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Impedance for a BJT stage, with collector-base feedback resistor

Discussion in 'General Electronics Chat' started by patroclus, Feb 6, 2012.

  1. patroclus

    patroclus New Member

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    BJT collector-base feedback impedance

    Hello.

    I'm getting trouble on getting (using PSPICE-like software) the input and output impedance of a common-emitter BJT amplifier with collector-base resistor, to try to see how it compares with a more common emitter degeneration feedback (or both). I attached schematic pictures of my set-up. I hope someone can help me see where am I missing something.

    As far as I understand, this is a paralell-parallel negative feedback (voltage sampling, current mixing), so, in theory, both impedances should decrease in a factor of (1+AB), being A: amplifier gain, B: feedback circuit gain. Thus, this amplifier should have lower input and output impedance than and emitter-degeneration one (generally speaking, of course).

    The first picture shows the circuit and simulation for input impedance, by doing an AC sweep and plotting:
    Zi = (voltage at Vi) / (current through R4).
    I get 186 ohms. This example is taken out of a book, where the author works out a very close 165 ohms Zi, so I think it's OK...

    But, the second picture shows the simulation for output impedance, by placing a 1A test signal at output and shorting amplifier input to ground.
    The book states that Zo = 495 ohm.
    I get Zo = 4,27 kohms (which is just the parallel of collector and feedback resistor, which I find to be something logic. I don't know how to get the close-to 500 ohm result...

    The values are significant. In fact, (1+AB) = 8,63 for this circuit.
    and the value 495 ohms is got from:
    Zo = 4,27 kohms / 8,63 = 495 ohms.

    4,27kohms is what I find.

    Any light?
    Thank you so much.
     
  2. audioguru

    audioguru Well-Known Member Most Helpful Member

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    You forgot that negative feedback actively reduces the output impedance, in addition to the resistors being paralleled.
     
    Last edited: Feb 7, 2012
  3. gilmanli

    gilmanli New Member

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    You have first check the DC biase at collector of transistor,it should be 1/2 of V2 then feed a signal ( 1kHz ) to check whether the output is disortion or not.If it is serious disortion at output,this is no meaning you measure those impedance.
     
  4. dave

    Dave New Member

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  5. crutschow

    crutschow Well-Known Member Most Helpful Member

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    To measure the output impedance you need to ground the input at V1 (or just leave V1 connected since it has a zero AC impedance to ground), not ground the base (which effectively turns the transistor off).
     
  6. patroclus

    patroclus New Member

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    Thanks for all replies, I almost got it!

    V1 is an AC source, so it isn't zero AC impedance to ground. But, following your tip, I shorted it (see attach) and now I get a Zo = 508.7 ohms, which is fairly close to the theorical 495 ohms !

    Yes, it's a little bit below 1/2 of supply voltage.

    I'm still missing something.
    If I use a test current of 1A at output, I measure 508,7 V (from where I get Zo = 508.7V / 1A = 508.7 ohm).
    BUT... I think that 508.7 V from collector to emitter is something crazy, unless maybe you are trying to destroy the transistor. At 508V output, this amplifier just isn't working. Am I wrong?
    If I use a test current of 1mA, I measure 508 mV (this seems reasonable), which leads to the same Zo = 508 ohms.

    Is it OK to use a 1A test current to find Zo in any circuit, even though that would make an impossible output?
     
  7. audioguru

    audioguru Well-Known Member Most Helpful Member

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    You should not feed 1000mA into the output of your circuit that idles at 1.3mA and has a peak output current of 2.6mA.
     
  8. crutschow

    crutschow Well-Known Member Most Helpful Member

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    It may work in an AC simulation, but it is not realiistic and would never work in the actual circuit.
     
    Last edited: Feb 7, 2012
  9. crutschow

    crutschow Well-Known Member Most Helpful Member

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    An ideal AC voltage source does have zero AC impedance to ground. An ideal current-source has infinite impedance.
     
  10. patroclus

    patroclus New Member

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    I see. In simulation software it is OK to use a 1A current to get Zo. Right now, I'm only interested in simulation.
    Thank you guys.

    True.
    But, if I don't short the AC source to ground, I get Zo = 3.6 kohms, which just isn't right (I can attach the LTSpice project if you want to quickly try it).
    To get the right Zo I need to set AC source to 0V or short it.
    According to "output impedance" definition I read in my book, it is the (output voltage) / (output current) with signal source = 0V. I understand that to be AC source short to ground.
     
  11. crutschow

    crutschow Well-Known Member Most Helpful Member

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    Certainly you have to set the AC voltage to 0 when you do the simulation. Otherwise the input signal interferes with the signal you are injecting at the collector.

    Even though the simulation may work at 1A it's still better to simulate under real conditions rather than impossible conditions.
     
  12. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    You dont short the base to ground. That makes the input impedance (at that point) zero but it also changes the DC bias point which is not allowed in a test like this.

    But there seems to be some confusion about what the input is and what it is not. Is it the AC input, or the base of the transistor? In an amplifier the input is the input beyond some other components, it's not usually the base. Therefore shorting the base doesnt make sense anyway because that's not going to give you a true result of what anything is, even if it did not affect the DC bias point.

    To short the actual input to ground simply set the AC input amplitude to some very low value, like 0.000001v or actually zero. If you need to place an AC short at any other place to ground, use a very large value capacitor, and in a transient simulation allow enough time for the cap to come up to the required DC operating point, then set the initial value of the cap to that value (or close to it) for the next simulations, still allowing a little time for it to come to the more exact value. The AC component will be shorted to ground, but the DC will be what it should be for that circuit.
    There is doubt as to what you are trying to do though, in shorting the base to ground. You shouldnt need to do that unless the base is really the input to the amplifier, which it doesnt look like it should be here.
    If you short the actual input to ground and you get a reading that seems too high, that may be what it actually is, but first you have to clear up what it is you really want to short out here, which really should be the input.
    Are you sure the input resistor is 10k?

    Also, you do not need to use a 1 amp current to test the output. You can use a lower value current. The output impedance would be the change in voltage divided by the change in current, for any reasonable current level.
    Try using a 1 milliamp peak test current on the output with the input AC source shorted to zero. Measure the output voltage total deviation Vdev in peak to peak volts. Divide this voltage by the test current and divide the result by 2:
    Rout=Vdev/(Itest*2)

    If you'd like to double check, with a small input AC test voltage of say 0.1v peak, measure the no load output AC peak to peak voltage, then load the output with a resistor of say 500 ohms in series with a 2 Farad capacitor and measure the output voltage again. Lower the 500 ohm resistor value until the output reaches 1/2 of the measurement with no load. The resistor then equals the output Rout approximately. This assumes a test frequency of 1kHz or better.
     
    Last edited: Feb 7, 2012
  13. patroclus

    patroclus New Member

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    Yes. I understand what you say, but the schematic in the book just calls R4 "R_signal" (the 1 ohm resistor is ommited, I added for testing purposes).
    That is why I just shorted the base to ground, though, obviouly, I forgot that this fact would just cut the transistor off.

    The book distinguishes between "Ro" and "Rout".
    "Ro" is measured with amplifier input shorted (so input resistor effectively "dissapears").
    "Rout" is measured with source shorted to ground (this time input resistor comes in).
    I find these concepts a bit odd yet. In fact, I don't see how to apply them to this particular amplifier as shorting the base just disables the amplifier.


    Why divide by 2?
    It works for me just as
    Rout = Vout / Iout
    So if Iout = 1mA, Rout is just Vout * 1000.

    Absolutely right. Just checked it.
    At first I forgot to place the cap, and I got very high Rout.
    It's important not to make this (maybe) common mistake. The cap is absolutely required.
     
  14. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,


    Divide by two if the voltage is measured in peak to peak volts and the input current is just peak amps. The divide by 2 factor is required. If you measure the voltage in peak volts, you dont need to divide by 2. I suggested measuring the output voltage in peak to peak volts because it is easier to measure than peak volts on a scope or in a sim plot.

    In my last post i mentioned that if you need to place an AC short at any other point in the circuit you can use a large value cap to ground. Thus, a large value cap like 1 Farad from base to ground would effectively short the base to ground for AC yet leave the DC bias point alone, given ample time to charge in the transient analysis. Once the DC operating point is known (after the first run) you can change the initial condition of this large cap to equal the measured DC value, then the simulation will run much faster because the DC level will come up to its exact value more quickly.
    If you look across the cap and see the voltage change by too much (AC) then make the cap value larger. 1 Farad is probably good enough for 1kHz but it also depends on the impedance at that point so a quick check is a good idea.

    If you want to post a snap shot of the book page with this problem that would be good too.
     
    Last edited: Feb 8, 2012
  15. Ratchit

    Ratchit Well-Known Member

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    patroclus,

    R2 is not a feedback resistor for AC. It is a collector-to-base bias resistor. In fact, you should connect two resistors, each half the value of R2 in series. Then connect a capacitor from the junction of the two resistors to ground. This will prevent AC degeneration of the signal.

    So you really don't have any AC feedback. The input impedance will be R4+ hie, and the output impedance is 1/hoe.

    Ratch
     
  16. audioguru

    audioguru Well-Known Member Most Helpful Member

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    No.
    R2 provides AC negative feedback because R4 prevents a very low impedance of the signal generator from shorting the AC negative feedback.
     
  17. Ratchit

    Ratchit Well-Known Member

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    audioguru,

    R2 exists for collector-to-base biasing. Its value for the correct bias might not be what is desired for feedback. It is better to split R2 and bypass it in order to suppress its AC feedback, as I suggested in my previous posting. Then if negative feedback is desired, an emitter resistor should be inserted.

    R4 is there to prevent the source voltage from being applied directly across the emitter-base junction, and thereby give nonlinear performance, or even damage the transistor.

    Ratch
     
  18. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi Ratch,


    I respectfully but emphatically have to disagree here, on several points.

    First, although it is true that R2 sets the DC bias, it also in part sets the AC voltage gain of the circuit. This should be easy to demonstrate with a circuit simulator.

    Second, if we divide R2 in half and use a capacitor to shunt the AC feedback to ground (so as to obtain a nearly pure DC feedback) we would end up cutting off most of the AC feedback, which would cause a very large uncontrollable AC gain which would probably cause a huge non linearity and hence great distortion in the intended application, but if we reduced the input to make up for that we would end up with a bandpass filter. That's because that feedback section is much like an op amps feedback section and would function in a similar manner.

    Third, R4 is partly what sets the AC gain in conjunction with R2. For example, with R2=50k and R4=10k we get an AC gain of about 5. With R2=100k we would get a gain of (roughly) 10. R2 also sets the DC bias so that's another consideration.

    If you still have a problem with any of the above i would suggest that you take another careful look at this and maybe do a circuit simulation.
     
    Last edited: Feb 9, 2012
  19. audioguru

    audioguru Well-Known Member Most Helpful Member

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    I simulated the transistor:
    1) With a 1k source impedance then the AC voltage gain is high and the distortion at high output levels is very high.
    2) With a 100k source impedance then the voltage gain is low, the distortion is low and the output level is high.
    3) With a 910 ohm emitter resistor the voltage gain is low, the distortion is low and the output level is a little lower than before.
     
    Last edited: Feb 9, 2012
  20. Ratchit

    Ratchit Well-Known Member

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    MrAl,

    That is the point I am trying to make. I believe R2 should only be used for bias and not AC feedback.

    If you want negative AC feedback, then use a emitter resistor.

    Yes, R4 will reduce the gain and protect the emitter-base junction, but not enter into feedback considerations.

    I don't have any problems with my analysis.

    Ratch
     
  21. Ratchit

    Ratchit Well-Known Member

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    audioguru,

    What is the point and conclusions of your simulations?

    Ratch
     

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