Small Signal Model Problem

Discussion in 'Homework Help' started by jegues, Apr 20, 2011.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    I've got some questions about the solutions given to part b) and c) to this particular problem.

    See figure attached for problem statement, given solution, and my solution.

    For part b,

    Everything makes sense up to the point where he obtains Vo = 5mV and I = 0.5uA and thus iD = 0.25uA.

    Now we want the diode currents iD to remain within 10% of their DC value. For this we note that the smallest value of I will be when it decrease by 10%

    So Ismall = 0.5uA - (0.1 * 0.5uA) = 0.45uA.

    Am I incorrect in thinking this way?

    Also, for part c it seems to me that the largest vo obtainable is 11V, 1v above its typical value of 10v.

    Why are they giving 1V as the max output signal? Are they referring to the largest deviation the output signal can have? Am I confusing something?

    Thanks again!
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Regarding part (b)....
    There's another way to argue for the minimum value of I that makes more sense to me (at least) & ends up with the same answer.

    The small signal source voltage = Vs
    The small signal diode current = id
    The small signal diode resistance = rd
    The diode DC current = Id
    rd=VT/Id
    Will assume VT=.025 at ambient temp

    The small signal load resistance Rload=10k

    The reasoning is that as Id decreases the ratio of id/Id increases. It is required that id/Id<=0.1 (10%) with Vs=10mV=0.01V ....

    So

    V_s=2i_dr_d+2i_dR_{load}

    V_s=2i_d\frac{V_T}{I_d}+2i_d10^4

    V_s=2i_d\frac{V_T}{I_d}+2i_d10^4\frac{I_d}{I_d}

    V_s=2\frac{i_d}{I_d}V_T+2\frac{i_d}{I_d}I_d10^4

    V_s=2(0.1)V_T+2(0.1) I_d 10^4

    V_s=2(0.1)(0.025)+2000I_d

    V_s=0.005+2000I_d=0.01

    I_d=\frac{0.01-0.005}{2000}=2.5uA

    hence the minimum current I is

    I=2I_d=5uA
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    As a means of gaining confidence in such calculations I tried a simulation using BAR99 silicon switching diodes as the model.

    At I=2.5uA bias per diode the ratio of rms signal current to bias current was 9.09%. A reasonable agreement.

    EDIT: Actually, I checked the result & it wasn't that good. I had set a small signal input level higher than 10mV peak. With 10mV peak the rms-to-DC ratio at I=1uA DC bias is only 6.4%.
     
    Last edited: Apr 21, 2011
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    It occurs to me that the voltage VT for diodes must be quite different to the often quoted 26mV rule-of-thumb value at room temp.

    I don't have equipment with which to take measurements but simulation models seem to confirm my "suspicions".

    I can easily model a 1N4184 diode with a DC bias and superimpose a small AC signal to extract the diode forward dynamic resistance.

    Here are some values leading to an implied VT value.

    At 1mA DC bias and 0.1 mA rms AC small signal,
    vac=4.25mV giving rd=42.5Ω or VT=0.0425

    At 10mA DC bias and 0.1 mA rms AC small signal,
    vac=0.437mV giving rd=4.37Ω or VT=0.0437

    Another diode - the BAV100 - shows up with VT=0.053 at 1mA DC bias. That's almost double the rule-of-thumb value.

    Interestingly if I substitute a 2N2222 BJT transistor for the diode in the simulation - base and collector joined as the pseudo anode and emitter as the cathode - I get values for VT of 0.027V at 1mA DC bias and 0.031V at 10mA bias. These are closer to the rule-of-thumb value adopted for small signal analysis.
     
  5. Adjuster

    Well-Known Member

    Dec 26, 2010
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    I see what you are saying, but VT is equal to kT/q, which has whatever value applies at a given temperature. I think that this discrepancy of slope can probably be explained by different values of an "ideality factor" n in the diode equation. The factor n typically varies between 1 and 2 for silicon junctions. At higher currents there can be effects due to ohmic resistance and (in practical measurements ) self-heating.

    So I = Is (e(VD/nVT) - 1), or approximately I = Is e(VD/nVT)

    Sorry that I can't get superscript to work so I cannot show the exponentials properly - see link:

    http://en.wikipedia.org/wiki/Diode_modelling#Shockley_diode_model
     
  6. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
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    I made another attempt at Part C) again without looking at the solution and here's what I've come up with.

    I can't see where I'm going wrong.

    I_{D} = \frac{I}{2}

    They give it as 1mA so,

    I_{D} = 0.5mA

    Now, i_{D} can only vary by to ±10% of this value so,

    0.45mA <= i_{D} <= 0.55mA

    For the largest output we would want iD to be as large as possible so take i_{D} = 0.55mA

    This results in,

    v_{o} = (2i_{D})(10K) = 11V

    If we want to know the input for this value of the output, simply take the value of the output and divide it by the gain measured at 1mA (found in part A).

    v_{i} = \frac{v_{o}}{\frac{v_{o}}{v_{i}}} = 11.055V

    Should there be a Δ symbol infront of the values they are giving because that's the difference of my two values,

    11.055 - 10 = 1.055V (indicated on solution)

    The way I preform my solution, is it correct?
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    No it's not.

    You are confusing the small signal AC and DC currents.

    The small signal model reduces to what was shown in the solution provided in your very first post. No DC current is taken into account in the AC analysis other than its role in determining the diode dynamic forward resistance value for the model.

    At bias current I of 1mA, each diode takes 500uA. Based on the presumed value of VT=0.025V this gives an individual diode resistance of 50Ω and the same total equivalent value (50Ω) for the series parallel combination of the four diodes configured as a bridge.

    So for the AC signal model you have an unknown signal source Vs and a voltage divider formed by the 50Ω equivalent diode bridge resistance and the 10kΩ load resistance.

    If the individual diode DC bias current is 500uA, the the allowable diode AC signal current would be 50uA, thereby giving a total allowable signal source current of 100uA (50uA+50uA). The 10kΩ load would have an AC signal current of 100uA and a AC signal voltage drop of 1V. The diode network would have a AC signal voltage drop determined from 100uA in 50Ω - i.e. 5mV.

    The AC signal source would therefore be a total of 1V + 5mV or 1.005V.
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Thanks for responding Adjuster. I agree the relevance of the factor 'n' has been lost somewhere in all of this.

    My point is really that one tends to blithely pluck values out of the air and these then become set in the stone (concrete?) of our corporate electronic knowledge. It's worthwhile asking the question sometimes - why is this rule of thumb "true"? The crazy thing is that these rules become embedded in the teaching of electronic circuit models and we (student & teacher(?) alike) think they automatically have a precise & verifiable physical reality.
     
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