Power Supply for mains circuit _linear or switch mode
Posted: Mon Nov 02, 2020 9:00 am
For Nuts and Volts Magazine
Power Supply Question: 11/2/2020
From: Andy_2020
I am building a power supply. The issue is understanding the ripple current for a capacitor. I am building a linear power supply for conversion of 110 AC 12 amps, to 110 DC 12 amps.
Question:
1) How do you select a capacitor when the ripple current appears to keep increasing? When I select a capacitor with lower ESR this causes the ripple current to increase? Again, how do you select the capacitor? Does anyone know where I can find the proper circuit diagram for or where I can purchase a linear power supply for the conversion of 110 AC 12 amps, to 110 DC 12 amps.
Below are my calculations and circuit diagram. I have neglected to include a fuse/breaker and a transformer for this circuit, but the basic explains the issue about selecting a capacitor for a linear power supply.
2) A switched Power supply is known to be very efficient. Does anyone know have a circuit diagram this circuit or where I can buy a switched mode power supply for the conversion of 110 AC 12 amps, to 110 DC 12 amps
From Nuts and Volts Website, we get that the equation for calculating ripple.
We are calculating for three scenarios
C= I load /Frequency *(voltage)
Change in voltage = delta Volts = I load /Frequency * C
I load = 12 amps, Frequency = 120 hertz, Voltage= 120 volts
Solving for two cases
C = 400 UF = .000400 Farad
1 C = 10,000 UF = 0.01 Farad
2 C = 20,000 UF = 0.02 Farad
3 C = 20,000 UF = 0.1 Farad
1) Calculating we get delta volts = 12 amps/(120 hertz) * (.0004 F) = 250 volts
2) Calculating we get delta volts = 12 amps/(120 hertz) * (.01 F) = 10 volts
3) Calculating we get delta volts = 12 amps/(120 hertz) * (.02 F) = 5 volts
4) Calculating we get delta volts = 12 amps/(120 hertz) * (.1 F) = 1 volts …. This a very low ripple
1) Change in voltage = ripple = 250 volts
2) Change in voltage = ripple = 10 volts
3) Change in voltage = ripple = 5 volts
4) Change in voltage = ripple = 1 volts
Using a higher capacitor reduces the ripple volts, reduces the ripple from 250 volts to 1 volt
Calculating ripple Current
I ripple = square root (P/ESR) = square root (1400watts/.062 ohm) = 150 amps and this value is larger than the stated value of 7 amps as stated in the table.
As the capacitor increase from .0004 F to .01 F, the ESR reduces about 61 micro ohms. That is great, but now I am concerned with the rated increase in ripple Current where at .01 uF, the ripple current 7.7 amps, and that appears contrary to reduce the overall ripple.
Power Supply Question: 11/2/2020
From: Andy_2020
I am building a power supply. The issue is understanding the ripple current for a capacitor. I am building a linear power supply for conversion of 110 AC 12 amps, to 110 DC 12 amps.
Question:
1) How do you select a capacitor when the ripple current appears to keep increasing? When I select a capacitor with lower ESR this causes the ripple current to increase? Again, how do you select the capacitor? Does anyone know where I can find the proper circuit diagram for or where I can purchase a linear power supply for the conversion of 110 AC 12 amps, to 110 DC 12 amps.
Below are my calculations and circuit diagram. I have neglected to include a fuse/breaker and a transformer for this circuit, but the basic explains the issue about selecting a capacitor for a linear power supply.
2) A switched Power supply is known to be very efficient. Does anyone know have a circuit diagram this circuit or where I can buy a switched mode power supply for the conversion of 110 AC 12 amps, to 110 DC 12 amps
From Nuts and Volts Website, we get that the equation for calculating ripple.
We are calculating for three scenarios
C= I load /Frequency *(voltage)
Change in voltage = delta Volts = I load /Frequency * C
I load = 12 amps, Frequency = 120 hertz, Voltage= 120 volts
Solving for two cases
C = 400 UF = .000400 Farad
1 C = 10,000 UF = 0.01 Farad
2 C = 20,000 UF = 0.02 Farad
3 C = 20,000 UF = 0.1 Farad
1) Calculating we get delta volts = 12 amps/(120 hertz) * (.0004 F) = 250 volts
2) Calculating we get delta volts = 12 amps/(120 hertz) * (.01 F) = 10 volts
3) Calculating we get delta volts = 12 amps/(120 hertz) * (.02 F) = 5 volts
4) Calculating we get delta volts = 12 amps/(120 hertz) * (.1 F) = 1 volts …. This a very low ripple
1) Change in voltage = ripple = 250 volts
2) Change in voltage = ripple = 10 volts
3) Change in voltage = ripple = 5 volts
4) Change in voltage = ripple = 1 volts
Using a higher capacitor reduces the ripple volts, reduces the ripple from 250 volts to 1 volt
Calculating ripple Current
I ripple = square root (P/ESR) = square root (1400watts/.062 ohm) = 150 amps and this value is larger than the stated value of 7 amps as stated in the table.
As the capacitor increase from .0004 F to .01 F, the ESR reduces about 61 micro ohms. That is great, but now I am concerned with the rated increase in ripple Current where at .01 uF, the ripple current 7.7 amps, and that appears contrary to reduce the overall ripple.