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Input and Output resistance(CE amplifier) 
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#1
Feb2308, 10:35 AM

P: 7

Hello people
I am trying to revise for my exam next week..I have searched everywhere but still I cant find how to calculate input resistance,stage input resistance and output resistance of a class A common emitter amplifier. any help will be highly appreciated. mathslova 


#2
Feb2308, 12:47 PM

P: 2,499

Input impedance depends a bit on how it is biased and if there is a bypass capacitor across the emitter resistor. For a class A CE amp with voltage divider bias and a bypass capacitor it will be:
 (r'E * beta) in parallel with the following quantity: (Bias resistors in parallel with each other).  You may think that the bias resistors would not be paralleled but they end up that way since the + power supply rail is considered AC ground.  Output impedance is simply the the resistance of the collector resistor. 


#3
Feb2308, 02:15 PM

P: 7

Thank you Sir
I am looking at these diagrams: http://www.zshare.net/image/8014898071a386/ I want to know how to find Rin, R'in and Ro for Figure A and r'in for Figure B, R1=120K, both when Ce1 is removed or not. cheers 


#4
Feb2408, 02:35 AM

P: 7

Input and Output resistance(CE amplifier)
anyone willing to help?



#5
Feb2408, 12:44 PM

P: 2,499

Figure A has no emitter bypass capacitor so the AC impecance looking into the base would be (r'E + RE1) * beta. I would say that is what the diagram call R'in.
 Parallel R1 and R2 to get the impedance that the biasing part of the circuit presents.  Finally, parallel the biasing impedance with the impedance looking into the base figured in the first step. This is a pretty good approximation of the input impedance of figure A.  Funny thing, I don't exactly recall how to arrive at the output imedance of an emitter follower (second stage figure A). I believe you have to know the output impedance of what is driving it and divide that by beta. You also have to figure in the RE and I don't recall exactly how. 


#6
Sep710, 08:08 PM

P: 60

Hello!
I know this is an old thread, but I have found myself here! My question is how you found Rin to be "(r'E + RE1) * beta" ? I keep getting Rb  r(pi) + RE  where is the beta coming from? Also, please know that I cannot view the image because the host of the image is evil (spam). Many thanks for any help here... 


#7
Sep710, 11:26 PM

Sci Advisor
P: 4,016

The emitter resistor carries a much magnified version of the input signal current and this produces a voltage which opposes the input current itself, making the input impedance a lot higher than you might expect.
For example, a transistor with a Hfe of 100 and a (internal) base resistance of 56 ohms and with a 220 ohm emitter resitor will have an input impedance of 22276 ohms. Quite a lot more than the 276 ohms (ie 56 + 220) you might expect. This is excluding any bias resistors. 


#8
Sep710, 11:48 PM

Sci Advisor
P: 4,016

In the above diagram, I have shown the base resistance of the transistor and the emitter resistor to be 56 ohms and 220 ohms respectively. The 56 ohms is inside the transistor. The transistor is assumed to be working with correct bias. Only signal currents are considered in the following: The transistor has a gain of 100 so there is a large current source shown which produces a signal current of 100 times the input signal current Ib. So the current in the emitter resistor is 101 * Ib (ie 100 Ib + Ib ) The voltage across the emitter resistor is (101 Ib * 220) or 22220 Ib The voltage across the base emitter junction is 56 Ib The total input voltage is 56 Ib + 22220 Ib or 22276 Ib (ie baseemitter voltage plus emitter voltage Input impedance = Input voltage / input current = 22276 Ib / Ib or 22276 ohms 


#9
Sep810, 05:39 AM

P: 60

Okay, thank you so much. I still don't get it. Later today, I will draw and post my question more clearly and see if you can point out my mistake.



#10
Sep810, 08:30 AM

Sci Advisor
P: 4,016

Can we go back to fundamentals first?
Suppose you have two batteries, one is 5 volts and the other is 4.9 volts. They are joined together at the negative end and there is a 1 K resistor between their positive terminals. There is 0.1 volts across the resistor, so the 5 V battery is delivering 5 volts at 0.1 mA. (0.1 V / 1000 ohms ) So, it is seeing an apparent load of 5V / 0.0001 amp or 50000 ohms. This is DC but it is similar to what happens in an emitter follower for AC signals. The emitter has a resistor to ground that has an almost exact replica of the input signal across it (produced by the transistor's current gain). So, the input impedance depends on the current gain of the transistor and the size of the emitter resistor. The closer the emitter signal voltage is to the base voltage, (like in our battery example), the higher the input impedance of the emitter follower. 


#11
Sep810, 07:05 PM

P: 60

Okay, I think I am starting to smell what you are stepping in.
I have two different CE amps that I worked out for my lab. I am pretty sure that I nailed everything except for the Rin for the CE amp with a resistor at the emitter node. There are definitely some basics that I am missing here and the complexity of my school work just went way up! Here is a link to the page I uploaded my work to. The file name is: CEamp.pdf. Thanks again for any help! 


#12
Sep810, 08:05 PM

Sci Advisor
P: 4,016

I looked at that link. I think it was OK until the second last line.
The input impedance of just the amplifier (without the bias resistors) is Vin / Ib. Notice (in my previous post) that Vin can also be expressed in terms of Ib and then the Ib's cancel out. This means you don't have to know what Ib is, to get the input impedance. Also, the input impedance is very close to Re * beta for an emitter follower. In the example I gave earlier it makes about 1% difference if you just use Re * beta. (Re is the emitter resistor) ie 220 * 100 = 22000 compared with 22276 I got as a more accurate answer. 


#13
Sep910, 07:31 PM

P: 60

Just to follow up...I think I get it now. My problem was with the definition of input impedance (vin/in). Some of these circuits can be very complicated...
Even though yours was very simple and easy to understand, I had trouble applying that to the circuit I was working on. Many thanks for your help. 


#14
Sep1210, 12:01 AM

P: 560

I've had a very simple method since HS.
Z seen at e is 26 ohm / ie (ma) + Re For example, if the emitter current is 2ma and the emitter ballast resistor is 50 ohms, then Ze = 13 + 50, or 63 ohms. What is seen at the base is Ze / Beta. Beta is the same as hfe Now, there are usually bias resistors to maintain the base at the correct operating voltage. These resistors are in parallel with each other and Ze/Beta, so you get Zin=R1R2Ze/B Some people just use the collector resistor as the output impedance. If you want to be picky, it's hoeRc 


#15
Sep1610, 10:50 PM

Sci Advisor
P: 4,016

What is seen at the base is Ze / Beta. Beta is the same as hfe
The input impedance of an emitter follower is equal (approximately, as described earlier) to the resistance of the emitter resistor TIMES hfe ......provided the emitter resistor isn't bypassed, of course. This is where the emitter follower gets its high input impedance. The output impedance of the emitter follower is equal to Rbase / Beta (where Rbase is the internal resistance of the base emitter junction) and this is where the emitter follower gets its low output impedance. 


#16
Oct1210, 08:14 PM

P: 60

Just wanted to say that I pretty much get this stuff now. Thanks for your help!



#17
Oct2010, 12:26 AM

P: 2

i need to calculate the input resistance of the CE amplifer
given Ic  1mA ,Ib  1uA at room temperature how to calculate Inp Resistance with given parameters.Somebody pls reply 


#18
Oct2010, 12:31 AM

P: 60

Is this a homework problem? If so, it doesn't belong here. 


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