Common Collector Class A Amplifier Load Resistance affect on Gain

Thread Starter

SamR

Joined Mar 19, 2019
2,412
Floyd threw me a curveball on this one.

Question:
A load resistance is capacitively coupled to the emitter in Figure 77. In terms of signal operation, the load appears in parallel with RE and reduces the effective emitter resistance. How does this affect the voltage gain?

Circuit:
1596768996474.png
So @ Vout I add a 1uF cap and RL to ground. The cap has an ESR of 0.09Ω so it is negligible. The gain is 0.998. RL is parallel so the only effect I see @ Vout is the coupling capacitor ESR? "load appears in parallel with RE and reduces the effective emitter resistance" Av= RE/er + RE So does that become Av = RE||RL/ re + RE||RL ? Apparently not (and I wouldn't think so) since the answer is given as 1% without even knowing what the value of RL is. Due to the coupling cap, the load resistance may be in parallel but has no effect on the transistor biasing is how I see it. The DC bias is all I am given here. I have no signal.

Sam
 

ZCochran98

Joined Jul 24, 2018
87
Electronic Devices, by T.L. Floyd, per chance? I have a version of it myself - I recognized his use of fuchsia in his annotations. Based on every example Floyd gives of this circuit, it appears that he uses the \(R_L = R_E\) assumption for his examples. By my calculations, if it is assumed that \(R_L = R_E\), then I see where he got the "1%" value in drop (It does, indeed, become \(A_v = \frac{R_E||R_L}{r_e'+R_E||R_L}\), like what you said, using Floyd's notation - I get \(r_e' = 12.2\Omega\) from \(I_E\approx 2.05\mathrm{mA}\) based on the method he outlines). I also get the gain as 0.988, not 0.998, for the unloaded buffer circuit.

For these kinds of circuits, the midband gain is usually what's being calculated, so the capacitors are treated to be shorts (and no ESR), so the only effect the coupling capacitors have (in the midband) is to eliminate the DC voltage offset (in this case, a 2.05V DC offset from across the 1kohm resistor). Usually, the only other effect these capacitors have is to set the low frequency cutoff point of the amplifier (the high frequency cutoff point being set by the internal parasitic capacitances).
 

sparky 1

Joined Nov 3, 2018
301
I have a circuit that bares some resemblance to what the TS posted above, it amplifies rather than attenuate.
In my circuit R1 is 3.26 times larger than R2. Vcc is 1.22 times larger than the quotient of (R1/R2)
I have a collector resistor R3 so that R3/ R4 = 0.16 or R4/R3=6 The circuit is not perfected
setting a load range between 14 Ohm to 1000 Ohm. 1000 Ohm gives roughly a gain of 1

If a potentiometer was used the direction would be opposite
What I am leaving out is the uA current at the base.
 
Last edited:

Thread Starter

SamR

Joined Mar 19, 2019
2,412
Copied directly by Snipping Tool from a PDF version of Floyd. K, he does like to throw a curveball from time to time to make sure you were paying attention and to integrate new knowledge with what you should have already learned. Yes the cap should act as a short and Xc is ignored. So that ends up as 500Ω/512.2Ω = ~.975 which /.988 yeah ~<1%. Got it. Wish he'd given a value for the load.

Thanks, Sam

EDIT: Hmmm... Which tells me for RE||RL divided by itself (plus a small re) that the value of the load has no real affect on the gain at all! Because it cancels out!
 
Last edited:

ZCochran98

Joined Jul 24, 2018
87
Copied directly by Snipping Tool from a PDF version of Floyd. K, he does like to throw a curveball from time to time to make sure you were paying attention and to integrate new knowledge with what you should have already learned. Yes the cap should act as a short and Xc is ignored. So that ends up as 500Ω/512.2Ω = ~.975 which /.988 yeah ~<1%. Got it. Wish he'd given a value for the load.

Thanks, Sam

EDIT: Hmmm... Which tells me for RE||RL divided by itself (plus a small re) that the value of the load has no real affect on the gain at all! Because it cancels out!
Sounds about right with Floyd, from my experience.

The output impedance of the common collector amp is \(R_{out} = R_E||r_e'\), so it makes sense that, until your load gets close-ish to \(12\Omega\), your gain is virtually unaffected by the load. That's why this makes such a great buffer stage - typically somewhat high input impedance (\(R_s + (\beta_{AC}+1)\left(r_e' + R_E\right)\)), low output impedance, and doesn't affect the high frequency cutoff too much. The MOSFET equivalent - common drain - is even better, as it has an even higher input impedance.
 

sparky 1

Joined Nov 3, 2018
301
Thomas L Floyd does not equal Floyd K.
I corrected the distortion in the waveform now I am correcting the Floyd distortion.
I did not include a picture because the work should be completed by the TS asking for understanding
he must have found what he wanted then left an unfinished circuit for posterity. Understanding for himself.
 

Thread Starter

SamR

Joined Mar 19, 2019
2,412
he must have found what he wanted then left an unfinished circuit for posterity. Understanding for himself.
Huh? Yes, it was resolved. Av = RE||RL/ (re + RE||RL) I have no idea what your circuit is without a diagram. Mine was a direct snip copy from the book.
Electronics Fundamentals Circuits, Devices and Applications Thomas L. Floyd 8th Ed. Pg. 823 Problem 23 dealing with Class A amps as the thread title states.
 

sparky 1

Joined Nov 3, 2018
301
On page 813 Fig 62 is being used to show trouble shooting procedures and analysis.
There is information in fig 62 that makes the lesson work. ( common only to some tutors / the textbook name edition and page )
It is understood in the book that it is ok to compare measurements with figure 62. The schematic in question and the reference circuit.

In figure 64 using the scope, No skipping ahead, skipping steps here going for the math test approach greatly adds to confusion.
The requirement of actually making a test point measurement could not have happened. The need to correct R2 is important
and obvious in many regards to the lesson when not hacked. A case of jumping ahead skipping important procedures.
Author throwing a curve ball, not really. The student would be lacking proficiency at a very important stage of skills development.
There could be more issues with the circuit fig 72. The input found in figure 62 labled TP1
I do see the dilemma posed by the two stage amplifier of fig 62 being a reference circuit is an exercise in trouble shooting procedures.
The real problem encountered last week; What book? what page? what copyright infringement ?
A huge instruction section on procedures gutted in order to get to these questions below. An awareness that the school system is in big trouble.
Curriculum is mangled and hacked the students grasping for fragmented disorganized cramathons.

Floyd 8th p825.JPG
 
Last edited:

ZCochran98

Joined Jul 24, 2018
87
On page 813 Fig 62 is being used to show trouble shooting procedures and analysis.
There is information in fig 62 that makes the lesson work. ( common to tutors / what textbook )
It is ok to compare measurements with figure 62. The schematic in question and the reference circuit.

In figure 64 using the scope, skipping ahead skipping steps here going for the math test approach.
The requirement of actually making a test point measurement could not have happened. The need to correct R2
is obvious in many regards. A case of jumping ahead of important procedures. Throwing Curve balls, not really.
There could be more issues. The input found in figure 62 labled TP1
I do see the dilemma posed by the two stage amplifier of fig 62 being a reference circuit to a trouble shooting section.
The real problem that was encountered last week was What book? what page? what copyright infringement ?
A huge procedures instruction gutted in order to get to this. Is the school system gutted also?

View attachment 214237
I have (a physical copy of) the 7th edition of the book. In that version of the book, the problem, with slightly different numbers, appears on page 811 and is problem 23, referencing fig. 17-77. Instead of two 47-ohm resistors in this edition, he uses a 10k and 4.7k resistor (R1 and R2, respectively). In this particular edition, the answer in the book is simply "Av is reduced slightly" - no reference to 1%. I would have imagined that the answer to that particular question would not have changed between editions, but who knows? Maybe he was looking for something more specific in later editions?

For the question specifically asked in this thread, there is no relation to any testing/troubleshooting methods, and as such is completely independent of figure 62-65, or any other figure other than figure 77, so I'm not sure why you brought those questions or sections up. This discussion was on question 23 and figure 77 in particular, which has no connection to the troubleshooting section (or any simulation/waveforms....) whatsoever, and all comments/discussions were in regards to how that particular question was worded, and explaining the answer (and commentary on Floyd's occasional semi-vague questions/answers - instances of "what kind of answer, exactly, is he looking for?").
 

Thread Starter

SamR

Joined Mar 19, 2019
2,412
? I haven't even gotten to page 813 much less fig. 62. This was for section 3 starting on pg. 786. What school? I got my BSc in the 70s and this is what I do to keep me busy in my retirement. I posted a question, a schematic, and I got my answer and also useful repsonses to my followup comments.
 

sparky 1

Joined Nov 3, 2018
301
Huh? Yes, it was resolved. Av = RE||RL/ (re + RE||RL) I have no idea what your circuit is without a diagram. Mine was a direct snip copy from the book.
Electronics Fundamentals Circuits, Devices and Applications Thomas L. Floyd 8th Ed. Pg. 823 Problem 23 dealing with Class A amps as the thread title states.
The book reads fine. The circuit is intentionally in need of trouble shooting. When I study electronics it is important to me to understand the material when there is supporting material available how the lesson plan was laid out can make a difference.
I am sorry for any misunderstanding.
 
Top