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Help!!!!!!!!!!!! How to find the values of resistance, and ohm's in this circuit board!!!

Discussion in 'Electronic Basics' started by Chris, Aug 6, 2003.

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  1. Chris

    Chris Guest

    I am a beginner in Electronics, My grandfather gave me a few books on
    electronics. I need to know the resistance, ohm's law. I know how
    to add 2 resistors together but. This circuit is a Common-base
    Amplifer. The book is called Collected basic circuits and this book is
    30 years old. Which values would be best to use if +vcc is 12 volts
    and there are 4 resistor values, R1, R2, R3, R4, 3 capacitors values
    C1, C2, C3, 1 NPN transistor value Q1, V out and V in Which formula
    will be best way to solve this problem. Thanks for your help!!!!!!

  2. Lord Garth

    Lord Garth Guest

    Here's a tickler:
  3. Lord Garth

    Lord Garth Guest

    More details here:
  4. Hello,

    Excuse me, but from what you specified no-one could probably assemble
    a working circuit. I don't know what your circuit looks like due to lack of
    schematics, but I'll try to guess a basic common-base amp scheme likely to
    meet your conditions.

    Use a fixed font to view it.

    ----------x--------------x------------------------ + Vcc
    | |
    - R1 - R3
    | | | |
    | | | |
    - -
    | | | | C3
    | x-----| |---------------- Output
    | |C | |
    | B /
    x-----x------¦ VT1
    | | >
    | | |E | | C2
    | | x-----| |---------------- Input
    | | | | |
    - R2 | C1 - R4
    | | ----- | |
    | | ----- | ¦
    - | -
    | | | (<-L)
    ----------x-----x--------x------------------------ Ground

    Now, using the component names from the schematics above (I hope, similar to
    yours), I'll try to explain the scheme.

    First, the capacitor values (all) are not really important. All 3 are used
    as coupling capacitors which means that all they have to do is to pass the
    entire frequency range through them with the least possible impedance. This
    is done best by choosing capacitors of sufficiently high values (too high
    is better than too low). The exact values depend on the frequency range.
    For audible frequencies (e.g. microprone or tape reading head amp) you can
    take 1 microfarad for each of them. Be aware that electrolytic capacitors
    must be connected with respect to polarity (For C1 this means negative
    ground, positive to base, C2 and C3 polarities depend on the rest of the
    circuitry with C3 most likely to be placed positive collector, negative
    output). Please also be aware that the initial charging of the caps at
    power-on will produce the stronger pulse the higher the caps values are, so
    consider using lower values for C2 and C3 if the amp is to handle very low
    voltages as from microphones or tape heads with this amplifier being the
    first stage in a multiple-stage complex amplifier. In this case, 0.1
    microfarad will be pretty much the maximum reasonable capacitance. If you
    plan to amplify radio frequencies (e.g. antenna preamplifier for listening
    to far-away or low signal AM radio stations), the caps' values will require
    a lot of math (which I can't stand nor use, sorry), but an estimate of 1 to
    5 nanofarad for C2 and C3 and 10 nanofarad for C1 will do in most cases for
    long and medium waves. Use less for short waves. Use very much less for FM.

    Now to the resistors. This is a more tricky part and will require you to
    understand what they really do before you can apply any formula
    successfully. First, the function of R3 and R4. These resistors AND the
    transistor are a voltage divider. For this reason the voltage at R3 divided
    by the resistance of R3 equals to the voltage at R4 divided by the
    resistance of R4. Assuming that the capacitors are chosen to provide
    minimal impedance, you can consider the input voltage to equal the voltage
    at R4 and the output voltage to equal the voltage at R3. Having that, we
    can summarize: output equals input times (R3 divided by R4), that is
    voltage gain is (R3 / R4). Please make sure to include a safety factor due
    to losses in the circuitry, e.g. if (R3 / R4) is 5, take 6 or 7 instead as
    there are losses in the caps. For successful and reliable operation of the
    entire circuit it is advisable to hold the (average) voltage between
    emitter and collector of VT1 at a level approximately half of the Vcc
    supply voltage. This is done by appropriately choosing the base voltage
    divider resistors R1 and R2. The emitter voltage is approximately equal to
    the base voltage minus 0.5 to 0.7 volts, and the voltages at R3, R4 and the
    transistor combined equal Vcc. That is if Vcc is 12V, you'll need 6V at VT1
    and 6V at R3 and R4 together. Let's assume, you want a voltage gain of 5 so
    that R3 is 5 times as high as R4, there will be 1 volt at R4 and 5 volts at
    R3. Further assuming that the transistor is to be operated at 5
    milliamperes, R4 equals 1V / (5 * 0.001)A = (1 / 0.005) ohm = 200 ohm. As
    R3 is 5 times R4, it will be R3 = 5 * 200 Ohm = 1k ohm. The base voltage is
    roughly the emitter voltage (here 1 volt) plus some 0.6 volts, that is 1.6
    volts in this case. Now, knowing that R1 and R2 is a voltage divider, you
    can calculate the ratio between R1 and R2 to get the 1.6 volts across R2:
    Vcc / (R1 + R2) equals 1.6 volt / R2 or 1.6V = 12V * R2 / (R1 + R2). Now
    use an estimated value of between 10k and 47k ohm for R1 (20-30 will
    probably be best) and the equation above to calculate R2.

    That's all, I hope for the voltages, just substitute the necessary voltage
    gain and current values in the calculations above to use your own values as
    desired. Please note that there are technical difficulties to achieve
    voltage gains greater than 100 with this method, in this case, you will
    probably have to lower the VT1 collector-emitter voltage to allow higher
    voltages on the resistors. Also note that if the voltage at R4 is low
    (below a quarter of a volt), the estimate of 0.6 volts at B-E will become
    pretty inexact and throw your whole calculation into trash. In this case
    and especially with higher voltage gains, it would be best to calculate
    some estimates, set up the whole thing, tweak it into more or less correct
    operation and determine the correct voltages and/or resistances

    The output voltage's phase is inverted in respect to the voltage at R3, but
    because there is one more inversion in the amplifier, the resulting output
    is not inverted in respect to the input (if phase shift in caps is

    Please don't rely on my estimates for values like 5 mA and 10 - 40 kOhm.
    They are not calculated, just taken out of experience, so that they may not
    apply for certain conditions, so a test is the best thing to verify them.
    The current largely depends on the transistor used and the maximal current
    output your input source can deliver.

    If the whole thing is to become an antenna preamplifier, I would recommend
    to connect an induction coil in series with R4 as indicated by (<-L) in
    the schematics, then connect C2 in parallel to R4 and apply the input
    voltage in any convenient way to the induction coil (e.g. from the antenna
    through a coupling capacitor of some 0.1 - 1 nanofarad or inductively by
    using the coil as part of a transformer). In this case, I would recommend
    to use a voltage gain of at the very least 20 and an emitter follower
    (common-collector amplifier) as a second stage after this circuit, then the
    resulting two-stage amp may well serve as an antenna preamp, especially for
    the AM range. Be aware that HF transistors must be used in and above the SW
    range (especially FM).

    If you have further questions, feel free to post or ask me by e-mail.


    PS. For the experts in this group: I know, the description is inexact,
    but if you find a real lot of errors, please let me know.
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