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NPN transistor position

alec_t

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Welcome to ETO!
The LED will still switch on, but will be slightly less bright because about 0.7V will be dropped between the positive supply rail and the transistor emitter. The voltage drop with the negative-switched version is less, because it is the collector-emitter saturation voltage (less than 0.1V).
 

Nigel Goodwin

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But worse than that, you're not 'switching', you're using the transistor as an analogue device so it will waste considerably more energy as heat.

The negative (sink) circuit not only switched properly, but also allows you to power the LED(s) from a higher or unregulated voltage.

Using an emitter follower is an often seen design flaw, common on Instructables etc. where they don't usually know what they are doing.
 

rjenkinsgb

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Most Helpful Member
Another serious drawback appears when you try to actually control the "switch".

With the emitter to 0V, you only need enough control voltage to provide adequate base current through the limiting resistor and transistor base and the transistor can switch any level of voltage; 5V, 12V, 600V or whatever is needed.

With the load connected to the emitter, the load voltage will never be higher than about 0.7V less than the control voltage, regardless of the supply voltage feeding the collector.
With eg. a 3.3V logic signal as the control, that means the maximum output voltage is around 2.6V or less.

As Nigel says, that is an "Emitter follower" configuration - it give current gain but no voltage gain.
 

gophert

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Most Helpful Member
If you just have a mechanical switch in series with the base resistor, then either circuit will work fine. You just have to be aware that the transistor in your "positive switch" circuit will drop 0.8v or so from c-e vs 0.1 to 0.2v in the other circuit.

since you MUST drop some voltage and limit current anyhow, NOTHING is wasted. The heat is the same whether it is at the transistor or at the resistor. In fact, if you use the same components on both circuits, and you are satisfied with the 20mA of LEd current brightness vs the 30mA of brightness for the other, hour battery will last longer and you would be using less power with the Positive Switch and, unlike what Nigel said, you would actually be wasting LESS power.

again, this statement from Nigel is FALSE.
it will waste considerably more energy as heat
The heat loss (required to limit current flow to the diode) is shifted to the transistor instead of the current limiting resistor.

ON THE OTHER HAND...
If you are trying to control this with a microcontroller, read post #4 - control of your positive switch is less straightforward than the "more correct" negative side switch.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
If you just have a mechanical switch in series with the base resistor, then either circuit will work fine. You just have to be aware that the transistor in your "positive switch" circuit will drop 0.8v or so from c-e vs 0.1 to 0.2v in the other circuit.

since you MUST drop some voltage and limit current anyhow, NOTHING is wasted. The heat is the same whether it is at the transistor or at the resistor. In fact, if you use the same components on both circuits, and you are satisfied with the 20mA of LEd current brightness vs the 30mA of brightness for the other, hour battery will last longer and you would be using less power with the Positive Switch and, unlike what Nigel said, you would actually be wasting LESS power.

again, this statement from Nigel is FALSE.

The heat loss (required to limit current flow to the diode) is shifted to the transistor instead of the current limiting resistor.
Sorry I didn't make that properly clear - you're wasting heat in the transistor, which is best avoided.
 

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