# Amplifier Comparison

by Physixs
Tags: amplifier, comparison
 P: 30 Hello, Thanks so much for all the previous help. I will probably be using this forum a lot this semester because my textbook is not very helpful. So thank you in advance. I am trying to create a comparison table of CE CC and CB amplifiers to help me study. However, I cannot find the "exact" information I am looking for online (concerning certain aspects of each). I am not sure if what I am trying to accomplish is too ambiguous (the characteristics of each type of amplifier changes too greatly based on the parameters of the circuit) or not. Please let me know if I did anything wrong. The "?" are things I am not sure about since several online sites I have read have said different things. Zin: CE is High (or very high?), CC is Very High, CB is very low Zout: CE is High (or very high?), CC is Very low, CB is High (or very high?) Voltage Gain: CE is medium (-RC/RE), CC is about 1, CB is (RC/RE?? I've read a few variations) Current Gain: CE is Beta, CC is Beta + 1, CB is about 1 The main one websites either don't mention or are contradicting about is power gain... CE the power gain is high (due to high voltage and current gains, P = VI, this makes sense but shouldnt power gain be Very high? not just high? This is confusing me because of the CC reason I listed below) CC the power gain is High (this didn't make much sense to me because when I reference the power gain in a CE, it seems off. If a CE's power gain is P=VI with an increase in current and voltage gains, how can that power gain be ony "high" if in a CC only the current is amplified and therefore in P=VI, since voltage does not change (voltage gain is slightly less than 1), how can the power gain be similar to a CE? Shouldnt the power gain of a CE be significantly larger than a CC in realtively similar circuits and conditions?) CB the power gain is a huge?? Some websites say it hase a power gain of 1. Some websites say it has a power gain that is large. Some say that is small. My textbook doesn't even mention it for a CB. I have no idea what to say here. Finally, my last question about all three amplifiers, I was reading ahead in my text. Would it be safe to say that they are all Class A power Amplifiers? I was a little lost on this idea too. Since they all amplify power, I thought it would be ok to say all single stage CC, CE, CB amplifiers could be classified as Class A. Where I thought this might not be true was with a CE. Since the CE inverts the output, does that still count as a "linear" relationship between Vin and Vout? If so it is a class A and if not, I am more puzzled. Thank you so much! This forum has be a huge help to me in all my classes
 Sci Advisor Thanks P: 1,915 Common Base is a low impedance emitter input to a high impedance collector output. The emitter input current is very similar to the collector output current. Voltage gain will be dependent on resistor ratio. CB can be used in “cascode” to isolate voltage in a high frequency current path. CC is a simple emitter follower, it is a current amplifier. The emitter output voltage will be very similar to the base input voltage. Current gain is dependent on resistor ratio. CC, CE and CB are all Class A amplifiers. The sign of the slope is not important, it only needs to be reasonably linear.
 Engineering Sci Advisor HW Helper Thanks P: 7,168 Don't tie yourself in knots trying to find the right words to describe the power gain. As you said, P = VI. The important facts are that CB has no current gain and CC has no voltage gain. The gain depends very much on the component values in the complete circuit, not just on the configuration. For example you can design a practical single-stage CC amp with a current gain of 2 or 2000 (using a Darlington transistor), so trying to use one word to describe "the gain" is a bit pointless. Also, note that CE amplifier stages are not necessarily used to amplify power. Often they are designed to amplify voltage, and the difference between the input and output currents is not important.
 P: 1,814 Amplifier Comparison Sometimes engineers become accustomed to using circuits in specific ways and don't always think of all the variations a type of circuit can have. To illustrate this I would like to show a common base amplifier that is contrary to what both Baluncore and AlephZero have said in the previous posts. In contrast to what Baluncore said, this common base amplifier has a higher input impedance than output impedance and the emitter input current is NOT similar to the collector output current. In contrast to what AlephZero said this circuit DOES have current gain. Ie = 1.032 uA p-p Ic = 105.2 uA p-p V(R1) = 1.032 mV p-p V(R2) = 10.55 mV p-p All of the above with 1kHz signal at 2 mV p-p. This simulation was done with LTSPICE. Attached Thumbnails
 Sci Advisor Thanks P: 1,915 skeptic2. Your AC analysis has some funny things happening there. R1 is not the input impedance. The input impedance is 10uF in series with emitter impedance divided by beta. Current gain of transistor is not specified. At 1 kHz, 10 uF has Xc is about 16 ohms. Your Ie and Ic do not add up, Ib is reverse biased.
 Sci Advisor Thanks P: 1,915 C2 should reference the base to ground, not to +12V rail. Because input voltage appears between ground and emitter, that will reject power supply noise, Back of the envelope calculations show... Base voltage is 12V * 18k / (18k+39k) = 3.8V Emitter voltage will be about 3.2 V DC emitter current will be 3.2 mA DC voltage across collector load R2 will be 0.32V. If beta is 100 then emitter impedance will be 1k / 100 = 10 ohms. Xc at 1kHz = -15.9 ohms reactive. Input impedance vector, Zin = 10 – j 15.9 = 18.8 ohms magnitude. Voltage gain will be R2 / Zin = 100 / 18.8 = 5.32 2 mv pp input will give 10.64 mv PP output. So we agree on that. It has voltage gain. We also see that Zin is 18.8 ohms and Zout is 100 ohms, contrary to your assertions. Ie = 3.2 mA Ib = Ie / 100 = 32 uA Ic = Ie – Ib therefore Ic = 3.168 mA We disagree about those currents.
 Sci Advisor Thanks P: 1,915 @ skeptic2. Your AC emitter current is not the AC current flowing through R1. The emitter current is the sum of the current through C1 and the current through R1. If you place a zero ohm resistor, R5, between the emitter and the junction of C1 and R1, you will find a similar emitter current in R5 to that flowing through collector resistor R2. Beginners are often at a disadvantage when compared to experienced engineers. A beginner does not know what can be safely ignored, and so drowns in complexity. An experienced engineer sees a circuit and without any need for analysis, understands immediately how it will behave.
P: 409
 Quote by Baluncore The input impedance is 10uF in series with emitter impedance divided by beta. .
Are you sure about that ? Zin = Xc1 + R1/β ??
Thanks
P: 1,915
 Quote by Jony130 Are you sure about that ?
I am never sure of anything when doing back of the envelope calculations. But it works as a first order approximation. skeptic2's AC analysis gave 10.55 mV PP output, my estimate gave 10.64 mv PP. That is close enough to be reassuring.

So which hat did I pull the Re/beta equation from? The output impedance of an emitter follower is approximately Re/beta. It seems appropriate that for small AC signals the input impedance to a CB amplifier will be similar.
See; http://en.wikipedia.org/wiki/Emitter...#Basic_circuit
 P: 2,529 The Zout of an emitter follower is Re in parallel with (the biasing resistor network on the base / beta). Re' might need to be added in. I always thought the Zin of a common base was Re in parallel with Re'. The base is at AC ground. No matter how you slice it though, the emitter is just about always considered low impedance.
 Sci Advisor Thanks P: 1,915 Averagesupernova. I agree, the differences in analysis seem to arise from the many possible interpretations of terms like “Re”.
 P: 2,529 The schematic skeptic2 posted is a good example of a transistor stage that should have been something else. The fact that Re' pretty much dictates a low Zin means that the 1K emitter resistor is rather pointless as it gives a voltage gain of .1 considering the 100 ohm collector resistor. We have built an attenuator in a most difficult way. All of the things we don't want in a transistor stage. A voltage gain much less than 1, a much lower Zin than Zout which implies a current gain much less than 1.
Thanks
P: 1,915
 Quote by Averagesupernova The fact that Re' pretty much dictates a low Zin means that the 1K emitter resistor is rather pointless as it gives a voltage gain of .1 considering the 100 ohm collector resistor.
No. The circuit is an impedance changing amplifier.
The small signal AC voltage gain is 5.3
The current gain is close to unity.

It is broadband and will handle VHF signals because little output current is fed back to the signal input through the Miller Effect capacitance, which would happen with a CE configuration.

The 1k emitter resistor is needed to DC bias the transistor so the small signal AC gain can be realised without clipping.
The majority of the input energy from capacitor passes to the emitter because the 1k does not significantly load the input.

I have designed circuits like that as front end amplifiers to time shock waves using gold leaf emitter resistors, heated by the wave front.
 P: 2,529 I had forgotten to include Re' in determining gain. Gain is pretty much the same as a common emitter with an AC bypass capacitor on the emitter resistor. Rc/Re'. So yes, gain would be on the order of 5 or so. I can also see how current gain would be close to 1 since there is about a 5:1 ratio of Zout to Zin as well as a voltage gain of 5. Generally I think of current gain referenced to a 1:1 voltage gain such as in an emitter follower. For those watching, please note that this amplifier is NOT well suited for anything other than fairly small signals. A low Z microphone amplifier or something of this nature would be appropriate. However, I do recall seeing common base amplifiers in larger signal applications where the previous stage biased the emitter resistor. So the signal went through the emitter resistor before it got to the transistor, which then also determined gain. I recall it being a wideband amplifier all the way down to DC, hence the direct coupling from the previous stage. I am wondering if it could have been in the chain of vertical deflection amplifiers in an oscilloscope.
 Sci Advisor Thanks P: 1,915 Oscilloscope electrostatic deflection amplifiers have high amplitude differential signals that would capacitively couple back from the collector into the base of a CE amplifier and so reduce bandwidth. By using a CB cascode stage, the high dv/dt slew rate is isolated from the base of the CE (differential amplifier / phase splitter / mirror) on the negative rail. Another high voltage advantage of cascode is that only the CB transistor need be rated for high voltage. The CE transistor can be selected from the wider range of low voltage transistors available.

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