Austin's Transistor Dossier: An Analysis of Transistor Design

Discussion in 'General Electronics Chat' started by ELECTRONERD, Nov 3, 2009.

  1. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    Hey Everyone,

    Firstly, let me introduce you to my simple yet seemingly extensive goal; to take any BJT transistor out there and bias it in order to make it a Class-A amplifier. In the future, I plan to move from one amp to the next, thus I am now designing Class-A amps and next it will be Class-AB. Then I will use MOSFETs and such, but first things first. In addition, I will use this particular thread whenever the need of asking questions is evident to me and also to look back on for review. My first design was a practical approach on how to get the BC547B in its linear region. Many of my Class-A amps are designated for audio amplifier operation. I encourage all comments, questions, and suggestions for the benefit of everyone; so please post when you would like to. I'm sure many would feel apprehensive at the length of this thread, but I beg that the knowledgeable come and read it anyway; posting with comments and suggestions.








    Attempt #1:
    1. Upon looking at the BC547 transistors specifications (http://www.fairchildsemi.com/ds/BC%2FBC547.pdf) I looked at the hFE (DC Current Gain) under "Electrical Characteristics". Under the "Test Condition" column, it advises the following conditions: Vce=5V, and Ic=2mA.
    2. That being said, I now looked at the saturation conditions of the transistor and noted that the base current should be 1/20th of the collector current (the provide examples in both the Vce(sat) and Vbe(sat), both of which say that Ib should equal 1/20th of the Ic).
    3. I then set forth the conditions I can now provide-Vce=5V, Ic=2mA, and Ib=0.1mA.
    4. Next, I remembered the equation Vc=Vcc/2 which means that Vc=6V (see schematic in attachment for details). Therefore, I can now choose the collector resistor (Rc). 6V/2mA = 3k
    5. The voltage divider can now be determined: R1=6k and R2=1k. Since I want the emitter voltage to be 1V (according to Vce. Vce=Vc-Ve-or so I think), the base voltage should be 1.7V which the voltage divider ensures.
    6. Now I need to calculate the emitter resistor (Re). Based upon the equation Ie=Ic+Ib, I arrived to the conclusion that Ie should equal 2.1mA. Therefore, 1V/2.1mA = 476Ω.
    That is my first transistor analysis design. Note that the transistor DC current gain (hFE) is Ic/Ib which is a gain of 20. Unfortunately, in the hFE characteristics, it says that the minimum gain should be 110 and the max should be 800. Would my design still provide a gain of 20 and would it be safe for operation? In addition, the voltage gain is Rc/Re which is a gain of 6.3. I am still unsure on how to verify that the base current is 0.1mA due to the voltage divider. I was advised in another thread that I shouldn't worry about that facet. Also note that I chose 0.001μF caps to help make the waveform more linear. In the future, a more extensive approach would be to add a capacitor in parallel with Re to increase the gain and also help with the microphone input. This will depend on the human voice frequency, and since It will be varying, would a varactor capacitor be a good idea? Again, comments, suggestions, and questions are all encouraged!

    Thanks,

    Austin
     
    Last edited: Nov 3, 2009
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Hi Austin,

    You seem to be confusing operation in the linear mode with the saturation mode.

    Your stated goal is to produce a design for a class A amplifier so you need to forget the saturation case - that's for operation as a switch.

    Say you want Ic=2mA you can do a design based on the minimum stated Hfe =110 as follows -

    Ib=Ic/110 = 2e-3/110=18.2uA

    For your divider bias set the divider current as ~10Ib = 182uA or thereabouts.

    For a 12V supply the total divider chain = 12/182e-6=66kohm

    If you want a small-signal voltage gain of 20 then

    Av=Rc/(Re+re) where re=26/Ie[mA]=13Ω with Ie approx Ic

    Say Rc=3.3kΩ, Re=150 [preferred values] - Giving Vce=5.1 V at Ic=2mA

    [Optionally, with Rc=2.7kΩ & Re=120Ω, Av=20 and Vce=6.36 @Ic=2mA]

    Then Av=3300/(150+13)=3300/163=20.25

    At 2mA Ve=150x2e-3=0.3V

    Hence Vb=Vbe+0.3=0.7+0.3=1.0V

    For the bias divider then the lower resistor R2= Vb/182e-6=5.5k

    Since the total chain R1+R2=66k (see above) then R1=60.5k

    These aren't preferred values - you could "tweak" them to 5.6k and 62k say.

    You could also consider the change in bias conditions when Hfe=800 at the upper limit. The difference should not be significant. In fact it should more closely match the design goals.

    I'm sure I've seen a post by a member in which he gives quite simple but practical design guides for this and several other amplifier cases. Do you plan to create some type of design tool - such as a spreadsheet?

    Rgds,

    t_n_k
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I notice on the BC547 data sheet that Vbe=0.66 at VCe=5V and Ic=2mA. I just used the old benchmark of 0.7V for a Si transistor.

    One of the issues with this design is that the bias [Ic] stability [against Hfe variation] is improved as the proportion of Vb due to Vbe is reduced. In this case Vbe dominates the total Vb value of ~1V. In any case, Hfe has to be quite low for this to be a real issue. This circuit will also afford Ic stability against variations in Vbe due to temperature effects with Vbe reducing by about 2.5mV per degree C of temp rise.
     
  4. Eduard Munteanu

    Active Member

    Sep 1, 2007
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    Here's how I do it, and you probably won't find it in electronics books.

    Specifications:
    Z_{in} \quad Z_{out} \quad G_{ac} \quad V_{cc}
    where G_{ac} = \frac{output \quad amplitude}{input \quad amplitude} is the AC gain, which is what you're interested in.

    Datasheet stuff:
    h_{FE_{min}}

    Ok, we know Z_{out}. Then R_3 = R_C = Z_{out}, simple as that, because the impedance looking into the collector is huge in the linear region.

    Next, choose a suitable V_B. It doesn't really matter as long as V_B <= \frac{V_{cc}}{2}. I'd rather choose V_B = \frac{V_{cc}}{2} because of simplicity and because you get the maximum input impedance.

    Now the quiescent collector current is I_Q = \frac{V_{cc} / 2}{R_C}. We choose the base divider such that I_Q < \frac{V_{cc}}{R_1 + R_2} h_{FE_{min}} at all times. This is also simple, because R_1 = R_2 under the previous assumptions. If the transistor is suitable, the input impedance requirement will also be met.

    We now go to the emitter side. In parallel with R_E = R_4 we put a capacitor and resistor in series, both going to ground. Let these two be C_3 and R_5. Under the previous assumptions we must set DC gain to unity (-1) and AC gain to G_{ac}.

    We know that:
    G_{dc} = - \frac{R_C}{r_e + R_4}
    G_{ac} = - \frac{R_C}{r_e + R_5}
    under the following assumptions: R_5 << R_4 and the impedance of the capacitor is negligible at the desired frequencies. r_e is the intrinsic emitter resistance and it is easy to approximate (Google it!).

    For now, ignoring the intrinsic emitter resistance, we can choose:
    R_4 = R_C = R_3
    R_5 = \frac{R_C}{G_{ac}}

    And we're done. The only thing left is to choose small enough impedances for C_1 and C_2.
     
  5. Ron H

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    Apr 14, 2005
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    Eduard, it appears you have chosen the bias point such that Vb=Vcc/2 and Vc=Vcc/2. The transistor will then have Vce≈0.7V.
    Vb=Vcc/2 does not allow enough signal swing at the collector.
     
  6. Eduard Munteanu

    Active Member

    Sep 1, 2007
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    Yes, sorry, I oversimplified it in the attempt to be short.

    ELECTRONERD, just set Vb lower (so that the transistor operates in class A) and calculate DC gain so the quiescent output voltage is Vcc / 2.
     
  7. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    Thank you everyone for the information-packed responses, I appreciate them!

    I'll start with t_n_k, you say:

    What is 12/182e-6=66kΩ? When you say to set the divider current as ≈10Ib, which is 182μA, why is that so? Is it a simple rule of thumb?

    Furthermore, what is "re"? Also, when you have re=\frac{26}{Ie}, how do you get 26? Is it a constant?

    I want to get those questions straightened out first, so don't be suprised if I have more questions regarding this post.

    Thanks,

    Austin
     
  8. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    Upon further search, Eduard already answered one of my questions:

    But, why should I have to approximate it? Shouldn't the specs tell me what the instrinsic emitter resistance is?

    Also a further note, I have the BC547B which means that the hFE can go from 200~450.

    Thanks,

    Austin
     
  9. hobbyist

    Distinguished Member

    Aug 10, 2008
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    Heres my go at it.


    Simple Design example:
    --------------------------

    PARAMATERS:

    VCC = 12v.
    RLoad = 20K
    Vout pk-pk. = 10v.
    Av = -10 (CE amp)
    Zoutgen.~ = 800 ohms, (audio generator output impedance)
    Vin pk-pk = 1v.
    Freq response = 100Hz - 100Khz
    ---------------------------------------------------------------------------------

    DESIGN:

    Choose a CE amp.
    RC = (Rload / 10) = 2K ohms
    RE = (RC / Av) = 200 ohms
    VC = (VCC / 2) = 6V.
    IC = (VC / RC) = 3mA.
    VE = (IC x RE) = 0.6V.
    Vbe ~= 0.7V.
    VB = (VE + Vbe) = 1.3V.
    R1 ~ = (5 to 20 x RE) = (chosen) 3.9K ohms. (base to ground)
    IB1 = (VB / R1) = 333uA.
    R2 = {(VCC - VB) / IB1} = 33K ohms
    ---------------------------------------------------------------------------------

    TEST MEASUREMENTS:

    STATIC:

    VC = 6.92V.
    VE = 0.51V.
    VB = 1.2V.
    ---------------------------

    DYNAMIC:

    Vin pk-pk = 1V.
    Vout pk-pk = 8.8V.
    Av ~= -8.8
    Fco (min) < 100 Hz.
    Fco (max) ~= 700 Khz. (- 3 db)
    Waveform Symetrical above and below zero reference.
    ---------------------------------------------------------------------------------

    Analysis:

    The VC is higher than calculated because R2 calculates out to around 32K ohms.
    So with higher R2 means less base current so transistor conducts less thereby allowing more collector voltage.

    With higher R2 also some base current loading the divider causing VB to be lower than calculated.

    And claculations also assumed 0.7v. for Vbe. (actual could be more or less)

    Av. suffers due to the low Zin of the 200 ohm emitter resistance, input waveform becomes distorted a little above the 1V. p-p threshold, also the Rload of 20K drops some of the output voltage due to RC being only 1/10 of Rload.
    Calculations did not incure the Rload in parrallel with RC when determining the RE value.

    In more tighter tolerances then, the RE would be calculated with the total Rout of the entire stage including the Load in parrallel with RC.

    And the peek Vin is quite large with respect to the bias base voltage, the base voltage would need to be made larger to handle a higher input peek signal.

    Further tweaking of bias voltages may be needed to gain maximum voltage gain.

    I have now changed the RE from 200 ohms down to 180 ohms, this is a calculation of using RC of 2K in parrallel with the RLoad of 20K, so , {(RC // RLoad) / Av. } = 180 ohms for RE.

    Av. now is running good with a symettrical 10V. pk-pk at the output with a nice 1V. pk-pk input wave.

    So Now I have the Voltage gain required for this design.

    Depending on the application this could be used with (loose or tight tolerances) will determine how well of a design this is.
    ---------------------------------------------------------------------------------
     
    Last edited: Nov 4, 2009
  10. Ron H

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    Apr 14, 2005
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    Intrinsic emitter resistance re≈.026/Ie ohms, where Ie is the emitter current in Amps. Note that, if the signal current is a significant percentage of the quiescent current, significant harmonic distortion will occur, because the instantaneous re, and therefore the gain, will change due to the signal current.
     
  11. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Hi Austin,

    182e-6 is my exponential form notation for 182x10^-6 i.e. 182 [uA] in this case.

    A reasonable rule of thumb is to make the bias divider current (i.e. through R1 & R2) at least 10xIb - it just ensures the resulting base voltage is reasonably easy to "fix" and stabilize with the unknown variation in Hfe (or Beta).

    Yes, the intrinsic emitter resistance is obtained by another reasonably consistent & widely used relationship derived from the emitter junction forward transfer function model conductance, gm=q*Ie/(kT) [unit A/V]. Where q is the charge of the electron, k is Boltzmann's constant and T is the absolute temperature. re is given by 1/gm. Since gm & hence re are temperature dependent then the approximation is usually taken at 300K - 27°C - a pleasant day. The "exact" value at 300K would be re=25.86/Ie[mA] - close enough to 26/Ie[mA].

    Because re is small, it can often be ignored when any included external Re is >>re. In this particular design case it does have some (albeit small) effect on the gain result. If there is no external Re then re obviously becomes a prime determinant of the gain.
     
  12. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    Hi t_n_k,

    Wouldn't that thumb rule simply make the gain 11 instead of 110?

    Austin
     
  13. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    Attempt #2:
    1. Looking once more at the BC547B transistors specs, if you take a look at page #2 of the datasheet, the first graph visible is Figure 1. "Static Characteristic". I will choose Ib to equal 50μA and therefore Ic is close to 12mA and Vce=5V.
    2. Vc=Vcc/2 = 7.5V/12mA = 625Ω. Since Vce=5V, 7.5V - 5V = 2.5V at the emitter. Thus, 2.5V + 0.66V = 3.16V at the base.
    3. Since Ib=50μA, 3.16V/50μA = 63.2kΩ resistor.
    4. Since Ie≈Ic = 12mA, Re= 2.5V/12mA = 208Ω.
    The first circuit in the attachment is without a voltage divider (notice that I use the 2N3904, I couldn't find the BC547B in the simulation program). Next I added a voltage divider: 63.2k\frac{15V}{3.16V} = 300k So R1 = 300k. R2 = (\frac{1}{63.2k} - \frac{1}{300k})^- ^1 = 80k. So R1=300k and R2=80k. R1 ensures the 50μA \frac{15V}{300k}=50uA and both the resistors combined ensures the voltage \frac{80k}{300k + 80k}15V=3.16V

    The hFE = 240 and the Av = 3. The linearity is fairly good, the distortion picks up at a little less than 1MHz which isn't a problem. The human voice ranges from 300Hz to 3400Hz (excluding harmonis I believe).

    It was interesting to see a phase change from the input signal to the output signal, is there any way I can calculate the difference? I assume I can subtract a point on the input signal from the same point on the output signal? I'm hoping I did the voltage divider compution correctly, if you compare the waveforms of the two circuits you will see there is a change.

    Any comments, questions, and advice is appreciated!

    Thanks,

    Austin
     
    Last edited: Nov 7, 2009
  14. Ron H

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    Apr 14, 2005
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    It appears that, in your simulation, you connected your signal source directly from the base to ground. Your input is symmetrical around ground, and your output is at 15V. The phase shift is -90°, because the input is coupling to the output through the collector-base capacitance. The transistor is actually OFF.
    You need to put a capacitor in series between the signal source and the base.
     
    Last edited: Nov 7, 2009
  15. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Hi Austin,

    With the new circuit, your voltage divider biasing network won't afford much stability in Ic against Beta variations. It seems you missed my point about making the divider current much greater (~10x) than the base current.

    Doing some DC operating point calculations for your circuit, if Beta varies from 110 to 800, the collector current will vary from around 3.2mA to 8.6mA (using Vbe=0.7V - see Fig2. on the data sheet for Vbe vs Ic). This gives about a 180% change in Ic over the Beta range. Notice you are also a long way from the predicted colector current of 12mA. You appear to have made an error in your Rc & Re value selections - step 2 of your post.

    Suppose you keep everything else the same and make RB1=22k (not 300k) and RB2=6.8k (not 80k). In that case Ic would vary from around 11mA to 13.2mA (a more modest 20% change). At Beta =240 you would have Ic=12.3mA which is about what you wanted.
     
  16. hobbyist

    Distinguished Member

    Aug 10, 2008
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    Austin,

    I don't mean to butt in on ya guys convo. here but...

    The reason you don't have the proper collector current, is because you don't have the 3.16v. at the base.

    I simulated your circuit,
    first without the transistor, and the volt. divider shows around 3.16v.

    then with the transistor base hooked up to the divider, and now the divider dropped to around 1.5v.

    That's why T N K was saying about the base current needs to be at least 10 times less than the divider current.

    The divider current is around 40uA.

    Because you used the base current figure to calculate the divider current. Wrong thing to do.

    If your base current is 50uA, then choose a voltage divider resistors that will give you at LEAST 500uA, preferably more, the whole reason this is done is because the base current will affect the base voltage, if the divider current is too small.

    So do you see what were trying to say to you on this now.

    A good rule of thumb is to make the bottom divider resistor no larger than around 20 times the emitter resistor, that will ensure a large enough divider current to keep the base current from upsetting the divider voltage.

    any where from 5 - 20 times RE is a good value for R3 in your circuit.
     
  17. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    There is a lot of good info in this thread.

    Have you read "The Art Of Electronics" by Horowitz and Hill? They show a few methods of design, with explanations similar to those by t_n_k, Eduard, and Hobbyist above.
     
  18. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    Thank you very much everyone for all the information! I realize my mistake now when I didn't do 10Ib. Tell me if I'm thinking about this correctly; the voltage divider can be explained as a power supply and in our case it can supply 500μA. However, when we had 50μA (according to hobbyist it was actually 40μA) the load from the base required more and so there was a voltage drop from the voltage divider. But, since we have 500μA, the voltage will remain the same until the load is far too high, but our load isn't very high.

    I will change everything in the simulation, and Ron, I added that cap you advised and there was a difference.

    Austin
     
  19. ELECTRONERD

    Thread Starter Senior Member

    May 26, 2009
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    Ok I changed the circuit as shown in the first attachment, and there was a significant amount of amplification. Since I supplied 500μA from the voltage divider, would the base actually be taking 50μA?

    Notice that when I added the voltage divider, the waveforms are different. Why? Shouldn't they be the same?

    Thanks,

    Austin
     
  20. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Add a couple of digits of precision (6.3158k, 3.1579V) to your Thevenin equivalents and you will see the differences disappear.
     
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