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Driver for Umax astra 4100 scanner?

Updated: 12/2/2022
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Q: Driver for Umax astra 4100 scanner?
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How much power is dissipated in the driver which drives a capacitor when its driven with a less than 50 percent duty cycle?

The simplified equivalent circuit of the setup would be a resistor R and a capacitor C in series. Where R is the output resistance of the driver circuit and C is the capacitor driven. We can assume the power is dissipated in the equivalent output resistance R of the driver circuit.___o---|___|---+-------oR C+--------->|Ur(t)The average power Pavg can be calculated from the effective value of the voltage on the resistor Ur,eff by the known simple formula for power:Pavg = Ur,eff * Ir,eff = Ur,eff2 / R (1)The effective value of the voltage on the resistor can be calculated by following formula.________________________/ TUr,eff = \/ (1/T) * integral ( ur(t)2 * dt ) (2)0In case of the above equivalent circuit and when assuming a voltage symmetrical square wave form drive signal ur(t) can be defined as_ __t__ T1 _ __t__ Tur(t) = ( 2 Umax * e R*C ) + ( -2 Umax * e R*C ) (3)0 T1where Umax is the amplitude of the square wave, thus 2 * Umax is the peak to peak amplitude of the symmetrical square wave, and T1 is time of the polarity switching in the square wave within the period.In praxis the positive and negative parts of the driving wave are provided through different branches of the driver circuit. Thus each half wave is dissipated on a different resistor. In this case the equation (3) can be split up into two parts_ __t__ T1 Tur1(t) = ( 2 Umax * e R1*C ) + ( 0 ) (3a)0 T1T1 _ __t___ Tur2(t) = ( 0 ) + ( -2 Umax * e R2*C ) (3b)0 T1For the positive side the integral of Ur1,eff in equation (2) solves to_____________________________/ _ __2_T1__Ur1,eff = 2 Umax \/ (1/T) * (R1*C/2) * ( 1 - e R1*C ) (4a)and from (1) follows the average dissipated power on R1 (positive side of the wave)2 _ __2_T1__Pr1,avg = 2 Umax (C/T) * ( 1 - e R1*C ) (5a)Analogous to (5a) the dissipated power on R2 (negative side of the wave) is2 _ __2_(T-T1)__Pr2,avg = 2 Umax (C/T) * ( 1 - e R1*C ) (5b)The total power dissipation is the sum of the two half wavesPagv,total = Pr1,avg + Pr2,avg (6)This is also valid if R1 and R2 are the same resistor R as assumed at the beginning.


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