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    Circuit and Biasing Technique

    There is so much to learn about the single-transistor amplifier, that this brief tutorial hardly scratches the surface. This discussion considers only the common-emitter configuration as applied to low level audio.

    History

    In the early days of solid state amplifiers, thermal stability was the big issue. The first devices available were leaky germanium PNP transistors. The collector to base leakage was often so excessive that it could cause thermal run-away because the leakage increased exponentially with temperature. The classic way of keeping this under control was the base divider-emitter swamping resistor topology. Early text books (including the one I used in 1963) had a detailed section on this and included a mathematical calculation for “stability factor.”

    Unfortunately, now (some 50 years later), we are still suffering from vestiges of this approach as we continue to see the same circuits popping up even though germanium transistors have been obsolete and unavailable for well over 30years, and the silicon bipolar NPN has been long the transistor of choice. Since leakage in silicon devices is so low that it can hardly be measured, we can make a fresh start.

    Self-Biased Circuit Schematic

    Single Transistor Amp Schematic Update

    A stable quiescent operating point (“Q” point) can be established simply by sourcing the base divider from the collector voltage. This dispenses with the emitter swamping resistor. While not perfect, it provides predictable results and simplicity. It is good for low power amplifier transistors that dissipate less than about 100mW. R1, 2 & 3 form the base divider. The juncture of R2 & 3 is bypassed to common via C2 to eliminate negative feedback from the collector—this negative feedback tends to reduce voltage gain. We will be covering negative feedback in the future. C1 is the input coupling capacitor and C3 is the output coupling capacitor—both pass the AC signal while blocking the DC component. To accommodate a wide range of hFE’s, the base divider current is in the range of 5 to 10 * base current.

    Operating point calculations (ohms law)

    1. Set collector voltage: My rule-of-thumb is to set it at about 40% of Vcc. In this case it is 5V.
    2. Calculate collector current: Ic = (Vcc – Vc) /R4 = (12V – 5V) /2.2K = 3.2mA.
    3. Calculate base current: Ib = Ic / hFE = 3.2mA /200 = 16uA (using the common 2N3904)
    4. Establish base divider current: Id = Ib * 5 = 16uA * 5 = 80uA (a factor of 5 is good)
    5. Calculate Ir1: Ir1 = Id – Ib = 80uA – 16uA = 64uA
    6. Calculate R1: R1 = Vbe / Ir1 = 0.65V /64uA = 10K
    7. Calculate R2 + R3: R23 = (Vc – Vbe) /Id = (5 – 0.65V) /80uA = 54K
    8. Calculate R2,3: R2 = R3 = 54K / 2 = 27K (may be unequal, but total must be 54K)

    Data

    Single Transistor Amp Part 1 Data Update

    The transistors used have hFE’s ranging from 58 to 414.
    Observe variations in Vbe: (0.615 to 0.708V) This is a greater range that I expected.
    Observe variations in Vc: (4.32 to 6.67V) This are the quiescent operating points at room temperature.)
    (Will drop significantly at higher temperatures.)
    Observe variations in Av: (170 to 234) Only the device with the lowest hFE had lower voltage gain.
    (44.6 to 47.4db) This was more stable than expected

    Update

    The gain data is significantly updated and as a result my observations also changed. The voltage gain (Av) is remarkably stable in regard to the transistor hFE—only the lowest hFE showed any reduction in Av.

    Previous data erroneously indicated the opposite because the amplifier was dangerously close to oscillation due to the phase shift caused by the low values of C1 & C2, and the low source resistance (10Ω). In the real world this would not have happened because the capacitor values would be selected for the full audio bandwidth (20 to 20,000hZ).

    Attenuator

    One mistake often made by amateurs concerns signal level. Few function generators put out signal levels that are low enough to be used as a low-level signal source. A 100:1 voltage divider is a necessary addition to your signal generator—it gets the signal level low enough to prevent distortion in a high-gain amplifier and provides a low impedance output that makes measurements easier. Furthermore, the AC voltage can be measured at the input to the attenuator thus simplifying instrumentation requirements.

    Graphical Visualization

    Single Transistor Amp Characteristic Curves

    Do not attempt to reconcile this with the circuit in figure 1—the parameters (Vcc, RL, Operating point etc.) were selected for best visual representation.

    The input current signal (right) is tilted slightly to match the tilt of the characteristic curves.
    The output current (left) indicates the input current times the current gain (hFE or ).
    The output voltage (bottom) indicates delta E = delta I * R (where delta = the change in value)
    It can be observed how signal clipping occurs—saturation on the upper left and cut-off on the lower right.

    Circuit replication

    Readers are invited to replicate my simple circuit and findings—hopefully you have as much fun and learn as much or more. Not much equipment is required—check out the following list:

    Protoboard

    • Small DC power supply or battery
    • Function generator
    • Signal Attenuator — make your own
    • DMM — preferably one that can read mV AC, or better yet, one that can display signal level in db or perhaps a classic vintage audio voltmeter (eBay search for HP 403B)
    • Oscilloscope — no need for high performance

    Photos

    For the future

    Part 2 will discuss the elusive input impedance (resistance)
    Part ?? will discuss an incredible, but simple means of greatly enhancing this simple circuit

    Find more projects
    

    11 Responses to "Single Transistor Amplifier Revisited – Part 1"

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    1. Boy does this bring back the memories! Stick a Darlington in there and see then what happens!

    2. Chispakkiu says: on September 17, 2012 at 9:11 am

      How do you set R4 value?

      • This is a legitimate question and still another nuance to the single transistor amplifier. I happen to believe (not actually know) that the maximum voltage gain occurs when the collector current is approximately 3 to 4mA, so I selected a common resistor value that would provide this current. Also, there are other factors that must be considered for given applications such as battery drain (if applicable) and AC load resistance requirement. Note that adding an AC load that is equal to R4 halves the voltage gain.

        I should have stated that a major goal in all this is to maximize both voltage gain and input impedance. Part 1 is the beginning.

        A good experiment would be to plot voltage gain vs Ic. for a number of transistors. This could be accomplished by setting R4 to 470, 1K, 2.2K, 4.7K and 10K and see what happens. I invite you to take this challenge.

        Also, I will be adding a few more high gain transistors to the experiment as well as a darlington as suggested by Mr. Stroebel.

      • Dear sir
        i have a positive energy and i want to learn more in life. i have a lot of interest in electronics but still i am struggling. what should i do..

    3. What is the point of adding R1 when R2 and R3 create a self-biased transistor?

      Just increase the value of R2 and R3 and the circuit will work perfectly and take less current and provide a higher gain.

      • Good question. Eliminating R1 does function as you indicate and a good many circuits here use that technique.

        For me, it is a personal preference, but it does offer some advantages. Adding base divider current makes the circuit less dependent upon hFE that is generally in the range of 2:1, and also makes it less temperature-dependent as hFE is a function of temperature. It avoids what I call “mushy biasing.” In subsequent parts of this series (if I ever get that far), I will attempt to demonstrate that the loading effect of R1 is not a serious concern.

        Thanks for asking.

    4. can u write on pic( lay man tutorial)

    5. See talkingelectronics.com for PIC projects and theory

    6. The loading effect of R1 IS a serious concern.
      You have already said the biasing is taking 5-10 times more current than a self biased stage with just a resistor between base and collector.
      This means the stage will need 5 to 10 times more energy from a previous stage to get the same output.

      • On the surface, it appears that the shunt resistance would be an issue, but the dynamic input resistance of the base to emitter junction is the major player and that is substantially lower than the shunt resistance.

        Also, while some propose making the base divider current 10x Ib, I do not propose exceeding 5x. This has been suggested as being the difference between European and US engineering practices.

        WOW! you certainly have a busy and interesting website.

    7. electron_revolution says: on October 8, 2012 at 5:20 pm

      i have made 2 formulas in this circuit. its a trial and error i want to prove it thru an xperiment but when i saw this circuit ,my first formula is right.. i am confused with the base current.This is my first formula: Vc=(Ir+Ib+Ic)Rc,
      Vc=[(((1+hfe)/hfe))(Ic)+(Vc/R1+R2))]Rc. please checked if im wrong..

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