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What are the transistors doing in this circuit? Watch

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    I was reading a book about electronics. It says that the two transistors form a complementary self-excited harmonic oscillator, and the oscillating frequency depends on the values of R1 and C1. The bigger the product of R1 and C1 the lower the frequency of oscillation.

    I don't understand why it is necessary to put in those transistors. Also, why is the frequency dependent on the values of resistance/capacitance? Please explain this.
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    Any idea?

    (Original post by Stonebridge)
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    (Original post by yangg)
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    I was reading a book about electronics. It says that the two transistors form a complementary self-excited harmonic oscillator, and the oscillating frequency depends on the values of R1 and C1. The bigger the product of R1 and C1 the lower the frequency of oscillation.

    I don't understand why it is necessary to put in those transistors. Also, why is the frequency dependent on the values of resistance/capacitance? Please explain this.
    I think the operation is as follows:

    1. Assume the switch is closed with the capacitor discharged i.e so that the potential at the base of the PNP transistor is 0 V. Then the b-e junction of the PNP transistor is forward-biased, and (conventional) current flows through this junction.

    2. This current charges the capacitor C_1 via resistor R_1. At some point, the voltage across C_1 reaches 0.65 V. This causes the e-b junction of the NPN transistor to be forward-biased, and it switches on.

    3. The length of time taken for NPN transistor to switch on depends on the time constant R_1C_1 of the resistor-capacitor pair.

    4. When the NPN transistor switches on, then

    a) current flows suddenly through the loudspeaker coil, causing a brief noise
    b) the capacitor discharges through the e-b junction of the NPN transistor

    5. Because the capacitor is discharging, at some point the voltage across the NPN transistor drops below 0.65 V, and it switches off. This happens more quickly than the charging of C_1 since there is no resistance (barring that of the e-b junction) between the capacitor and 0 V.

    6. No current now flows through the loudspeaker coil, and the e-b junction of the PNP transistor is again forward-biased. Note that the final potential on the capacitor after steps 1-5 is not very important (we need'nt assume that it's 0 V), since the battery will ensure that the e-b voltage > 0.65 V regardless.

    Now repeat steps 1-5, and you have an oscillator, with period determined by time constant R_1C_1.
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    (Original post by atsruser)
    I think the operation is as follows:

    1. Assume the switch is closed with the capacitor discharged i.e so that the potential at the base of the PNP transistor is 0 V. Then the b-e junction of the PNP transistor is forward-biased, and (conventional) current flows through this junction.

    2. This current charges the capacitor C_1 via resistor R_1. At some point, the voltage across C_1 reaches 0.65 V. This causes the e-b junction of the NPN transistor to be forward-biased, and it switches on.

    3. The length of time taken for NPN transistor to switch on depends on the time constant R_1C_1 of the resistor-capacitor pair.

    4. When the NPN transistor switches on, then

    a) current flows suddenly through the loudspeaker coil, causing a brief noise
    b) the capacitor discharges through the e-b junction of the NPN transistor

    5. Because the capacitor is discharging, at some point the voltage across the NPN transistor drops below 0.65 V, and it switches off. This happens more quickly than the charging of C_1 since there is no resistance (barring that of the e-b junction) between the capacitor and 0 V.

    6. No current now flows through the loudspeaker coil, and the e-b junction of the PNP transistor is again forward-biased. Note that the final potential on the capacitor after steps 1-5 is not very important (we need'nt assume that it's 0 V), since the battery will ensure that the e-b voltage > 0.65 V regardless.

    Now repeat steps 1-5, and you have an oscillator, with period determined by time constant R_1C_1.
    Why does the transistor switch on at the specific value of 0.65V?
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    (Original post by yangg)
    Why does the transistor switch on at the specific value of 0.65V?
    That is due to the standard operation of a transistor. The b-e junction is essentially a diode. A diode will not conduct until this junction is forward-biased by about 0.65 V (this value varies but 0.65 V is a decent rule of thumb). This is the voltage needed to give charge carriers enough energy to cross the so-called "depletion layer" at the p-n junction.

    The precise value for the 0.65 V is determined by the energy levels of the electrons in the p and n-type materials in the semiconductor material, the temperature of the material, and so on.

    You probably need to read up on transistor operation, I guess.
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    (Original post by atsruser)

    You probably need to read up on transistor operation, I guess.
    I did, actually. It's really confusing! Thanks for the help anyway.
 
 
 
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