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Step 2Inductor characteristics
Though very nice, the Microchip app note seems a little backwards to me. It begins by determining the required power, then chooses an inductor charge time without concern for available inductors. I found it more useful to choose an inductor and design the application around that.
The inductors I used are "C&D Technologies Inductors RADIAL LEAD 100uH" (Mouser part 580-18R104C, 1.2 amp, $1.40), (Mouser part 580-22R104C, 0.67 amp, $0.59). I chose these inductors because they are very small, very cheap, yet have decent power ratings.
We already know the max continuous rating of our coil (0.67 amps for the 22R104C), but we need to know how long it will take to charge (rise time). Rather than use a fixed charge time (see equation 6 in TB053) to determine the required coil amps, we can interrogate equation 6 and solve for rise time: (note: equation 6 in TB053 is wrong, it should be L, not 2L)
(Volts in/Inductor uH)*rise_time=Peak Amps
-becomes-
(Inductor uH/Volts in) * Peak Amps = rise time.
-using the 22R104C with a 5 volt supply gives the following-
(100/5)*0.67=13.5uS
It will take 13.5 uS to fully charge the inductor coil at 5 volts. Obviously, this value will vary with different supply voltages.
As noted in TB053:
"The current in an inductor cannot change instantaneously. When Q1 is switched off, the current in L1 continues to flow through D1 to the storage capacitor, C1, and the load, RL. Thus, the current in the inductor decreases linearly in time from the peak current."
We can determine the amount of time it takes the current to flow out of inductor using TB05 equation 7. In practice this time is very short. This equation is implemented in the included spreadsheet, but will not be discussed here.
How much power can we get out of a 0.67 amp inductor? Total power is determined by the following equation (tb053 equation 5):
Power=(((rise time)*(Volts in)2)/(2*Inductor uH))
-using our previous values we find-
1.68 Watts=(13.5uS*5volts2)/(2*100uH)
-convert watts to mA-
mA=((Power Watts)/(output volts))*1000
-using an output voltage of 180 we find-
9.31mA = (1.68Watts/180volts)*1000
We can get a maximum of 9.31 mA from this coil with a 5 volt supply, ignoring all inefficiencies and switching losses. Greater output power can be achieved by increasing the supply voltage.
All of these calculations are implemented in "Table 1: Coil Calculations for High Voltage Power Supply" of the spreadsheet included with this instructable. Several example coils are entered.
The inductors I used are "C&D Technologies Inductors RADIAL LEAD 100uH" (Mouser part 580-18R104C, 1.2 amp, $1.40), (Mouser part 580-22R104C, 0.67 amp, $0.59). I chose these inductors because they are very small, very cheap, yet have decent power ratings.
We already know the max continuous rating of our coil (0.67 amps for the 22R104C), but we need to know how long it will take to charge (rise time). Rather than use a fixed charge time (see equation 6 in TB053) to determine the required coil amps, we can interrogate equation 6 and solve for rise time: (note: equation 6 in TB053 is wrong, it should be L, not 2L)
(Volts in/Inductor uH)*rise_time=Peak Amps
-becomes-
(Inductor uH/Volts in) * Peak Amps = rise time.
-using the 22R104C with a 5 volt supply gives the following-
(100/5)*0.67=13.5uS
It will take 13.5 uS to fully charge the inductor coil at 5 volts. Obviously, this value will vary with different supply voltages.
As noted in TB053:
"The current in an inductor cannot change instantaneously. When Q1 is switched off, the current in L1 continues to flow through D1 to the storage capacitor, C1, and the load, RL. Thus, the current in the inductor decreases linearly in time from the peak current."
We can determine the amount of time it takes the current to flow out of inductor using TB05 equation 7. In practice this time is very short. This equation is implemented in the included spreadsheet, but will not be discussed here.
How much power can we get out of a 0.67 amp inductor? Total power is determined by the following equation (tb053 equation 5):
Power=(((rise time)*(Volts in)2)/(2*Inductor uH))
-using our previous values we find-
1.68 Watts=(13.5uS*5volts2)/(2*100uH)
-convert watts to mA-
mA=((Power Watts)/(output volts))*1000
-using an output voltage of 180 we find-
9.31mA = (1.68Watts/180volts)*1000
We can get a maximum of 9.31 mA from this coil with a 5 volt supply, ignoring all inefficiencies and switching losses. Greater output power can be achieved by increasing the supply voltage.
All of these calculations are implemented in "Table 1: Coil Calculations for High Voltage Power Supply" of the spreadsheet included with this instructable. Several example coils are entered.
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