my emitter-follower calculations don't line up with Art of Electronics

Discussion in 'Homework Help' started by nubelube, Jan 12, 2010.

  1. nubelube

    Thread Starter New Member

    Jan 3, 2010
    16
    1
    I was reading the beginning of Ch.2 of Art of Electronics, and I thought I understood how transistors behaved. The first example they gave was a 10 V source that connects to 1) a 100 Ohm lamp that goes to the (npn transistor) Collector and 2) a 1 kOhm resistor that goes to Base. (Emitter at ground.) You assume the Base-Emitter junction drops 0.6 V, so you have 9.4 V through the 1k resistor, hence 9.4mA. Assuming Beta=100, the collector then tries to drive 940 mA, but it can't because the 100 Ohm lamp limits current to 0.1 A for a 10 V source. Cool.

    They reinforce this by going on to describe the Transistor Man that looks at the Base-junction circuit acting independently and then sets the effective resistance of the Collector in order to achieve I_C = Beta * I_B. (Up to transistor saturation at 0 Ohms, or rather 0.2 V drop.) Also fine.

    The next thing they discuss is the emitter-follower. 10V at the Collector, Base receives variable V_in=V_base, and Emitter passes through resistor R to ground. V_out=V_emitter measured between Emitter and resistor.

    My initial analysis (using mostly their equations):
    V_out=V_in-0.6 (their equation)
    dV_out = dV_in (theirs)
    dI_out = dV_out / R = dV_in / R (theirs)
    Depending on how you look at it, I either implicitly assumed that I could treat the Base-junction circuit separately or, as I describe below, I may have implicitly assumed that the current:voltage relationship across the Base junction continued to hold such that the voltage drop was 0.6 regardless of current. Anyway, I then did:
    dI_in = d(V_in-0.6) / R = dV_in / R

    Of course, if dV_out = dV_in (theirs) my assumption implies bad things like dI_collector=dI_out - dI_in = 0, which would be a good indication that I shouldn't do transistor analysis this way... :)

    By their calcuation,
    I_out = I_in + I_collector = (1+Beta)*I_in
    Hence, dI_in = dI_out / (1+Beta)
    Since dI_out = dV_out / R and dV_in = dV_out,
    dI_in = dV_in / [(1+Beta)*R]
    (rather than my dI_in = dV_in / R)

    This was confusing the *crap* out of me. After banging my head against the wall for awhile, I realized that my assumption about the fixed 0.6 V drop may be the problem. A diode has an exponential I:V relationship, so you can usually just assume a 0.6V (or 0.7) drop and then treat it like a short circuit. However, maybe that's not the case here; if I treat the collector as a super-diode that has I:V relationship 100 times that of the base (and is magically dependent on the Base-Emitter voltage drop), then it seems like that might somehow be able to save me.


    At the outset, I thought I could just calculate i_Base by converting the Base junction to a diode and independently calculating the current of that circuit assuming a 0.6V drop across the diode. Then take the current multiplier, and the Collector either delivers that current or delivers as much as it can under saturation. (As with the lamp-example.) But if I treat the Base-junction voltage drop as fixed 0.6V, then I seem to get bad answers.

    If it isn't fixed at 0.6V, what voltage drop should I be using in order to calculate the Base current?? Something dependent on the Beta multiplier? Would that approach deal properly with transistors in saturation?

    Or did I make some other silly mistake and I'm off in left field?

    Sorry for the long post...thanks for reading it.
     
  2. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    The base current must be much higher than a beta calculation for a transistor to saturate. Look at the datasheet for any transistor and it shows the max saturation voltage drop (Vce) when the base current is 1/10th to 1/20th of the collector current regardless of the beta. Beta is shown when the transistor is linear (not saturated) with plenty of collector-emitter voltage.

    Don't worry about the very small difference between a base-emitter voltage of 0.6V or 0.7V in the base current calculation.
     
  3. nubelube

    Thread Starter New Member

    Jan 3, 2010
    16
    1
    Thanks for the reply. Pardon my ignorance, but I'm afraid I don't quite follow what you're saying. Is your response addressing my main question or perhaps some sub-part of it? In particular, I'm not sure what saturation has to do with it.

    In case my question got lost in my really long post, I'm essentially wondering why my initial calculation about the effective resistance (dV_in / dI_in) of an emitter follower is wrong (it's supposed to be (1+Beta)*R, while I got just R). It seems to me that this has to do with an unexpected relationship between base-current and voltage-drop across the base, but perhaps I'm mistaken. Your response seems to relate to dI_in, but that's about as far as I got.
     
  4. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    Start with a schematic. It is the reference for all else. Given this circuits basic simplicity it should be no big deal.
     
  5. hobbyist

    Distinguished Member

    Aug 10, 2008
    764
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    The reason the resistance is ((B+1) x R), is only because the emitter resistor is part of the base curcuit, as well as the collector circuit,

    If the collector was not connected, than the I_b = {(Vin - Vbe) / R}

    But with the collector connected than part of that current through R is now branching to the collector circuit as well, so now R has a effective resistance seen at the base as being much LARGER due to more current branching to the collector and less current flowing to the base.

    While initially ALL current flowed through the base junction. (collector NOT connected)

    BUT NOW (collector connected), PART of THAT total current branches to the collector by a value of Beta, so now less current in the base means higher input resistance.

    The current flowing through R = IE. (emitter current)

    So IE = (Ib + Ic)

    but Ic = (B x Ib)

    therefor IE = { Ib + (B x Ib)}

    which simplifies to {Ib x (B + 1)}

    final equation, IE = {ib x (B + 1)}
     
    Last edited: Jan 13, 2010
  6. nubelube

    Thread Starter New Member

    Jan 3, 2010
    16
    1
    You're right; sorry about that...

    Crap, I made the diagram, but spaces disappeared. I'm sure there's a better way, but I have to run in a few minutes, so I just replaced spaces with periods. Ignore all periods.


    .............................10 volts
    .................................|
    ................................./
    .......................Base |/ Collector
    .............V_in ---------|
    ..............................|\ Emitter
    .................................\
    ..................................|
    ..................................|-------- V_out
    ..................................|
    ..................................>
    ..................................> resistor R
    ..................................>
    ..................................|
    .......................0 volts Ground
     
  7. nubelube

    Thread Starter New Member

    Jan 3, 2010
    16
    1

    Thanks for all the thoughtful responses! I did eventually figure things out in a way that makes sense to me. It's basically what people have been saying, but for me the key was in hashing out where my diode-based model of the Base-Emitter junction failed. I've written it up all in one place, in case it might benefit others who land on this page.

    All references to Art of Electronics 2nd edition.

    On p. 63, the diagram shows (ignore periods; they are just for spacing)

    ---Vdd---
    |..........|
    R.........L
    |..........|
    |.........C
    |------B
    ..........E
    ..........|
    .....Ground

    Where Vdd=10 Volts, L=10 Ohm lamp, R=1000 Ohm resistor, and C-B-E are Collector-Base-Emitter of an NPN transistor.

    They calculate the current through the Lamp.
    V_Base - V_Emitter = 0.6 So there is a 10-0.6=9.4 Volt drop across resistor R. Hence I_Base=9.4mA.
    The transistor has Beta=100; that is I_Collector=Beta*I_Base=940mA. (I have actually modified their example slightly; they had lamp L=100 Ohm, which corresponded to the transistor being in saturation, but that's not relevant here.)

    On p. 64, they introduce the Transistor-Man model of a transistor. In this model, the Base-Emitter junction is a diode. And the Collector-Emitter junction is a variable resistor, whose resistance is set so as to maintain I_Collector = Beta*I_Base.

    On p. 65, they introduce the Emitter follower:
    .......+Vdd
    ...........|
    ...........|
    ...........C
    V_in---B
    ...........E
    ...........|--- V_out
    ...........R
    ...........|
    .......Ground

    Where Vdd is some positive voltage, C-B-E are Collector-Base-Emitter of a transistor, V_in is some input voltage applied to the Base (V_in=V_Base), V_out is the voltage at the Emitter (V_out=V_Emitter), and R is some resistor.

    Given V_Base (=V_in), what are I_Base, V_Emitter, I_Emitter?

    Adapting the approach taken previously, I proceeded as follows:
    Treat the Base junction like a diode with a fixed voltage drop, assign the remaining voltage drop to whatever resistor(s) are in the path, and calculate the Base current as that voltage drop over that resistance. In equations:
    V_Emitter = V_Base - 0.6
    I_Base = V_Emitter / R = (V_Base - 0.6) / R
    By this reasoning,
    dI_Base = dV_Base / R

    But this is incorrect. dI_Base = dV_Base / [R*(1+Beta)]

    One or more of the following must be incorrect:
    1) V_Emitter = V_Base - 0.6, or
    2) I_Base = V_Emitter / R

    #1 is true, and #2 is not. Thus,
    V_Emitter = V_Base - 0.6, but
    I_Base does not equal V_Emitter / R [= (V_Base - 0.6)/R ]


    --------- CONCLUSION:
    In other words, YOU TREAT THE BASE-EMITTER JUNCTION LIKE A DIODE FOR THE PURPOSES OF DETERMINING VOLTAGE DROP: V_Emitter = V_Base - 0.6. However, YOU DO NOT CONTINUE TREATING THE BASE-EMITTER JUNCTION LIKE A DIODE FOR THE PURPOSES OF DETERMINING THE BASE-EMITTER CURRENT!!

    V_Emitter = V_Base - 0.6
    I_Emitter = V_Emitter / R = (V_Base - 0.6) / R
    I_Emitter = I_Base + I_Collector
    I_Collector = Beta*I_Base
    I_Emitter = (1+Beta)*I_Base
    (V_Base - 0.6) / R = (1+Beta)*I_Base
    dI_Base = dV_Base / [(1+Beta)*R]

    This nuance is not particularly clear in the Transistor-Man diagram, where the Base-Emitter path looks like a diode circuit that would follow both voltage and current rules of a diode. Rather, not only does Transistor-Man decrease the effective Collector-Emitter resistance so that I_Collector=Beta*I_Base (for a given Collector-Emitter voltage), but he also increases the effective resistance of Base-Emitter, so that dV_Base / dI_Base = R*(1+Beta) rather than R.

    This nuance doesn't come up in the lamp example because there is no resistor at the Emitter. Thus R=0, and the difference between R and R*(1+Beta) doesn't matter.
    ----------------------------

    Thanks again everyone for all your help!
     
  8. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    The Art Of Electronics confirms what I said before that the base current must be much higher than the collector current divided by beta for a transistor to saturate well.

    Most transistors have a Max Saturation Voltage Loss spec when the base current is 1/10 the the collector current regardless of beta.
     
  9. nubelube

    Thread Starter New Member

    Jan 3, 2010
    16
    1
    Agreed. Sorry, I didn't mean to imply anything by referring to saturation not being relevant to the problem. I just meant that it wasn't the source of my confusion. I was confused about how they managed to calculate the effective resistance dV/dI of the Base-Emitter junction for a transistor in the active region, and I was perfectly willing to stipulate that the transistor was not yet in saturation. Of course, if your goal was to force the transistor into saturation, then the Beta current multiplier would not hold there.
     
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