Class A Push-pull power amplifier
During positive half cycle \(Q_1\) conducts,So \(I_{c1}\) flows
During negative half cycle \(Q_2\) conducts,So \(I_{c2}\) flows
\( R^{,}_{L}=\left( \frac{\frac{N_1}{2}}{N_2}\right)^2R_L\)
\( =\left( \frac{\frac{N_1}{2}}{N_1}\right)^2R_L=\frac{R_L}{4}\)
\( V_{cc} \) is center tapped to \(N_1\) in first half cycle ,only \(Q_1\) conduct.So effective primary winding is \(\frac{N_1}{2}\) same for \(Q_2\)
Output current
\(I_{c1}=B_0+B_1cosωt+B_2cos2ωt+.... \)
\(I_{c2}=B_0+B_1cos(ωt+180°)+B_2cos2(ωt+180°)+.... \)
\(I_{c2}=B_0-B_1cos(ωt)+B_2cos(2ωt)+.... \)
Total current \(I_{c}=k[I_{c1}-I_{c2}] \)
\(=2kB_1cos(ωt)+2B_3cos(3ωt)+.... \)
Thus even harmonics are eliminatedMathJax example
During positive half cycle \(Q_1\) conducts,So \(I_{c1}\) flows
During negative half cycle \(Q_2\) conducts,So \(I_{c2}\) flows
\( R^{,}_{L}=\left( \frac{\frac{N_1}{2}}{N_2}\right)^2R_L\)
\( =\left( \frac{\frac{N_1}{2}}{N_1}\right)^2R_L=\frac{R_L}{4}\)
\( V_{cc} \) is center tapped to \(N_1\) in first half cycle ,only \(Q_1\) conduct.So effective primary winding is \(\frac{N_1}{2}\) same for \(Q_2\)
Output current
\(I_{c1}=B_0+B_1cosωt+B_2cos2ωt+.... \)
\(I_{c2}=B_0+B_1cos(ωt+180°)+B_2cos2(ωt+180°)+.... \)
\(I_{c2}=B_0-B_1cos(ωt)+B_2cos(2ωt)+.... \)
Total current \(I_{c}=k[I_{c1}-I_{c2}] \)
\(=2kB_1cos(ωt)+2B_3cos(3ωt)+.... \)
Thus even harmonics are eliminated
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