Quick transistor question

[Dan East]

Hi. I've got a simple question about general pnp transistor use. Normally the load is placed on the collector. Why can't the load be attached to the emitter, so that current flowing into the base would also pass through the load?

When I try that the amperage through the load is not nearly as high. Why aren't the results the same?

Dan East

 

[Grant Fleming]

Hi Dan,

I would say that the resulting lower current observed is because the base/emitter current used to "turn on" the transisistor is much lower because of the resistance of the load is in this circuit now. If this current is low then the collector/emitter current is lower- (transistor not turned on fully).

Grant

 

[akg]

Hi. I've got a simple question about general pnp transistor use. Normally the load is placed on the collector. Why can't the load be attached to the emitter, so that current flowing into the base would also pass through the load?

When I try that the amperage through the load is not nearly as high. Why aren't the results the same?

Dan East

yes u can connect the load at the emitter (say Re) , then it is called an 'emitter follower' , it will reflect the voltage change in the base , hence the name.
usualy used where high impedence i/p is needed , in this case the Zin = Beta*Re

 

[Dan East]

Thanks guys. I figured out part of my problem. I have a 1k resistor on VDD of the PIC, which further reduced the amount of current I could send to the base of the transistor. I removed that resistor, but the PIC no longer oscillates at the right frequency (powered by 6V). So I moved the resistor to VSS. Now the frequency is correct, and the output of the transistor is acceptable.

Dan East

 

[akg]

Thanks guys. I figured out part of my problem. I have a 1k resistor on VDD of the PIC, which further reduced the amount of current I could send to the base of the transistor. I removed that resistor, but the PIC no longer oscillates at the right frequency (powered by 6V). So I moved the resistor to VSS. Now the frequency is correct, and the output of the transistor is acceptable.

Dan East

why shld u power the PIC at 6V.? , It's the absolute maximum!!(assuming F series)

 

[mstechca]

Normally the load is placed on the collector.

In real terms, that is an output of a common emitter amplifier.
When I do circuits, I don't go with the normals, I go with creativity, and decide where I want the load to be (provided that the circuit doesn't blow up in DC mode).

Why can't the load be attached to the emitter

It could, and if it was done that way, you have a common collector amplifier (which is also known as an emitter follower amplifier).

 

[Nigel Goodwin]

Normally the load is placed on the collector.

In real terms, that is an output of a common emitter amplifier.
When I do circuits, I don't go with the normals, I go with creativity, and decide where I want the load to be (provided that the circuit doesn't blow up in DC mode).

There are clear advantages to using specific configurations for specific purposes, in your case 'creativity' usually means choosing a poorer option!.

By placing the load in the collector you have considerably less voltage drop to the load, and also considerably more base current available for the transistor - BOTH of which are important.

 

[hyedenny]

in your case 'creativity' usually means choosing a poorer option!.

I disagree. Lets call a spade terminal a spade terminal: It ALWAYS means choosing a poorer option.

 

[audioguru]

Since there is a worry about the circuit blowing up, then a simple max current calculation and glance at a datasheet were not done.