Common emitter amplifier output question - Page 1
Forum section: electronics / beginners

author: Lucky-Luka
author_rank: Unknown
timestamp: December 15, 2019, 05:32:06 pm
content: Hi allCan somebody explain to me why I get increased NPN transistor collector voltage when I decrease the resistor value in the current mirror?When I decrease that value the current mirror can give more current to the NPN transistor but I cannot understand why my OP voltage increases.Do I have to see the current mirror as its ouput equivalent resistor? Doesn't that equivalent ouput resistor value decrease when the output current increases? I'm confused.Cheers

author: ferdieCX
author_rank: Regular Contributor
timestamp: December 15, 2019, 05:55:31 pm
content: When you increase the transistor collector current, you also increase the gm parameter of the hybrid pi equivalent circuit.

author: Lucky-Luka
author_rank: Unknown
timestamp: December 15, 2019, 06:00:05 pm
content: Quote from: ferdieCX on December 15, 2019, 05:55:31 pmok, but how does this effect affects Vo?

author: Zero999
author_rank: Super Contributor
timestamp: December 15, 2019, 06:03:17 pm
content: Well if you increase the collector current, then you're effectively pulling the collector up harder, so it intuitively makes sense the collector voltage will rise.Obviously, this is not a practical circuit and won't work in real life, even though a lot can be learned by playing with it in SPICE.

author: magic
author_rank: Super Contributor
timestamp: December 15, 2019, 06:08:44 pm
content: Output impedance of the mirror decreases, but this only means that it takes less variation in Q8 collector current to produce larger variation in Q4 collector voltage.However, your Q8 collector current is constant, determined by R1 and R2, as far as operating point analysis is concerned. Therefore the additional Q4 collector current is not sunk by Q8 until Q8 collector voltage rises a bit, which allows the additional current to flow into Q8 due Q8's output impedance.Q4 and Q8 are both set up as current sources and they are fighting each other trying to drive different currents through the same wire between them. Output voltage will be determined by the difference between their collector currents multiplied by paralleled output impedances of the two transistors.

author: ferdieCX
author_rank: Regular Contributor
timestamp: December 15, 2019, 06:09:53 pm
content: Vo is approximately equal to gm * vb'e * (parallel of R4 and the current mirror equivalent resistor)The equivalent resistor of the mirror doesn't vary too much if you increase its IcSo if you increase gm, you also increase VoThis method of shifting the operating point of the BJT is used in AGC systemsMhhmm, it seems that I misunderstood your question, you are asking about DC analysis,and I understood that you where asking about the output signal.For DC I agree with magic's explanation

author: Cyril Mechkov
author_rank: Newbie
timestamp: December 17, 2019, 06:09:51 pm
content: Quote from: magic on December 15, 2019, 06:08:44 pm@magic, is that your insight? If so, my congratulations! I ask because I have been using for a long time this functional approach to explain the idea ofdynamic load... but no one has accepted it so far. For the first time I see someone using such an explanation. Here are some related links:https://www.circuit-fantasia.com/my_work/conferences/cs_2005/paper1.htmhttps://www.circuit-fantasia.com/my_work/conferences/cs_2005/paper2.htmhttps://www.researchgate.net/post/Can_we_apply_input_current_to_the_collector_of_a_BJT_whose_base_is_held_at_a_constant_voltage_and_to_take_the_collector_voltage_as_an_outputhttps://www.researchgate.net/post/Where_is_the_dynamic_load2Structure.The circuit represents an AC common-emitter stage with dynamic load. For the purposes of understanding, we can think of this arrangement as of two "fighting" current sources (current-stabilizing non-linear resistors)... or a source (Q4) and a sink (Q8). They are "incorrectly" connected in series so each of them tries to pass its current through the common path. As a result, acurrent conflictappears between them, and the voltage of the middle node vigorously changes. Similarly, we can think of a BJT differential pair (akalong-tailed pair) as of two "fighting" voltage sources that are "incorrectly" connected in parallel so each of them tries to set its voltage at the common emitter node. As a result, avoltage conflictappears between them, and the current vigorously steers between transistors.Operation.Initially, we have to adjust the currents produced by the two sources to be equal so the output voltage at the common point to be V2/2. We can control the output voltage by changing the current of the one of sources (while keeping the other constant) or both. We can do it by means of the voltage divider R1-R2 (DC) and V1 (AC) for the sink Q8... and by the current-setting resistor R3 (DC) for the source Q4.We can imagine the circuit operation in terms of static (instant, chordal) collector-emitter resistances instead of currents flowing through them. This means to think of the two collector-emitter junctions (CE4 and CE2 in the attached picture) as of two partial resistances (RCE4 and RCE2) of a potentiometer as shown in the attached picture. In the graphical representation, static resistances are represented by lines starting from the origin of the coordinate system (the picture shows the more sophisticated case when both elements change their resistances in a differential manner).Initially, the static collector-emitter resistances are equal... i.e., the potentiometer slider in the analogy is in the middle. When the input base-emitter voltages (VBE4 and VBE2) change differentially - e.g., the magnitute of VBE4 increases while of VBE2 decreases, RCE4 decreases but simultaneously RCE2 increases... like the two partial resistances of the potentiometer when moving the slider to right. But the total resistance RCE4 + RCE2 remains constant... so the common current flowing through the network remains constant as well.... only the output voltage VA vigorously changes.In the graphical representation, RCE4 IV curve rotates clockwise... and RCE2 IV curve simultaneously rotates in the same direction... so the operating point A vigorously moves to right along the horizontal blue line. When VBE4 and VBE2 change differentially but in the opposite direction, the processes are reversed...

author: magic
author_rank: Super Contributor
timestamp: December 17, 2019, 07:09:57 pm
content: I think I have seen this behavior explained in very similar language before but I seldom write down where I got any random tidbit of information soIt is certainly well known (to those who know, at any rate) what happens when you connect two current sources in series and the only question, I presume, is is that of what wording is used to explain it in textbooks. I have no idea, I don't read textbookseditOkay, I will tell you what happened. I have certainly heard about "contention" when two current sources try to drive the same net, the usual result being that the output slams into one rail and the amplifier doesn't work. So that part certainly exists in popular folklore.But of course OP established a bionic feedback loop going through his eyes, fingers and keyboard and ensured that the circuit works. So I added another sentence which described the behavior of such circuit when it is biased into operating area by feedback, which in the real world of course would be provided by some other means in order to compensate drift

author: Cyril Mechkov
author_rank: Newbie
timestamp: December 17, 2019, 08:19:36 pm
content: Exactly, @magic... In practice, the output voltage can be held somewhere between the two rails only by means of negative feedback (servo). A mechanical analogy of this can be a hypersensitive weighing scale that has no stable state. And of course, the best "analogy" is the CMOS stage when we try to keep its output voltage somewhere in the middle. I assign this task to my students in the lab and promise them to pass the exam if they can do it. And if they can't do it, I suggest they make itself CMOS do it by connecting its output to the input.Off the topic, I am glad to see here people like you with technical flair, imagination and sense of humor... who not only know specific circuit implementations but also understand the basic ideas behind them...