Upcoming SlideShare
×

# Olano

427
-1

Published on

0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total Views
427
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
1
0
Likes
0
Embeds 0
No embeds

No notes for slide

### Olano

1. 1. NATIONAL COLLEGE OF SCIENCE AND TECHNOLOGY Amafel Bldg. Aguinaldo Highway Dasmariñas City, Cavite ASSIGNMENT 1 CLASSES OF AMPLIFIEROlaño, Reymart September 01, 2011Electronics 3/BSECE 41A1 Score: Engr. Grace Ramones Instructor
2. 2. A class A power amplifier is defined as a power amplifier in which output current flows for the full-cycle(360°) of the input signal. In other words, the transistor remains forward biased throughout the inputcycle.A schematic circuit of a series fed class A large signal amplifier using resistive load Rc is shown below.The term “series fed” is derived from the fact that the load Rc is connected in series with the transistoroutput. The only difference between this circuit and the small-signal amplifier circuits consideredpreviously is that the signals handled by the large-signal circuit are in the range of volts and thetransistor used is a power transistor capable of operating in the range of a few watts. This circuit isseldom used for power amplification because of its poor collector efficiency but will give clearunderstanding of class A operation to the readers. The output characteristics with operating point Q arealso shown. ICQ and VCEQ represent no signal collector current and collector-emitter voltagerespectively. When ac input signal is applied, the operating point Q shifts up and down causing outputcurrent and voltage to vary about it. The output current increases to Ic max and falls to Ic min. Similarly,the collector-emitter voltage increases to Vce max and falls to Vce min.Power Distribution in Class A power amplifiers.Input power from the collector supply VCC, Pin(dc) = VCC ICQ The power drawn from the collectorsupply is used in the following two components• Power dissipated in collector load as heat, PRC (dc) = (ICQ)2 RC
3. 3. • Power supplied to the transistor, Ptr (dc) = Pin(dc) – PRC (dc. Power supplied to the transistor, Ptr(dc) is further subdivided into ac power developed across the load resistor constituting ac power outputand is given as P out (ac) = (IC)2 RC = (Vce)2/ Rc where Ic and Vce are the rms values of collector currentand Collector- emitter votage = {(Ic max)2÷ √2}RC = V2 CE max / 2 Rc(I2 c (peak-to-peak) Rc)/ 8 = V2CC (peak-to-peak) / 8 RCPower dissipated, in the form of heat, by the transistor itself. The cause of power dissipation intransistor is explained below :Consider an N-P-N transistor. The potential difference across the depletion layer formed near thecollector junction is called the barrier potential. This potential gives the P-region (base) slightly moreenergy than N-region (collector). Thus when electronics emitted from emitter cross the base junctionand enter the collector region, they give up energy in the form of heat and it is this energy that thetransistor has to dissipate to the surrounding. With zero signal applied at the input of the class A poweramplifier, ac power developed across the load reduces to zero and therefore all the power fed to thetransistor is wasted in the form of heat. Thus, a transistor dissipates maximum power under zero-signalcondition. Thus the device is cooler when delivering power to a load than with zero-signal condition.Since in class A operation, maximum power dissipation in the transistor occurs under zero-signalcondition, the power dissipation capacity of a power transistor, for class A operation, must be at leastequal to the zero-signal rating.Collector Efficiency: The collector efficiency of a transistor is given asEfficiency = Average ac power output, Pout (ac) / Average dc power input to the transistor Ptr (dc),Power Efficiency: A measure of the ability of an active device to convert the dc power of supply into theac (signal) power delivered to the load is called the power or conversion or theoretical efficiency. Bydefinition the efficiency isEfficiency = AC power delivered to the load, Pout (ac) / Total power drawn from dc supply Pin (dc),Now ac power delivered to the load,Pout (ac) = (Ic (peak-to-peak) *Vce (peak-to-peak)) ÷ 8Maximum Power and Efficiency. If the operating point Q is set at the midpoint of themaximum signal swing, the resulting maximum power condition may be achieved.
4. 4. Maximum VCE(peak-to-peak) = VccMaximum ICE(peak-to-peak) = Vcc ÷ RCwe have maximum ac power developed across the load resistor,.Pout (ac) max = 1/8 * Vcc/Rc * Vcc = V2cc/8 RcFor the quiescent point Q, ICQ = (Vcc/Rc) ÷2and dc power drawn from dc supply, Pin (dc) max = Vcc ICQ = V2cc/2RcSo maximum efficiency of an amplifier (class A power) is given asEfficiency = Pout (ac) max / Pin (dc) max = (V2cc/8 Rc) ÷ (V2cc/2Rc)This is the maximum percent efficiency for a series-fed class A power amplifier. Since this maximumefficiency will occur only under ideal conditions and for the maximum ac signal swings, most series-fedclass A power amplifiers have power efficiencies much less than 25%.
5. 5. CLASS-B-POWER-AMPLIFIER-OPERATIONIn class B operation the transistor is so biased that zero-signal collector current is zero. Hence class Boperation does not need any biasing system. The operating point is set at cut-off. It remains forwardbiased for only half cycle of the input signal.i.e its conduction angle is 180 degree.As illustrated in figure, during the positive half cycle of the input ac signal, the circuit is forward biasedand, therefore, collector current flows. On the other hand, during negative half cycle of the input ac“signal, the circuit is reverse biased and no collector current flows.Power and Efficiency Calculations of class B operation. Input dc power, Pin (dc) = VCC ICCwhere ldcis the average or direct current taken from the collector supply.If Ic max is the maximum or peak value of collector or output current, thenldc = lc max / ∏RMS value of collector current, Ic rms = Ic max/√2RMS value of output voltage, Vrms = VCC/√2Hence output power during half cycle, Pout (ac) = (Ic max VCC)/ 4In above expression factor 1/2 is used because power is developed during one half cycle only.DC power loss in load, PRc(dc) = (I2 dc Rc = Ic max/∏) Rc . ..DC power loss in collector region or transistor, = Pin dc - PRc(dc) – Pout (ac)Overall efficiency, noverall = Pout (ac)/ Pin dc = 0.785 or 78.5%
6. 6. CLASS-AB-POWER-AMPLIFIER-OPERATIONIn class AB power amplifiers, the biasing circuit is so adjusted that the operating point Q lies near thecut-off voltage. During a small portion of negative half cycle and for complete positive half cycle of thesignal, the input circuit remains forward biased-and hence collector current flows. But during a smallportion (less than half cycle) of the negative cycle”‘ the input circuit is reverse biased and, therefore, nocollector current flows during this period. Class AB operation needs a push-pull connection to achieve afull output cycle.In AB diagram, a small amount of bias current is flowing through the valve. For the output valves in atypical class AB guitar amplifier, this would amount to around 30-40mA, with peaks of approximately250-300mA.In the push-pull output stage, there is a little overlap as each valve assists its neighbour during a shorttransition, or crossover period.Many larger guitar amplifiers are class AB, and well find out why a little later on.
7. 7. PUSH-PULL AMPLIFIERSOne use of phase splitters is to provide input signals to a single-stage amplifier that uses twotransistors.These transistors are configured in such a way that the two outputs, 180º out of phase witheach other,combine. This allows more gain than one transistor could supply by itself. This "push-pull"amplifier isused where high power output and good fidelity are needed: receiver output stages, publicaddressamplifiers, and AM modulators, for example.The circuit shown in figure 1-29 is a class A transistor push-pull amplifier, but class AB or class Boperations can be used. Class operations were discussed in anearlier topic. The phase splitter for thisamplifier is the transformer T1, although one of the phasesplitters shown earlier in this topic could beused. R1 provides the proper bias for Q1 and Q2. The tappedsecondary of T1 develops the two inputsignals for the bases of Q1 and Q2. Half of the original inputsignal will be amplified by Q-1, the otherhalf by Q-2. T2 combines (couples) the amplified output signalto the speaker and provides impedancematching.Figure 1-29.—Class A transistor push-pull amplifier.