The image shows the formula to calculate out power of a class B amplifier. Altho
ID: 3349286 • Letter: T
Question
The image shows the formula to calculate out power of a class B amplifier. Althogh all the parameters in the formula are constant, this gives an ac function. Would you explain the reason?
Signal Swing and Output AC Power Fiure 12.9 shows the voltage and current signal swings from the circuit of Fig. 12.6 From the signal variations shown in ig, 12.9, the values of the peak-to-peak signal swings are 692 POWER AMPLIFIERS ee (V) (A) lb) FIG. 12.9 Graphical operation of transformer-coupled class A amplifier The ac power developed across the transformer primary can then be calculated using (12.13) The ac power calculaled is that developed across the primary of the transformer. Assuming an ideal transformer (a highly emcient transformer has an efficiency of well over 90%), we find thal the power delivered by the secondary lo the load is approximalely that calculaled using Eq. (12.13). The output ac power can also be delermined using the voltage delivered to the load. For the ideal ransformer, the voltage delivered to the load can be calculated using Eq. (12.9): The power across the load can then be expressed as Vrms)Explanation / Answer
A DC signal in the input range of the transistor will be amplified, the transfer function in the operating range (for an amplification application!) is a bijection: for 1 given input voltage, there exists 1 and only 1 output voltage. For any output voltage, there can be 1 and only 1 input voltage leading to the result. You can have a DC input and a DC output.
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