Parallel-resonant class-D RF amplifiers in .NET Printing QR Code 2d barcode in .NET Parallel-resonant class-D RF amplifiers

9.2.1 Parallel-resonant class-D RF amplifiers using barcode drawer for .net vs 2010 control to generate, create qr bidimensional barcode image in .net vs 2010 applications. Internatioanl Orgnization for Standardization The class-D qr bidimensional barcode for .NET amplifier of Figure 9.6 consists of a square-wave current source driving a parallel LCR circuit.

In this circuit, a large inductor (RF choke). Radio-frequency electronics: Circuits and applications Figure 9.6. Class-D parallelresonant RF amplifier operation. + RL Square wave drive Figure 9.7. Practical class-D parallel-resonant RF amplifiers. + Vdc RFC RFC + RFC Vdc V1 V2 0 V1 Q4 RL Q3 V1 Q1 Q2 Q1 (a) C Q2 Cds (b) RL V2 V2 Vg1 Vg2 provides the constant current. A DPDT switch commutates the load, effectively forming a square-wave current source. Two practical versions of this circuit are shown in Figure 9.

7. In (a) the peak sine-wave voltage on the load, RL, will be Vdc/2 and the dc supply current will be 2Vdc/(8RL). For the appealing circuit in (b), these quantities are Vdc and 2Vdc/(2RL).

(See problem 9.4.) But the beauty of the parallel class-D amplifier is that the voltage on the parasitic capacitances of the transistors (drain-tosource capacitance, shown in dotted lines in Figure 9.

7b) is zero at the instants the switches open or close. Thus, there is no lossy abrupt charging or discharging of these parasitic capacitors. (Refer to the waveforms in Figure 9.

7b. Note that the value of C is effectively increased by Cds, the parasitic capacitance of one transistor first one, then the other.).

9.3 The class-E amplifier We have seen that a class-C amplifier can be built with a single transistor and that its efficiency is high, relative to class-B and class-A amplifiers. The class-D switching amplifier, on the other hand can, in principle, achieve 100% efficiency, but requires two transistors. The class-E amplifier [4], has both virtues: single transistor operation and up to 100% efficiency.

Figure 9.8 shows the circuit..

Class-C, D, and E power RF amplifiers Figure 9.8. Class-E amplifiers. Vdc + RF choke High-Q resonant LC C1 Square wave drive Figure 9.9. Class-E amplifier equivalent circuit. Idc Vdc Vdc 0 L1 V1(t). Voutpk 0 Vout (t). IC1(t). The transist visual .net QR-Code or is used as a switch: fully on or fully off. An equivalent circuit is shown in Figure 9.

9, where the transistor is represented as a switch. The switch operates with a 50% duty factor, so the voltage at the switch, V1(t), is forced to be zero for half the cycle (or almost zero if we consider the switch to have a small series resistance, r). During the other half-cycle, V1(t) consists of a rounded positive pulse.

The power supply is connected through L1, an RF choke (a high-value inductor). The choke and the switch form a flyback circuit in which the power supply pumps energy into the inductor while the switch is closed and the inductor pumps energy into the rest of the circuit while the switch is open. Since there can be no dc drop across the choke, you can see that the pulses in V1(t) at the switch must have an average amplitude equal to twice the power supply voltage.

The key to the efficiency of this circuit is that it can be designed so that the voltage V1(t) has fallen to zero at precisely the instant the switch closes, i.e., the capacitor C1 is shorted out without a sudden lossy discharge.

Note that C1 includes the transistor s parasitic capacitance. The RF choke, L1, is large enough to ensure that the dc supply current, Idc, has essentially no ac component, i.e.

, the inductor current s increase and decrease during the flyback cycle are much smaller than the average current. A simplified description of the circuit operation is as follows. The pulses at C1 are not square, but must have an average voltage of 2Vdc.

The waveform is a rough approximation to a sine wave with a peak voltage of Vdc plus an equal dc offset. The bandpass filter formed by L2 and C2 passes a good sine wave to the load, RL. Let us now look in detail at the circuit operation and design.

. Radio-frequency electronics: Circuits and applications 9.3.1 Class-E amplifier design procedure We have seen that we can expect this amplifier to furnish the load with a sine wave whose amplitude is something like Vdc, so the RF power output will be approximately Vdc2/(2RL). If a different power is desired, the resistance of the load, RL, can be transformed to a different value using any of the techniques of 2. Next, we can simply pick a value for L2 that will give a reasonably high Q, maybe 10, so that the waveform at the load will be a good sine wave, i.

e., have minimal harmonic content. We are left with finding values for C1 and C2.

So far, the only constraint on the circuit is the condition that V1(t) have fallen to zero at /2, the instant of switch closure. It turns out that this can be satisfied over a range of combinations of C1 and C2 and that this provides a way to set the output power. Combinations with lower values of C1 reduce the output power.

However, it is possible and beneficial to impose a second constraint which will require a unique combination of C1 and C2. This constraint is that dV1/dt, as well as V1(t), be zero at the instant the switch is closed. If this condition is met, the frequency of the amplifier can be shifted somewhat without seriously violating the condition that V1( /2) be zero.

Moreover, at high frequencies, where the transistor s switching time is relevant, V1(t) will at least remain close to zero during the switching process. Before outlining the analysis, we summarize the resulting design procedures as follows: 1. Pick a value for Q, say Q = 10.

Then L2 = QRL. Pick L1 to be; say; 100RL : 2. Calculate C1 as follows: 1  2  RL 5:45RL : 1 4 C1 2 3.

Calculate C2 as follows:  2 1 1 1 1 8 2  L2  L2 0:212 : 2 C C2 C1 1 1 4 4. The amplitude of the output sine wave (the voltage on RL) is 2 Voutpk Vdc q 1:07Vdc : 2 1 4 5. The output power is given by Pout Voutpk 2 V2 1:154 dc : 2RL 2RL (9:10) (9:9)  (9:7) (9:6).

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