NPN question- simple?

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Hi

In the attached diagram I am measuring very different currents going through the LEDs. Can someone please tell me why this is the case? and how I can get the current in the left hand circuit up to the same magnitude as in the right hand?

Thank you in advanced.
a.

 

[panic mode]

to operate, transistor needs certain forward voltage base-emiter.
in first circuit transistor is connected in so called common collector configuration. (high current gain but voltage gain is less than 1).
in second circuit transistor is in common emiter (high current gain and high voltage gain). you can use separate power source (higher voltage than rest of the circuit) for the base circuit or you can simply flip it over and use PNP transistor.

 

[arrow]

Hi Panic Mode

I tried to use higher voltage (8V) on the emitter- left hand side circuit. While it increased the current to about 20mA it was not enough for my application.

Should I increase the Voltage on the emitter further?
(I have to keep the base voltage at 5V).

Is it possible for you to upload a sketch of PNP please?
Or give me a step by step description of what you have in mind?

Thank you for your help.
a.

 

[Nigel Goodwin]

The right hand diagram is the preferred way to do it, although the base resistor should be MUCH higher than 10 ohms - you can't get them the same, because they work in completely different ways - in particular, the left hand one will always have less voltage available.

 

[audioguru]

The preferred right circuit inverts the control signal. If you don't want it inverted then use a second inverting transistor stage to feed it.

Ordinary LEDs have an absolute max continuous current rating of 30mA. If you need them to be brighter then use brighter LEDs or use many more strings of them.

 

[mstechca]

From what I understand, a common emitter amplifier is out of phase with respect to the input signal (which can be an inverter).
The other two amplifiers (common base and common collector) are in phase with the input. Am I right?

 

[audioguru]

From what I understand, a common emitter amplifier is out of phase with respect to the input signal (which can be an inverter).
The other two amplifiers (common base and common collector) are in phase with the input. Am I right?

Correct.

 

[ljcox]

As I told you in response to your previous thread, you don't need the base resistor in the left hand diagram. I also gave you the formulae to calculate the emitter current.