Common Emitter Amplifier

Thread Starter

Madgesture

Joined May 2, 2010
5
A technician installs a transistor, in a common emitter layou, with the Collector and Emitter terminal reversed. What would be seen a cross the load resistor?

Any help would be greatly appreciated. On the last page of the homework and my brain is already fried.

Mike
 

hobbyist

Joined Aug 10, 2008
892
NPN transistor to conduct:
Collector more positive than base reversed biased
base more positive than emitter forward biased

PNP transistor to conduct:
collector more negative than base reversed biased
base more negative than emitter forward biased.
 

studiot

Joined Nov 9, 2007
4,998
Well this is still homework, with or without fries, so we shouldn't tell you directly without you say what you think first.

So here are some hints.

Firstly the question asks what is the voltage across the load resistor, not across the transistor.

So where is the load resistor in a CE amp?

OK so now is the transistor on or off?

So is it passing current?

So what does that tell you about the current in the load resistor?

So what does that tell you about the voltage across the load resistor?

Does it make any differernce to the answer if the transistor is NPN or PNP?
 

Thread Starter

Madgesture

Joined May 2, 2010
5
What I see with the transistor installed backwards is an NPN, Common Collector Amplifier. The Load resistor will now be on the emitter. The Output on Load Resistor will basically be the input voltage, as there is very little or no voltage gain in in a common collector amplifier, just power and current gain.
 

Thread Starter

Madgesture

Joined May 2, 2010
5
The load resistor on an NPN CE Amp is on the collector. The transistor installed backwards it will be on the Emitter making the cicuit a NPN CC Amp

There is 1.08V at the base so the Base Emitter junction will be forward biased.

With the BE forward biased there will be current passing and also current and voltage on the load resistor.

Since the this is now a CC Amp the Output Voltage will be the roughly the same as the input as there is no voltage gain and the Output will be in phase with the input.
 

Thread Starter

Madgesture

Joined May 2, 2010
5
So where is the load resistor in a CE amp?
On the Collector

OK so now is the transistor on or off?
On 1.08V at Base before reversal

So is it passing current?
Yes

So what does that tell you about the current in the load resistor?
It will be higher than input current

So what does that tell you about the voltage across the load resistor?
There will be some voltage

Does it make any differernce to the answer if the transistor is NPN or PNP? Yes
 

Thread Starter

Madgesture

Joined May 2, 2010
5
So, basically with reversing the Transistor it is going from a CE Amp to an CC Amp.

The major differences would be:

No Voltage Gain in CC Amp
No Phase Shift in CC Amp

Thanks
 

hobbyist

Joined Aug 10, 2008
892
A technician installs a transistor, in a common emitter layou, with the Collector and Emitter terminal reversed. What would be seen a cross the load resistor?

Any help would be greatly appreciated. On the last page of the homework and my brain is already fried.

Mike

Here is a hint:

draw 2 circuits using a CE layout properly biased for a NPN and a PNP.

Then redraw them with the transistors backwards with there respective circuits.

Then using the info in above posts analyse what you see.

The key is don't change battery polarities when you change the transistors to be backwards.

To change from CE to CC with backwards transistors the polarities of the battery needs to be reversed as well.
 
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Audioguru

Joined Dec 20, 2007
11,248
When the emitter-base junction of a silicon transistor is reverse-biased then it has avalanche breakdown (like a zener diode) at about 7V.
 

Ron H

Joined Apr 14, 2005
7,063
An inverted transistor (collector and emitter swapped) still acts like a transistor, except the beta is low, and generally unspecified. Also, as AG stated, Vbe is now Vce, so the breakdown voltage will be low.
There is insufficient information in the statement of the problem to give a quantitative answer.

The circuit is NOT now a CC amp. It is still CE, but with the characteristics mentioned above.

EDIT: From Wikipedia:
The bipolar junction transistor, unlike other transistors, is usually not a symmetrical device. This means that interchanging the collector and the emitter makes the transistor leave the forward active mode and start to operate in reverse mode. Because the transistor's internal structure is usually optimized for forward-mode operation, interchanging the collector and the emitter makes the values of α and β in reverse operation much smaller than those in forward operation; often the α of the reverse mode is lower than 0.5. The lack of symmetry is primarily due to the doping ratios of the emitter and the collector. The emitter is heavily doped, while the collector is lightly doped, allowing a large reverse bias voltage to be applied before the collector–base junction breaks down. The collector–base junction is reverse biased in normal operation. The reason the emitter is heavily doped is to increase the emitter injection efficiency: the ratio of carriers injected by the emitter to those injected by the base. For high current gain, most of the carriers injected into the emitter–base junction must come from the emitter.
 
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