CHAPTER 5: MPPT ALGORITHMS TEST AND DATA ANALYSIS 58-80
5.4 Verification of the Conditions of the Algorithms 67
1.
dP=P(12)-P(11)= 153.5308-79.3584=74.1724>0 Now, dV=V(12)-V(11)=331-300=31>0
So, D=D-dD=0.2048-.0038=.2010
Here,The difference of power is greater than zero.The difference of voltage is also greater than zero.So, according to the algorithm,the duty cycle measured one step before will get decreased by the fixed duty cycle change 0.0038 and the new duty cycle will be that of the current step.The previously measured duty cycle was 0.2048 and if it is decreased by 0.0038,then the new duty cycle will be 0.2010 .
2.
dP=P(11)-P(10)= 79.3584-37.8=41.5584>0
dV=V(11)-V(10)=300-300=0,which is surely not less than zero So, D=D-dD=0.2010-.0038=.1972
Here, The difference of power is greater than zero. The difference of voltage equals to zero, i.e., it is surely not less than zero. So, according to the algorithm, the duty cycle measured one step before will get decreased by the fixed duty cycle change 0.0038 and the new duty cycle will be that of the current step. The previously measured duty cycle was 0.2010 and if it is decreased by 0.0038, then the new duty cycle will be 0.1972.
3.
dP=P(32)-P(31)= 589.86-612.598=-22.738<0 Now, dV=V(32)-V(31)=348-343=5>0
So, D=D+dD=0.2048+.0038=.2086
Here, The difference of power is less than zero. The difference of voltage is greater than zero.
So, according to the algorithm, the duty cycle measured one step before will get increased by
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the fixed duty cycle change 0.0038 and the new duty cycle will be that of the current step.
The previously measured duty cycle was 0.2048 and if it is decreased by 0.0038, then the new duty cycle will be 0.2086.
4.
dP=P(49)-P(48)= 245.874-179.036=66.838>0 Now, dV=V(49)-V(48)=258-313=-55<0 So, D=D+dD=0.2162+.0038=.2200
Here, The difference of power is greater than zero. The difference of voltage is less than zero.
So, according to the algorithm, the duty cycle measured one step before will get increased by the fixed duty cycle change 0.0038 and the new duty cycle will be that of the current step.
The previously measured duty cycle was 0.2162 and if it is increased by 0.0038, then the new duty cycle will be 0.2200.
5.
dP=P(25)-P(24)= 1397.9-697.533=700.367>0 Now, dV=V(25)-V(24)=350-321=29>0 So, D=D-dD=0.2162-.0038=.2124
Here, The difference of power is greater than zero. The difference of voltage is also greater than zero. So, according to the algorithm, the duty cycle measured one step before will get decreased by the fixed duty cycle change 0.0038 and the new duty cycle will be that of the current step. The previously measured duty cycle was 0.2162 and if it is decreased by 0.0038, then the new duty cycle will be 0.2124.
5.4.2 Incremental Conductance Algorithm
1.
dV=V(12)-V(11)=331-300=31,not equal to zero dI=I(12)-I(11)= .463-.264=.199
dI/dV=.199/31=.0064
-I/V=-.463/331=-330.537,not equal to dI/dV
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Now, dI/dV>-I/V
So, D=D+dD=0.2352+.0038=.2390
Here, The difference of voltage is not equal to zero. The quotient of difference of current and difference of voltage is .199 and it is not equal to but greater than the negative value of the quotient of current and voltage. So, according to the algorithm, the duty cycle measured one step before will get increased by the fixed duty cycle change 0.0038 and the new duty cycle will be that of the current step. The previously measured duty cycle was 0.2352 and if it is increased by 0.0038, then the new duty cycle will be 0.2390.
2.
dV=V(25)-V(24)=350-321=29,not equal to zero dI=I(25)-I(24)= 3.994-2.173=1.821
dI/dV=1.821/29=.0628
-I/V=-3.994/350=-.0114,not equal to dI/dV Now, dI/dV>-I/V
So, D=D+dD=0.2238+.0038=.2276
Here, The difference of voltage is not equal to zero. The quotient of difference of current and difference of voltage is .0628 and it is not equal to but greater than the negative value of the quotient of current and voltage. So, according to the algorithm, the duty cycle measured one step before will get increased by the fixed duty cycle change 0.0038 and the new duty cycle will be that of the current step. The previously measured duty cycle was 0.2238 and if it is increased by 0.0038, then the new duty cycle will be 0.2276.
3.
dV=V(49)-V(48)=258-313=-55,not equal to zero dI=I(49)-I(48)= .953-.572=.381
dI/dV=.381/-55=-.0007
-I/V=-.953/258=-.0037,not equal to dI/dV Now, dI/dV<-I/V
So, D=D-dD=0.2238-.0038=.2200
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Here, The difference of voltage is not equal to zero. The quotient of difference of current and difference of voltage is -.0007 and it is not equal to but less than the negative value of the quotient of current and voltage. So, according to the algorithm, the duty cycle measured one step before will get decreased by the fixed duty cycle change 0.0038 and the new duty cycle will be that of the current step. The previously measured duty cycle was 0.2238 and if it is decreased by 0.0038, then the new duty cycle will be 0.2200.
4.
dV=V(5)-V(4)= 242-259=-17,not equal to zero dI=I(5)-I(4)=0-0=0
dI/dV=0/-17=0 -I/V=-0/242=0 Now, dI/dV=-I/V So, D=D=.2466
Here, the difference of voltage is not equal to zero. The quotient of difference of current and difference of voltage is zero and it is equal to the negative value of the quotient of current and voltage. So, according to the algorithm, the duty cycle measured one step before will not change and will be that of the current step. The previously measured duty cycle was .2466 and so the new duty cycle will be 0.2466.
5.
dV=V(11)-V(10)= 300-300=0
dI=I(11)-I(10)=0.264-.126=0.138,greater than zero So, D=D+dD=0.2390+.0038=0.2428
Here, the difference of voltage is equal to zero. But, the difference of current is greater than zero. So, according to the algorithm, the duty cycle measured one step before will get increased by the fixed duty cycle change 0.0038 and the new duty cycle will be that of the current step. The previously measured duty cycle was 0.2390 and if it is increased by 0.0038, then the new duty cycle will be 0.2428.
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Fig 5.3: Duty Cycle vs Power Curves of P&O & InC Algorithms.