Capacitor coupled investing amplifier basic circuit
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A positive-going ac signal at the base of Q1 produces an increase in IE1 and this results in an IC2 increase. Thus, the voltage drop across RC is increased, producing a decrease in the output voltage. So, there is an ac phase inversion between the input and output. The input impedance at the emitter of Q2 constitutes the load for the collector of Q1. Therefore, the voltage gain from the circuit input to Q1 collector is, With an input stage gain of -1, there is no significant Miller effect at the circuit input.
The voltage gain produced by the Q2 CB circuit is, Converting to CE parameters, the overall voltage gain for the cascode amplifier is the same as that for a CE circuit; Design Calculations: The design procedure for a cascode amplifier is easily derived from the CE design process.
The emitter voltage for Q2 is, The transistor base voltages are, When the voltage and current levels are selected, the resistor values are easily calculated. The base bypass capacitor C2 in Fig. This ensures that the impedance looking into Q2 emitter terminal is not significantly affected by R1 and R2.
Also, that no portion of the input signal at Q2 base is developed at Q1 base. Differential Amplifier as a High-Frequency Amplifier: The differential amplifier discussed already can have the same high-frequency performance as a CB circuit. The differential amplifier circuit shown in Fig. The output impedance of an inverting amplifier is calculated in exactly the same way as for a noninverting amplifier, using Eq. Like the case of the noninverting amplifier, the output impedance is very low for an inverting amplifier.
The input impedance of an inverting amplifier is easily determined by recalling that the right side of R1 in Fig. Therefore, Design of an inverting amplifier is very simple. Voltage divider current I1 is selected very much larger than the op-amp input bias current. Resistors R1 and R2 are calculated from Eqs. Capacitor-Coupled Inverting Amplifier: A capacitor-coupled inverting amplifier circuit is shown in Fig.
In this case, the bias current to the op-amp inverting input terminal flows from the output via resistor R2, so that the input coupling capacitor does not interrupt the input bias current. A voltage drop IBR2 is produced at the inverting input by the bias current flow, and this must be equalized by the IBR3 voltage drop at the noninverting input.
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