This is a trivial question but I thought I'd ask it anyway:

If you're using an NPN transistor in a switch-configuration to power a small devices (drawing about 5mA), is it better to leave the Vdd-end of the device connected directly to the supply and have the Vss-end switched by the transistor or vice-versa? Or does it not make ANY difference?

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hi nj,

I try to have the Vdd end of a device to +Vsupply and switch the Vss to 0V thru a npn transistor, this is a common way to do it.

The reasoning behind it is, if the npn is the low end of the device, the low end is held within about 0.1V of 0v [gnd], when the npn is saturated.

If the npn transistor is in the high end of the device, you will always 'lose' at least 0.8V from the Vsupply line level,
because of the Vbe required by the npn, to keep it saturated.[ON]

The converse is true for a pnp transistor.

If its important that the device's low end HAS to be a the actual ground 0V, due to say the device having an input referred to ground then I would use a pnp in the high end of the device and just 'lose' 0.1V

Hope this makes sense.

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Eric
"Good enough is Perfect"

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NPN's are much more efficient at switching earths ericgibbs stated above, perhaps this simulation video would help illiterate that?

Just remember that it’s a simulation, and you should use a resistor in series with each segment

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Spency.

PIC Micro's - Your mind is the limit

PIC's and interfacing with other devices - a PIC Basic Guide @ digital-diy.net

 

I don't understand why your transistor has 3.5V across it.

You don't need a base resistor for the emitter follower in the lower configuration (unless the base drive voltage is greater than the collector voltage).

You can saturate the emitter follower in the lower configuration if the base drive voltage is greater than the collector voltage, in which case the base resistor is required. This eliminates most of the undesirable drop across the transistor. It is generally the case, however, that the higher base drive voltage is not available.

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Ron

 

They are not fully driven on

Point taken - it was a quick throw together, the fact remains though - NPN's are more efficient for switching earths and PNP's for switching supplies

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Spency.

PIC Micro's - Your mind is the limit

PIC's and interfacing with other devices - a PIC Basic Guide @ digital-diy.net

 

If you switch VSS with an NPN, then the entire device is floating near VDD when it's supposed to be off. This is fine, if nothing is connected to the I/O (or if all I/O circuits are also switched off in the same manner.)

But if the circuit drives any ground-referred I/O, those circuits will see a 'high' input.

I always use a PNP (or p-channel) on the Vdd side.

 

The transistor has 3.5V across it because it is driving an LED segment that is 1.4V (kind of low for an LED) and the supply is only 5V. The transistor would saturate properly if the LED has a current-limiting resistor in series with it.

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Uncle $crooge

 

True. Both are probably smoking. Gramo did point out that resistors should be in series with the transistors. It shouldn't have been to difficult to include them in the sim, thereby avoiding stupid questions like mine (and questionable answers like Gramo's).

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Ron

 

I knock up that many cicruits for people, I tend to lean toward shortcuts

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Spency.

PIC Micro's - Your mind is the limit

PIC's and interfacing with other devices - a PIC Basic Guide @ digital-diy.net

 

That's what I was fearing. Because the device is a radio RX module and it seems to be sending out something from the data-out pin EVEN THOUGH it's supposed to be off. And yeah, it still has some voltage across it. Explains a lot...

Now the same things probably happening with the TX module (which also has a NPN switch to ground), so surely if thats partially-on when its not supposed to be it will interfere with the TX signal thats supposed to be received by the RX module?

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