Hey everyone
I designed a SEPIC converter, and in simulation everything worked perfectly. Once I went to PCB however I experienced massive issues.
First the SEPIC parameters:
Vin= 36V 68V
Vout = 48V
Iout= 10A
Fsw = 20kHz
Like i said in simulation it worked fine, however in practical I experience the following:
The load in simulation is 2.5 ohms, in the first practical test I used a 2.8 power resistor. I also used scaled down input voltages for safety purposes....
(1) Rload = 2.8 ohms
At a 50% duty cycle, I am expecting Vout = Vin  Vdiode  V_otherlosses
My practical results:
Vin = 5V
Total current drawn from source = 3A
Voltage measured at the Load = 4.26V
Duty Cycle = 50%.
Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 4.26V/2.8 ohms = 1.5A, half of what is being drawn from the power supply.
For my 2nd test I used a 4.7 power resistor:
At a 50% duty cycle, I am expecting Vout = Vin  Vdiode  V_otherlosses
My practical results:
Vin = 5V
Total current drawn from source = 1.2A
Voltage measured at the Load = 4.98V
Duty Cycle = 50%.
Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 4.98V/4.7 ohms = 1.05A
For my 3rd Test i used a 22ohm power resistor:
At a 50% duty cycle, I am expecting Vout = Vin  Vdiode  V_otherlosses
My practical results:
Vin = 5V
Total current drawn from source = 270mA
Voltage measured at the Load = 5.72V
Duty Cycle = 50%.
Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 5.72V/22 ohms = 260mA, this seem to be working
For my 4th Test i used a 22ohm power resistor:
At a 55% duty cycle, I am expecting Vout = Vin  Vdiode  V_otherlosses
My practical results:
Vin = 5V
Total current drawn from source = 400mA
Voltage measured at the Load = 6.84V
Duty Cycle = 55%.
Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 6.84V/22 ohms = 310mA
For my 4th Test i used a 22ohm power resistor:
At a 60% duty cycle, I am expecting Vout = Vin  Vdiode  V_otherlosses
My practical results:
Vin = 5V
Total current drawn from source = 600mA
Voltage measured at the Load = 8.51V
Duty Cycle = 60%.
Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 8.51V/22 ohms = 380mA
For my 5th Test i used a 22ohm power resistor:
At a 65% duty cycle, I am expecting Vout = Vin  Vdiode  V_otherlosses
My practical results:
Vin = 5V
Total current drawn from source = 1A
Voltage measured at the Load = 10.4V
Duty Cycle = 65%.
Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 10.4V/22 ohms = 472mA
For my 6th Test i used a 22ohm power resistor:
At a 70% duty cycle, I am expecting Vout = Vin  Vdiode  V_otherlosses
My practical results:
Vin = 5V
Total current drawn from source = 2A
Voltage measured at the Load = 12.75V
Duty Cycle = 70%.
Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 12.75V/22 ohms = 580mA
For the final test I increased the duty cycle to 71% and then all of a sudden the current draw from my source jumped to 3A.
The measured values were:
Vin = 5V
Total current drawn from source = 3A
Voltage measured at the Load = 13.16V
Duty Cycle = 71%.
Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 13.16V/22 ohms = 600mA
I really don't know what is causing so much loss in power..
I designed a SEPIC converter, and in simulation everything worked perfectly. Once I went to PCB however I experienced massive issues.
First the SEPIC parameters:
Vin= 36V 68V
Vout = 48V
Iout= 10A
Fsw = 20kHz
Like i said in simulation it worked fine, however in practical I experience the following:
The load in simulation is 2.5 ohms, in the first practical test I used a 2.8 power resistor. I also used scaled down input voltages for safety purposes....
(1) Rload = 2.8 ohms
At a 50% duty cycle, I am expecting Vout = Vin  Vdiode  V_otherlosses
My practical results:
Vin = 5V
Total current drawn from source = 3A
Voltage measured at the Load = 4.26V
Duty Cycle = 50%.
Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 4.26V/2.8 ohms = 1.5A, half of what is being drawn from the power supply.
For my 2nd test I used a 4.7 power resistor:
At a 50% duty cycle, I am expecting Vout = Vin  Vdiode  V_otherlosses
My practical results:
Vin = 5V
Total current drawn from source = 1.2A
Voltage measured at the Load = 4.98V
Duty Cycle = 50%.
Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 4.98V/4.7 ohms = 1.05A
For my 3rd Test i used a 22ohm power resistor:
At a 50% duty cycle, I am expecting Vout = Vin  Vdiode  V_otherlosses
My practical results:
Vin = 5V
Total current drawn from source = 270mA
Voltage measured at the Load = 5.72V
Duty Cycle = 50%.
Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 5.72V/22 ohms = 260mA, this seem to be working
For my 4th Test i used a 22ohm power resistor:
At a 55% duty cycle, I am expecting Vout = Vin  Vdiode  V_otherlosses
My practical results:
Vin = 5V
Total current drawn from source = 400mA
Voltage measured at the Load = 6.84V
Duty Cycle = 55%.
Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 6.84V/22 ohms = 310mA
For my 4th Test i used a 22ohm power resistor:
At a 60% duty cycle, I am expecting Vout = Vin  Vdiode  V_otherlosses
My practical results:
Vin = 5V
Total current drawn from source = 600mA
Voltage measured at the Load = 8.51V
Duty Cycle = 60%.
Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 8.51V/22 ohms = 380mA
For my 5th Test i used a 22ohm power resistor:
At a 65% duty cycle, I am expecting Vout = Vin  Vdiode  V_otherlosses
My practical results:
Vin = 5V
Total current drawn from source = 1A
Voltage measured at the Load = 10.4V
Duty Cycle = 65%.
Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 10.4V/22 ohms = 472mA
For my 6th Test i used a 22ohm power resistor:
At a 70% duty cycle, I am expecting Vout = Vin  Vdiode  V_otherlosses
My practical results:
Vin = 5V
Total current drawn from source = 2A
Voltage measured at the Load = 12.75V
Duty Cycle = 70%.
Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 12.75V/22 ohms = 580mA
For the final test I increased the duty cycle to 71% and then all of a sudden the current draw from my source jumped to 3A.
The measured values were:
Vin = 5V
Total current drawn from source = 3A
Voltage measured at the Load = 13.16V
Duty Cycle = 71%.
Now, when I apply ohms law to calculate the current going through my load resistor, I = V/R = 13.16V/22 ohms = 600mA
I really don't know what is causing so much loss in power..
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