Biasing a BJT with a voltage divider, differences in expected vs measured results

Discussion in 'General Electronics Chat' started by tinker95, Jan 10, 2018.

  1. tinker95

    Thread Starter New Member

    May 15, 2017
    11
    0
    I built the amplifier circuit shown, and performed measurements to compare with the values I calculated.
    [​IMG]
    Of particular interest to me is that I expected the base voltage to be about 1.97 volts (I assumed that this was calculated by simply finding the voltage dividers output voltage), but measured a voltage of 1.57 volts. This must mean I have to take some loading of the voltage divider into consideration, but I am not sure what is loading it.
    Any explanation would be lovely, thank you.

    Also, on a side note, I noticed the gain increased significantly, and then fell off once I passed a certain frequency. Could this be due to some resonance? This was built on a breadboard if that makes a difference.
     
  2. ElectricSpidey

    Member

    Dec 2, 2017
    453
    96
    It's the base.
     
  3. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
    2,181
    410
    Yes, the additional loading is provided by the small emitter-base current of the transistor, also in series with R4. It should be a SMALL change from the "unloaded" voltage divider for best stability....however, you need to consider the input impedance too; you don't want R1/R2 to draw too much current.
     
  4. crutschow

    Expert

    Mar 14, 2008
    20,223
    5,718
    The base loading is due to the Q1 base bias current.
    The base current equals the collector current divided by the transistor current gain (hFE or Beta) which is typically about 100.

    At what frequency does the gain rise and then fall?
    The reduction in gain at higher frequency is likely due to stray capacitive loading of the collector output circuit.
     
  5. Audioguru

    Expert

    Dec 20, 2007
    10,605
    1,182
    The collector current is about 3mA so the base current is about 3mA/100= 30uA. The voltage divider should have a current that is 10 time if you are using the low gain 2N2222. Then the total of the voltage divider is 18V/300uA= 60k ohms then the divider is 16V/330uA= about 51k at the top and 2V/300uA= about 6.8k at the bottom.
     
  6. tinker95

    Thread Starter New Member

    May 15, 2017
    11
    0
    Thank you all.
    How do I go about calculating what the actual base voltage would be, with the loading? I am not sure how I would calculate collector or base current without knowing base voltage.

    Also, I believe the peak was somewhere around 20 kHz.
     
  7. Veracohr

    Well-Known Member

    Jan 3, 2011
    662
    100
    Consider a resistor in parallel with R2 that is equal to R4 times the DC current gain.
     
  8. WBahn

    Moderator

    Mar 31, 2012
    23,160
    6,973
    Since you don't know what the actual transistor beta is, you can't calculate it exactly. But you should be able to come close and, with a bit more effort, but some reasonable bounds on it.

    You can go about this a number of ways. One is to analyze the circuit as just any old circuit and see what you get. At the other end are some loose estimates. Let's try the latter first.

    Let's assume the transistor beta is 100. That means that to get the same voltage across that 220 Ω emitter resistor from the base current that you get due to the emitter current, it would have to be 100 times as large (actually 101, but 100 is good enough). So roughly speaking that resistor, as far as the voltage divider is concerned, looks like a 22 kΩ resistor. We are ignoring the base emitter voltage when we make this approximation. So that 22 kΩ resistor is approximately in parallel with the 27 kΩ resistor. Due to the base-emitter voltage, it actually looks like something larger than 22 kΩ, so let's call them about equal and say that the parallel combination of them is about 13 kΩ. Using that as the bottom resistor of the voltage divider would give us 1.0 V at the base.

    This doesn't match too well with your measured result of about 1.6 V, but remember that this is an ultra quick and dirty estimate. However, the question of whether or not the result is off by so much due to the method being too dirty or our assumption of a beta of 100 being wrong is up in the air. What would the beta need to be in order for this approach to yield the observed measurement? Well, by inverting the voltage divider formula and solving for an unknown bottom resistor with a known output voltage, we find that the bottom resistor would need to effectively be 21 kΩ. To get that by putting a resistance in parallel with 27 kΩ we would need a resistance of 94.5 kΩ and to get that from a 220 Ω emitter resistor we would need beta to be 429. Not completely out of line for many NPN transistors, but asking a bit much for 2n2222, especially since under certain conditions it has a spec'ed max beta of 300.

    However, remember that I said that the base-emitter voltage drop makes the effective resistance looking into the base higher? Well, the less the total drop is across the across the base-emitter and emitter resistor combination, the greater this affect. With a base voltage of just 1.57 V, the voltage across the resistor is only about 55% of this. That makes the resistor look almost twice as large as we estimated. So now we would only need a beta of around 200 to get the result we got, which is quite possible for a 2m2222.

    We'll knowing that we will have a fairly small voltage across the base/resistor combo (the unloaded voltage divider is putting out less than 2 V), let's do the analysis a bit more carefully.

    Let Vb be the base voltage. That makes the emitter voltage roughly Vb - Vd where Vd is a diode voltage drop. For a transistor with only 4 mA of collector current, we should probably use something a bit less than o.7 V, so we'll use 0.65 V, but we'll keep it symbolic until the end.

    The emitter current will be (Vb-Vd)/Re and the base current will be reduce from this by a factor of (β+1). We'll keep the +1 for now because we are operating in a low current regime where the beta is reduced. We can always approximate it away later if it really makes things easier.

    That means the total current leaving the base junction if Vb/R2 + (Vb-Vd)/[(β+1)Re], while the total current into it is (Vcc-Vb)/R1. Set these equal and solve for Vb. Using a beta of 100 again, we get Vb = 1.28 V. Asking again what the beta would have to be in order to get 1.57 V, we see that it would need to be about 250.

    That's quite a bit of variation from owing to the variation in beta and that's generally not a good thing. The problem is that the base current is a significant fraction of the current in R1. So make R1 and R2 smaller by a factor of 10. Now, using a beta of just 50, we would expect Vb to be about 1.74 V, which compares pretty favorably with the unloaded voltage of 1.97 V.

    But remember that quick and dirty approach that gave us a really bad result of only 1 V? Let's use it with the new values for R1 and R2. With a beta of 100 we get 1.77 V and with a beta of 50 we get 1.61 V. So our bias circuit has become a lot more insensitive to variations in beta.
     
    tinker95 likes this.
  9. Audioguru

    Expert

    Dec 20, 2007
    10,605
    1,182
    You do not start with the base voltage. Don't you want the output swing to be symmetrical? Then set the collector voltage near half the supply voltage making allowance for the emitter voltage reducing the output voltage swing. Select a supply voltage that will allow the amplitude of the maximum output swing and calculate a collector resistor that works well with your load resistance. Make the emitter resistor value around 1/10th the collector resistor value.
    Now calculate the emitter and base voltages then lookup the transistor's typical hFE at the collector current you have chosen. Make the base voltage divider current 10 times the typical base current so the transistor biasing will work well with a transistor with minimum or maximum hFE.
     
    tinker95 likes this.
  10. crutschow

    Expert

    Mar 14, 2008
    20,223
    5,718
    You can add a diode in series with the added dummy resistor to give more accuracy to the calculated value.
     
  11. tinker95

    Thread Starter New Member

    May 15, 2017
    11
    0
    Thank you all, that was very helpful.
    As for the process of designing the amplifier, It is my understanding that the collector resistor's value is chosen based on the collector current desired. How do you know what this collector current should be?
    For example, If I wish to drive an 8 ohm, .5 Watt speaker, does that determine my collector current?
     
  12. ian field

    AAC Fanatic!

    Oct 27, 2012
    6,252
    1,131
    From the point of view of the emitter resistor; you're dealing with an emitter follower that should follow the voltage from the base divider. R4 should have 0.7V less than the divider. you probably don't want that to be any more than 1/3 Vcc.

    From the collectors point of view; its a constant current generator. The current is set by Vb minus Vbe and R4. Calculate the collector resistor to drop half the remainder of Vcc at that current.

    R4 needs a bypass capacitor or you'll get an awful lot of AC nfb.
     
  13. crutschow

    Expert

    Mar 14, 2008
    20,223
    5,718
    Of course.
    For a Class-A amp the collector current needs to be more than 350mA if you want to drive the 8 ohm speaker @ 0.5W.
    (The collector current needs to be greater than the peak sinewave speaker current of 350mA @ 0.5w).
    The high collector current is why Class-A amps are typically only used for low power loads.
     
    tinker95 likes this.
  14. ian field

    AAC Fanatic!

    Oct 27, 2012
    6,252
    1,131
    Not a great idea without a transformer - the standing current offsets the cone and you can get slamming on alternate half cycles.

    The Philips EE kits used a 150R speaker as the collector load. The EE8/EE20 used either AC126 or 128, the EE1003 had, AFAICR: a BC148. Both needed a heat sink that didn't look like it did much - especially on the lockfit BC148.
     
  15. Audioguru

    Expert

    Dec 20, 2007
    10,605
    1,182
    Transformers, the germanium AC128 and the lockfit BC148 were used in amplifiers made by Philips 54 years ago when I worked for Philips. An output coupling capacitor can block the DC instead of a transformer.
    Why would anybody want to heat their home with a class-A audio amplifier today??

    Today, most audio amplifiers are class-AB push-pull but digital switching class-D will soon be the most used type.
     
  16. tinker95

    Thread Starter New Member

    May 15, 2017
    11
    0
    Great. How was the 350mA arrived at?
     
  17. crutschow

    Expert

    Mar 14, 2008
    20,223
    5,718
    Of course not.
    That's why the speaker is typically capacitively coupled.
     
  18. crutschow

    Expert

    Mar 14, 2008
    20,223
    5,718
    Basic Ohm's law for 0.5W into 8 ohms.
    0.5W = I²R
    I² = 0.5/8 = .0625
    I = √(.0625) = 250 mArms
    Ipk = 1.4 * 250 mArms = 350 mApk.
     
    tinker95 likes this.
  19. ian field

    AAC Fanatic!

    Oct 27, 2012
    6,252
    1,131
    Some people don't completely trust class AB for no crossover distortion.
     
  20. Audioguru

    Expert

    Dec 20, 2007
    10,605
    1,182
    Class-B has severe crossover distortion. Class-AB designed by smarty-pants does not.
    It takes only a little idle current to eliminate crossover distortion but the designers of the LM324 and LM358 were not smart enough so they used almost no idle current.

    The very hot class-A amplifier idles at 350mA even when it is not playing sounds. A class-AB amplifier with the same output power idles at about 20mA.
     
Loading...