Power via low-dropout transistor

[nigelmercier]

I'm constructing a device with a PIC and an opto sensor [**broken link removed**]. This has an operating voltage range of 4.75V to 5.25V, and in total draws less than 50mA.

My original design had the sensor powered directly from the 5V rail, but it occurred to me that I could power it down when the PIC is in sleep mode. This is where I hit the problem, to keep the voltage above 4.75V I need a transistor to supply the power to the sensor with a voltage drop Vce of less than 0.25V.

Any suggestions?

 

[betwixt]

I'm not sure this makes sense. How is the opto sensor connected to the PIC?

If the LED side is driven by the PIC, simply turn it off before going into sleep mode.
If you are trying to turn off the sensor side, why not power it from one or more of the PIC pins and make them low before sleeping. If you do use the transistor method you shouldn't have any trouble finding a transistor with VCEsat below 0.25V or you could use a FET instead. You can connect them in the ground side of the sensor if it's easier but you have to be careful the output doesn't look like logic high when disabled if you do that.

Brian.

 

[nigelmercier]

If the LED side is driven by the PIC, simply turn it off before going into sleep mode.
If you are trying to turn off the sensor side, why not power it from one or more of the PIC pins ...

I want to power down both the sensor and the LED, the PIC can not source the current for either.

 

[betwixt]

The PIC must have at least one pin spare for use as the 'power down' signal. On most PICs the pins can provide 25mA maximum and you should be able to use one as a power source given that you only need about 10mA maximum for the sensor.

If you have to use a transistor, connect it in the common ground path of the sensor and the LED current and drive it's base from the 'power down' pin. Even a humble BC237 has a maximum VCEsat of 0.2V and at 10mA it is specified to be no more than 0.07V.

Brian.

 

[nigelmercier]

On most PICs the pins can provide 25mA... you only need about 10mA maximum for the sensor... Even a humble BC237 has a maximum VCEsat of 0.2V and at 10mA it is specified to be no more than 0.07V.

Hi Brian,

Thanks again. Some confusion here I think: the sensor may take 10mA, but the LED needs 30mA, so a total of 40mA.

From what you say, is there any advantage in using a MOSFET over a BJT? I need be sure the voltage is never below specification.

I assume you don't mean a BC237 (NPN) as a choice; I need a PNP here don't I?

 

[betwixt]

Confusion.... it's, my middle name 8-O

From what you describe it should be possible using an NPN transistor. Don't think in terms of interrupting the + supply which would present other problems, do it by interrupting the ground side. As I see it, you can power the sensor directly from the PIC pin and also use the same pin to provide a minimal bias current to an NPN transistor. Connect the emitter to ground and the collector to the LED via whatever current limiting resistor is needed.

When the PIC pin is high, the sensor receives power and the transistor conducts to make the LED operate. When it's low, the sensor has no supply and the transistor has no bias so no LED current flows. As I mentioned earlier, you could leave the sensor + supply permanently connected and return it's ground, together with the LED through the transistor. The drawback to this is that when powered down the sensor supply and ground sides both go to 5V so whatever it is driving needs to be aware that the output will also be high in powered down state.

Clear as mud ?

Brian.

 

[nigelmercier]

Hi Brian,

Thanks for the suggestions, but I don't want to power the sensor from the PIC. The main reason is that the PIC is on one PCB, and the sensor and PSU are on another; about 1m away, and next to a huge DC motor! I'm also a bit wary of switching the ground, as the PCB has a large ground plane.

So my choice is PNP BJT, or P-MOSFET, yes? Can anyone recommend a guide to using MOSFETs as switches?