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low frequency response of transistors

Discussion in 'Electronics Homework Help' started by max_torch, Feb 9, 2014.

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  1. max_torch

    max_torch

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    Feb 9, 2014
    Im having trouble understanding how to determine the 'R' that you have to exactly apply when using the formula f = 1/(2(pi)(R)(C)). For instance look at this screenshot from a page of "Electronic Devices and Circuit Theory" by R. Boylestad.
    help me.png
    It says:
    For the network of Fig. 11.16, the ac equivalent as “seen” by CE appears in Fig. 11.22.
    The value of Re is therefore determined by
    Re = RE|| ((R's/beta) + re), where R's = R1||R2||Rs
    He just says the formula but he never says how he determined the formula to use. This is a big problem for me since there are problems in the book that have you analyze the low freq response of BJT and FET amplifiers which are of different configurations from the examples given in the discussion. Because the examples shown in the book are that of a voltage divider for BJT and a Self-bias for FET but the problems im facing in the book and in school feature fixed bias, collector feedback, common base, emitter follower, etc..
    Basically, so what if i will deal with other circuit configurations, how do i determine what constitutes the 'R' in the above formula for frequency?
    I'm trying to study by myself and not depend on the teacher but this is killing me
    EDIT: In the rules of rewriting an amplifier circuit into its ac equivalent, you are supposed to short out capacitors but here in the above page Fig.11.22 is supposed to be the localized ac equivalent yet it has a capacitor.. why?
     
    Last edited: Feb 9, 2014
  2. mursal

    mursal

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    Dec 13, 2013
    Not sure of the rules about helping with homework?
    So I better thread carefully :)

    If you draw a diagram of a NPN transistor in an AC circuit on a piece of paper. You will have the emitter resistor RE and the base resistors, in this case called Rs. You will also have a little resistor on the base emitter junction called re` and the gain called beta.

    So if you now imagine looking back into the emitter circuit as if you were the capacitor, what do you see?
    You see RE in parallel with (Rs/beta + re`)
    That is your ac emitter resistance Re
    If I remember correctly, Capacitor Ce is there to short out the emitter resistor to increase the ac gain (Rc/re`)

    Remember capital letter = DC values
    lowercase letters = ac values

    Hope I didn't break any rules and it helps you understand what is, a difficult topic
     
    Last edited: Feb 9, 2014
  3. Harald Kapp

    Harald Kapp Moderator Moderator

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    Mursal, you didn't break any rules :)
    Good explanation.

    Max_torch:
    Whether you can apply this rule or not depends on what you are going to calculate. And probably you've gotten this rule a bit wrong. I don't have the book, so I can't check.
    What you do is you short capacitors for AC considerations if they have a large capacity and therefore an negligible AC impedance Xc=1/(2*pi*f*C) (with respect to the frequency you're dealing with).
    Examples for "negligible" capacitors are the electrolytic buffer capacitors across the power supply. 1000µF are equivalent to ~8Ohm at 20 Hz, if the power supply is buffered with 4700µF (not uncommon) this results in Xc ~ 1.7Ohm and is therefore mostly ignored (shorted) when considering the AC transfer function of an amplifier.

    In your case the situation is different. The capacitors within the signal path are small compared to the power supply buffer capacitors, Therefore their influence on the transfer characteristic (versus frequency) is not negligible.
    Consider filters and tone control circuits where the frequency dependence of capacitors is intentionally used to form the frequency response of the circuit.
     
  4. max_torch

    max_torch

    98
    1
    Feb 9, 2014
    Let me flesh out my problem with this topic more specifically and in greater detail. Here are a bunch of things I don't get:

    After you said that Harald I checked everything again and I realized that bypass and coupling capacitors are not negligible at low frequencies so they have to be present in the ac equivalent circuit, and the reason they are shorted when calculating for gain at midband frequencies is that their reactances are negligible for midband, and finally in highband frequencies they are still shorted for the same reason as in the midband but the interelectrode, wiring, and miller effect capacitances are beginning to affect your gain and therefore must not be neglected and will be present in your ac equivalent circuit.
    So basically really small capacitances affect high frequency and large capacitances affect low frequency.

    Despite knowing this, I am still unsure how to determine what constitutes R's in the formula f=1/(2(pi)RC) in the case when the emitter resistor is split into two parts: one part bypassed and one part unbypassed.
    Here is a circuit that shows what i mean:
    split emitter.png
    Here is my attempt to redraw this into an ac model(re approximation) at low frequency inputs, hence inclusion of the capacitors:
    equiv.png
    1.) So how do I determine the 'R' for solving the frequency filtered by the reactance of CE?
    Normally if there is only one bypassed emitter resistor the R would be
    R = RE||{[(R1||R2||RS)/BETA] + re} ...
    2.) Also a problem point for me is I don't know what will constitute the 'R' in the frequency filtered by CS. Normally if there was only one bypassed emitter resistor the R would be
    R = [R1||R2||(beta)*(re)] + RS, but with this kind of config would it change to
    R = [R1||R2||(beta)*(re + RE1)] + RS ? *RE1 as in the part that isn't bypassed.
    So how am I supposed to "look back" with this kind of configuration wherein the emitter resistor is split? How do i generate a formula? How do i deal with all situations?

    For all I say below when I say gain i mean voltage gain and not current:
    You know guys, I had my exams in this subject today and I'm pretty sure that I failed. We had 3 problems. The first was a common base amplifier and we have to solve for the midband gain and the upper and lower cutoff frequencies - I was totally screwed because I was assuming that the common base config wouldn't show up so I had no idea.. The second was an amplifier wherein we were given the bandwidth and the values of all the components and capacitances except for the wiring capacitance at the output, where we were supposed to determine how much wiring capacitance at the output would result in a bode plot exhibiting the given bandwidth. I think I knew how to solve it. The third was a two stage amplifier where we had to solve for the midband gain and the upper and lower cutoff frequencies, and I thought I had it down but i forgot to take into account the loading effect of the input impedance of the second stage to the voltage gain of the first stage as well as the effect of the load resistor to the gain of the second stage so pretty much all the calculations were wrong... It's hard not to be depressed with failing this subject when education is so expensive.

    and dear moderators,
    as this isn't exactly a homework assignment per se, maybe there doesn't have to be so much restriction to the help that people can give? it's too late anyway i already failed my exams..
    And does anyone know any good resource material specifically for this, ones that are not vague and very enlightening? Do most electronics engineers really make amplifiers this way? or any advice you have to give me in general about this please it will be appreciated.
     
  5. Harald Kapp

    Harald Kapp Moderator Moderator

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    Not so easy. Any capacitance influences a signal at any frequency. Whether you can neglect the effect depends on the impedance of the circuit around the capacitance. Generally small capacitances have a stronger effect on the frequency response than large capacitances - at least for ideal capacitances.
    When it comes to real capacitors, other, non-ideal parameters like series resistance, series inductance etc. complicate the behavior of a capacitor.

    For this circuit the simple f=1/(2*pi*R*C) equation is insufficient. Taking into consideration only RE1, RE2 and CE the frequency response of the voltage across CE with respect to the voltage across RE1+RE2 is
    V(ce)/V(re1,re2)=1/(1+Re2/re1*(1+j*w+Re2*ce))
    But: this is not the frequency response of the amplifier as it describes only the frequency response of the voltage across CE but you're interested in the output voltae which depends on the collector current and collector resistor.

    Read this analysis for the frequency response of a common emitter amplifier. or Google "common emitter amplifier frequency response" - you'll find quite some information and tutorials.
     
  6. max_torch

    max_torch

    98
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    Feb 9, 2014
    @ harald
    Thank you very much. From first glances I like the link you gave me^. I will read it all.
    As to searching through the internet, what I'm looking for is something that is very comprehensive yet accessible, which i must say is hard to find (that, or i suck at searching for things online) regarding this topic, even with the books in our library i have difficulty finding something that is comprehensive enough and with enough examples.. like a book electronics for idiots will get a 'check' for accessibility and examples but not comprehensiveness, while the book 'electronic devices and circuit theory' will get a 'check' for comprehensiveness but not accessibility in getting to the root of how to attack all the problems IMO..
     
    Last edited: Feb 11, 2014
  7. mursal

    mursal

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    Dec 13, 2013
    Malvino Electronic Principles (fifth edition) ISBN 0-07-113480-8

    Might be an idea to get your head around first approximation of the circuits, then build more and more detail in as your studies progress over the coming months.
    For example, first approximation of a common emitter amplified (as above) any capacitor on the emitter will be a short circuit at the working frequency of the amplifier. Therefore ac gain approximates to Rc/re`. Likewise the only time you need Re will be in a common collector (emitter follower) circuit. Where Re will be the output impedance.

    Good luck with your exams
     
  8. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    I really do have to spend more time in this section of the forum. Lots of esoteric but absolutely necessary subject matter for engineers.

    Sorry to hear that you failed the exam, as you do seem to be a genuinely studying the subject matter. On the up side you have Spice at your fingertips. Bow your head in great reverence to the engineering students that came before you in a Spice-less world. And if you really want to grovel at the feet of greatness, lets go back to the time of a slide-rule in the pocket of every engineering student.

    So when things seem daunting and you need inspiration... just think of them. ;)

    Chris
     
    Last edited: Feb 13, 2014
  9. max_torch

    max_torch

    98
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    Feb 9, 2014
    Okay you just gave me an idea. I should simulate the different circuits using SPICE and then try to work out the formulae that matches the readings. thank you. I will update if it works. This way I'll be able to verify what exactly goes on.
     
  10. Harald Kapp

    Harald Kapp Moderator Moderator

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    I beg to disagree. Imho SPICE is a tool to verify that your design, your calculations are correct. You can then use theh simulation to tweak parameters of the circuit in detail.
    But starting with SPICE simulations to actually design a circuit is the wrong paradigma.

    You should understand how (and why so) a circuit operates. SPICE alone can't give you that understanding.
     
  11. davenn

    davenn Moderator

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    That's true

    Also you need to remember or understand that circuit simulators like LT Spice, Multisim etc tend to work with circuits and components in ideal situations.

    In the real world with real components, circuits rarely perform the same way

    so as Harald hinted at .... experiment with real components and learn to understand their abilities and limitations :)

    cheers
    Dave
     
  12. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    While I don't disagree with the views of the role Spice should play I can't help but feel that seasoned engineers (you old farts :p) have a pony in this race. Humans instinctively devalue any technology that threatens to replace them. The old world master carpenter who used nothing more than his hands and a handled sharp piece of steel most assuredly rejected the first Block Plane. Likewise, the Block Plane guy must have thought those confounded rotary planers would never replace his hand skills either.

    My first employment as a Jr. Tech was a classic example. The tech and engineering staff were basically divided into two groups.. Those who had a working grasp of Transistor design and those who didn't. I was in the "didn't" group and relegated to vacuum tube design. Those of us in the "didn't" group thought transistors were nothing more than a novelty that would never replace the vacuum tube.

    Spice is in its infancy. It will only become more powerful over time. That said I don't think it should or could replace an understanding of what it's simulating.

    Chris
     
  13. Harald Kapp

    Harald Kapp Moderator Moderator

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    Chris, I don't devalue SPICE. I've been using it for many years. But SPICE has its shortcommings. Not so much in the math behind i´t, but in the difference between the models and reality. One should be aware of these differences. How could you be aware of differences, if you don't understand the circuit?

    Take this thread for example. The lady designed a voltage divider to get 1.4V from a 10V supply. She did ist using 10 Ohm + 60 Ohm in a resistive divider circuit. Works like a charm - consumes lots of unnecessary power. Her divider biases an NPN transistor with a base current of ~80µA. Therefore 1.8kOhm + 10.7kOhm do the same job at greatly reduced power.

    SPICE wouldn't tell you the difference. It would show you the bare numbers, but without some solid background knowledge you wouldn't know how to improve the original circuit.
     
  14. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    Harald, no argument there. Spice is a tool and like any tool it's only as good as the hands that holds it. ;)

    Chris
     
  15. Harald Kapp

    Harald Kapp Moderator Moderator

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    I totally agree :D
     
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