particular transistor circuits?

Note: LTspice files attached.
 

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crutschow

Joined Mar 14, 2008
34,285
Both circuits look okay. If you want to save a resistor then use the second design. I see no particular reason to prefer one over the other otherwise.

If the circuit is to work at an elevated temperature then you want to put a 1k resistor between the base and emitter of Q2 to sink any leakage current which could turn Q2 partially on.

Such a circuit is generically called a high-side switch.

You could also use a logic-level P-MOSFET for Q2. With that you could use a much larger drain-source resistor to bias Q2 off when Q1 is also off then the 1kΩ required by the BJT. This reduces the current needed through Q1 to turn on Q2. But that apparently is not a significant factor in your short duty-cycle requirements.
 
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