All About Circuits Forum Common collector amplifier
 Register Blogs FAQ Members List Today's Posts Search Today's Posts Mark Forums Read

 General Electronics Chat Discussion forum for general chat about anything electronics related, including asking questions about material in the All About Circuits E-book, Worksheets, and Videos.

#1
10-19-2007, 11:58 AM
 mentaaal Senior Member Join Date: Oct 2005 Location: Ireland Posts: 451
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!
#2
10-20-2007, 08:48 PM
 thingmaker3 Super Moderator Join Date: May 2005 Location: Rural, Oregon GMT -8 Posts: 5,072 Blog Entries: 6

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.
__________________
"I want to establish in your mind very clearly that you must not think I deny all that I do not admit. On the contrary, I think there are many things which may be true, and which I shall receive as such hereafter, though I do not as yet receive them; but that is not because there is any proof to the contrary, but that the proof in the affirmative is not yet sufficient for me"
#3
10-21-2007, 12:40 AM
 wireaddict Senior Member Join Date: Nov 2006 Location: E. Central MI [Temp.] Posts: 133

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.
__________________
Dave

If it ain't broke, learn how it works & improve it.
#4
10-21-2007, 07:59 AM
 GS3 Senior Member Join Date: Sep 2007 Posts: 268

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.
#5
10-21-2007, 04:04 PM
 mentaaal Senior Member Join Date: Oct 2005 Location: Ireland Posts: 451

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?
#6
12-27-2007, 03:00 PM
 mentaaal Senior Member Join Date: Oct 2005 Location: Ireland Posts: 451

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.
#7
12-27-2007, 04:43 PM
 Audioguru Banned Join Date: Dec 2007 Location: Ontario, Canada Posts: 9,411

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.
#8
12-27-2007, 05:17 PM
 mentaaal Senior Member Join Date: Oct 2005 Location: Ireland Posts: 451

Quote:
 Originally Posted by Audioguru 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
#9
12-27-2007, 06:00 PM
 Pootworm Junior Member Join Date: May 2007 Posts: 29

Quote:
 Originally Posted by mentaaal 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.
#10
12-27-2007, 09:19 PM
 mentaaal Senior Member Join Date: Oct 2005 Location: Ireland Posts: 451

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?

 Tags amplifier, collector, common

 Related Site Pages Section Title Worksheet Class A BJT amplifiers Video Lecture Digital Applications - Transistors and Transistor Circuits Video Lecture JFET Amplifiers - Transistors and Transistor Circuits Video Lecture Amplifier Configurations - Transistors and Transistor Circuits Video Lecture Transistor Biasing - Transistors and Transistor Circuits Textbook Current mirrors : Bipolar Junction Transistors Textbook Biasing calculations : Bipolar Junction Transistors Textbook The common-collector amplifier : Bipolar Junction Transistors Textbook The common-emitter amplifier : Bipolar Junction Transistors Textbook Bipolar junction transistors : Solid-state Device Theory

 Similar Threads Thread Thread Starter Forum Replies Last Post tmac General Electronics Chat 2 11-21-2012 09:35 AM donut General Electronics Chat 3 06-11-2012 03:44 PM Neyolight Embedded Systems and Microcontrollers 5 03-07-2012 12:16 PM strantor Math 6 09-22-2011 05:27 AM zorbzz The Projects Forum 9 04-28-2010 05:06 AM

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
 Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Electronics Forums     General Electronics Chat     The Projects Forum     Homework Help     Electronics Resources Software, Microcomputing, and Communications Forums     Programmer's Corner     Embedded Systems and Microcontrollers     Computing and Networks     Radio and Communications Circuits and Projects     The Completed Projects Collection Abstract Forums     Math     Physics     General Science All About Circuits Commmunity Forums     Off-Topic     The Flea Market     Feedback and Suggestions

All times are GMT. The time now is 04:00 PM.