A discussion about maximum power transfer

Transcription

1 基本电路理论课程论文 第一学期 A discussion about maximum power transfer 夏海颖, 包雪娜, 李光北, 张引玉, 杨阳 Outline: In this record, we will discuss the problem of maximum power transfer in resistor circuit and sinusoidal steady-state circuit And for the cases of three-phase circuit, we will compare it with other circuits to show that it is the most efficient circuit for maximum power transfer Finally, we will give some applications in maximum power transformation to verify that this issue is quite useful in real life Key words: maximum power transfer, resistor circuit, sinusoidal steady-state circuit, three-phase circuit, Applications Chapter One: inear Circuit e evenin equivalent is useful to find the maximum power a linear circuit can deliver to a load As shown in the figure (1), the power delivered to the load is p i + 1 Figure (1) Figure () For a given circuit, and are fixed By varying the load resistance,the power delivered to the load varies as sketched in figure()we can see that when is equal to,the maximum power occurs We take differential about p in Eq 1 with respect to and set the result to zero en we can get, and the maximum power transferred is p max 4 But if the domain of is limited and can t reach (ie > or < ), the result will change 1

2 基本电路理论课程论文 第一学期 Case 1: [, ] 1 Where 1 > When, the power reaches to the maximum and p 1 max Case : [, ] 1 where 1 < max + When,the power reaches the maximum p e two cases above can also be solved from the figure ()In a word, the smaller the difference between &,the the power is larger ere is another situation when the circuit is Norton equivalent, then we can use the similar process to solve the problems As shown in figure () When G G,the power of reaches the maximum N G Figure () Example: Find the maximum power transferred to resistor in the circuit of figure (4) Figure (4) We need the evenin equivalent across the resistor o find, consider the circuit below kω v 1 10 kω + v o 40 kω 0 kω v o 1mA

3 基本电路理论课程论文 第一学期 Assume that all resistances are in k ohms and all currents are in ma , and v o (v 1 /0) + (v 1 /0) (v 1 /15) v o v 1 But v o (8/0)v 1, hence, x(8v 1 /0) v 1, which leads to v v 1 /1-166 k ohms o find, consider the circuit below 10 kω v o kω v v o 40 kω 0 kω v o + (100 v o )/10 (v o /40) + (v o v 1 )/ (1) [(v o v 1 )/] + v o (v 1 /0) () Solving (1) and (), v 1-46 volts p /(4 ) (46)/[4(-166)] watts Chapter wo: Sinusoidal steady state Draw the evenin equivalent circuit like Figure 1, then P 1 I t DC Figure 1 l So I + ( + ) + j( + ) P ( + ) + ( + ) 1 ( + ) + ( + ) 1 e common situation ( no restriction ) 1 et then P ( + ) From what has been discussed in chapter one,we can simply let sure that P reach its maximum So * o make

4 基本电路理论课程论文 第一学期 And the maximum average power is P MA where is the peak amplitude of the evenin 8 equivalent circuit voltage source When the load is a resistor At this situation, 0, then P P 1 ( + ) + ( ) ( ) ( + ) ( ) When and Adjust as near to and P dp d are restricted to a limited range of values 1 ( + ) + ( + ) ( + ) 0 ( ) ( ) ( ) ( ) ( ) + ( + ) So in order to have the max power,we should adjust as near to, and as near to + ( + ) 4 When φ can be changed but its phase angle cannot, which means 4

5 基本电路理论课程论文 第一学期 φ is a constant, then I ( + cosφ ) + j( + sinφ ) ( ) P I e So ( + cosφ) + ( + sinφ) dp S S cosφ 0 d + + at is Example 1: As in the figure below, rms 1 0,Under the following 4 conditions, ask what value of can make the maximum power of, and P max? a) no restriction; b) is a resistor; c) (1, ), (-15, -1); d) the phase angle of is fixed to 45 but the magnitude can be changed k Sol: 1<0 ma k J4k a) jkω P oc max Figure 1 0 j4 + + j ( + ) j4 + jk + + j4 45 oc W 4 4 5

6 基本电路理论课程论文 第一学期 b) 0, Ω c) d) + jω 15Ω + ( + ) Ω tanφ tan jω Chapter ree: ree-phase Circuits: en we discuss the power in the special balanced three-phase Circuits Since all the balanced circuits can be converted to wye-connections, here we only discuss the Y-connected load For a Y-connected load: e phase voltages are: AN p cos( wt) cos( π BN p wt ) cos( + π CN p wt ) ia cos( wt θ ) 6

7 基本电路理论课程论文 第一学期 π i cos( θ b Ip wt ) π i cos( θ + c Ip wt ) e instantaneous power: PPa+Pb+Pc ANia BNib CNic I p pcosθ en we can see that the instantaneous power remains constantby this, we can also get other powers(total): e total everage power: P p pcos e total reactive power: Q I θ I cosθ Q p pipsinθ I cosθ e total complex power: SP+jQ I θ After so much has been done, we can discuss the advantage of the three-phase system by comparing it with the single-phase system First, for the two-wire single-phase system, suppose the resister of each line is, then we have: I P/ I So the power loss: P I P loss / en for the three-wire three-phase system, we also suppose the resister of each line is : I Ia Ib Ic P/ So the power loss: loss ( ) / / P I P P For pl / π r,for the same material and length of line, we have: Ploss / Ploss / r / r, If the two types of system have the same power loss: Material for single-phase/material for three-phase r / r 1 From the above we can see that if the same power loss is tolerated in both system, then using three-phase system can reduce the material that we actually need And if we use the material with the same, the three-phase system can reduce half the power loss! All in all, the three-phase circuit can save not only the power but also the material, that s why three-phase circuit is so popular in many fields today Applications: 1 Microphone e impedance of each microphone will be listed as a specification According to the 7

8 基本电路理论课程论文 第一学期 Maximum Power ransfer eorem, when the load impedance matches the microphone impedance, the power transferred will reach the maximum However, in most cases, the two impedances don t match each other Despite of this, the microphone still can be used If the micro s impedance is lower than 600 Ω, it is considered as low impedance; when its value is in a range from 600 to 10,000 Ω, it is considered as middle impedance; if the value is greater than 10,000, it will be regarded as high impedance A model of pocket headphone e A ere are several situations, where the pocket amp could benefit from a higher voltage power supply - when driving high impedance headphones, when the amplifier is being fed from a high gain equalizer or when the listener just wants more volume With very high impedance headphones (600 ohms or more), the amp may not be able to develop sufficient voltage across the load for maximum power transfer If the amp is fed from an equalizer or tone control with a high boost, the output of the pocket amp could be driven into clipping (efer to the net Power consumed in Bluetooth e bluetooth has low power consumption longevity for battery-powered devices During data transfer the maximum current drain is 0mA However during pauses or at lower data rates will be lower eferences [1] Fundamentals of Electric Circuits, Charles K Alexander, Matthew N O Sadiku, sing Hua University Publishing [] [] 8