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  #1  
Old 10-19-2007, 11:58 AM
mentaaal mentaaal is offline
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Default Common collector amplifier

Hey again guys. I find that i am still having alot of trouble understanding how the common collector works.
In the allaboutcircuits site this example is used:


In the text, the author says that the voltage across the be pn junction will always be around .7 volts with the rest of the input signal across the 5kohm resistor. If this is the case then at the peak, the current through the resistor should be: 3.1/5000 = .62 milliamps. I know this is wrong but howcome this .62 milliamps doesnt go through the base? Shouldnt all the input current go through the base?

Even if as the circuit describes, current is beta plus 1. Then the current through the base should be emitter current/beta+1 For this to occur hie or imput inpedance would have to be very high which it is. But if this is the case that the input inpedance was very high then surely most of the voltage would be dropped across hie and not Rload? As these two "resistors" are connected in series as far as the input is concerned, and the load resistor being only 5 kohms should drop a tiny voltage with respect to hie.

I know i am not doing a very good job of explaining myself but i just dont get how this circuit does what it does.

Help!
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  #2  
Old 10-20-2007, 08:48 PM
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thingmaker3 thingmaker3 is offline
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My math was a little different than yours. I get peak load current of (1.5*1.414+2.3-0.7)/5000 = 0.744 mAmps

Input current does indeed go through the base. Not all of the load current is input current, though. Most of the load current is collector current.

Divide collector current by base current and add 1 to calculate current gain.

[quote=mentaaal;43189]For this to occur hie or imput inpedance would have to be very high which it is. But if this is the case that the input inpedance was very high then surely most of the voltage would be dropped across hie and not Rload? /quote]Remember that semiconductor resistance is dynamic. Hie will be whatever value it has to be in order for the base-emitter junction to drop 0.7V.

Perhaps a review of http://www.allaboutcircuits.com/vol_3/chpt_2/6.html will help.
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Old 10-21-2007, 12:40 AM
wireaddict wireaddict is offline
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Another consideration in "real world" common collector amps is that the biasing is done with resistors & not a second battery. One of the resistors would be connected between the collector & the base [usually between 50K & several Megohms] to apply a slight forward bias. Also, since the collector is the common element, a bypass cap is connected between it & the circuit common [or ground] to keep the charge on it constant; the input is applied between the base & the circuit common. As such, no input current flows through the load resistor; the input controls the current flowing from the emitter to the collector causing a varying voltage drop across the load resistor. Hope this helps.
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Old 10-21-2007, 07:59 AM
GS3 GS3 is offline
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To put it in the most simple terms: The voltage on the load is the same as the input voltage but the current is multiplied by the gain of the transistor. That means the power has also been multiplied even though the voltage has not changed.

Another aspect of this circuit is that the input impedance is much higher than the output impedance so it can be used as an impedance matcher.
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Old 10-21-2007, 04:04 PM
mentaaal mentaaal is offline
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Hey guys, thanks to all of you for replying.
I think if i can just understand this one thing the rest will fall into place:
Hie will change to ensure that .7 volts will be dropped across the base emitter junction. Thats ok. So for a common collector circuit hie is typically huge like over 100 Kohm.So if this is the case, the current going through the base will be tiny, in the order of a few microamps. So as this input current goes through the load resistor, wouldn't it drop an even smaller voltage across it than it would for the massive hie as in this case the load resistor is only 5kohm? So if the voltage is even less than .7 volts across the load resistor where did the rest of the input voltage get dropped?
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Old 12-27-2007, 03:00 PM
mentaaal mentaaal is offline
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MY HEAD IS GOING TO EXPLODE! I have read and reread your replies and the information on this website regarding the common collector amplifier and its just not making any sense to me!

Say for example this circuit:



As was said to me earlier, the voltage across the 5kohm resistor will be always around .7 volts less than the input voltage. If i just accept that for the moment and try to work out the base current: As the voltages and current here are obviously dynamic lets assume that the instantaneous voltage is the maximum of 2.3 + 1.5 volts. So the voltage across the load resisor will be 2.3 + 1.5 - .7 which is 3.1 volts. So the current going through the resistor is 3.1/5000 = .62mA. So if i use below:



If i assume beta to be something like 100 then Ibase will be .62mA/101 = 6.138uA. Then if this is correct then the voltage across the load resistor from this base current will be 6.138uA * 5000 = .03 Volts. If as i know already .7 volts approx will be dropped across the base emitter junction then where did the rest of the input voltage go? What am i not doing correctly here? Help my exam is next week!!!!

Sorry thingmaker3 i know you must be blue in the face trying to answer this for me but i just cant grasp it! I know i have essentially just repeated my first question above but i dunno maybe you could to and rephrase it for my altogether dense brain!

Last edited by mentaaal; 12-27-2007 at 03:04 PM.
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  #7  
Old 12-27-2007, 04:43 PM
Audioguru Audioguru is offline
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The peak input voltage is 3.8V and the emitter voltage is 3.1V if the 1.5V input signal is measured as a 1.5V peak signal (it might be 1.5V RMS or 1.5V p-p).
Then the emitter voltage is 3.1V and its current is 620uA.
If beta is 100 then the base current is 6.2uA.
The collector current is 620uA - 6.2uA= 613.8uA.

That is all there is to it.
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Old 12-27-2007, 05:17 PM
mentaaal mentaaal is offline
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Quote:
Originally Posted by Audioguru View Post
The peak input voltage is 3.8V and the emitter voltage is 3.1V if the 1.5V input signal is measured as a 1.5V peak signal (it might be 1.5V RMS or 1.5V p-p).
Then the emitter voltage is 3.1V and its current is 620uA.
If beta is 100 then the base current is 6.2uA.
The collector current is 620uA - 6.2uA= 613.8uA.

That is all there is to it.
Ok thanks for the reply but this is what i dont understand:
If the base current is as you say 6.2 uA then this base current will drop 6.2ua * 5000 = .031 volts across the load resistor right? And as this current flows through the emitter base junction then the voltage dropped across that junction is round .7 volts right? Then these two voltages would sum to .731 volts. Where did the other 3.069 volts get dropped? This is the problem i am having. There is probably something very obvious that i am just overlooking! thanks for bearing with me
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Old 12-27-2007, 06:00 PM
Pootworm Pootworm is offline
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Quote:
Originally Posted by mentaaal View Post
Ok thanks for the reply but this is what i dont understand:
If the base current is as you say 6.2 uA then this base current will drop 6.2ua * 5000 = .031 volts across the load resistor right? And as this current flows through the emitter base junction then the voltage dropped across that junction is round .7 volts right? Then these two voltages would sum to .731 volts. Where did the other 3.069 volts get dropped? This is the problem i am having. There is probably something very obvious that i am just overlooking! thanks for bearing with me
The voltage at the emitter end is 3.1V. That's what gets dropped across the resistor. That voltage plus the b-e junction's 0.7V drop sum back to the original peak voltage.
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  #10  
Old 12-27-2007, 09:19 PM
mentaaal mentaaal is offline
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Yes i am aware of that but the 3.1 volts dropped across the load resistor isnt all from the base current. Treating the base and collector current as two different entities or thinking about this with the substitution method, the base current is 6.2 uA this base current would drop the .031volts mentioned earlier across the load resistor. Wouldnt it? And then the collector current would drop the remaining 3.069 volts across the load resistor. So if the collector current was zero for example say the collector terminal is open circuited, the base current would have to increase to satisfy KVL? So the Collector current drops the larger voltage across the load resistor which reduces the base current in order to satisfy KVL?
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