Flyback Transformer design for PV microinverter
5:20 AM
The designing of the flyback transformer requires to list the following quantities ahead of design.
Vdc_mpp | 40 V | |
---|---|---|
Vdc_min | 30 V | somewhere below the Vdc_mpp |
Vdc_max | 50 V | somewhere above the Vdc_mpp |
Vgrid_peak | 311 V | |
Vgrid_rms | 220 V | |
ω | 100 rad/sec | It will be 120 rad/sec for 60 Hz |
Po=Prated | 200 W | |
Pin | Po/η=222 W | |
fgrid | 50 Hz | |
η | 0.9 | |
fs | 200 kHz |
Now we have Vdc_mpp for solar modules, but it varies with the change in environmental conditions. Therefore, an approximate is required for the voltage range for whole day operation is required. For figure below the values taken are from 30-50 V.
Now using the above mentioned data we will find the important parameters of the flyback transformer [1].
- Find the amplitude of the output current
- IA= (2*Po)/Vgrid_peak (1)
- = 2*220/311 =1.286 A
- Find the turn ratio "N" by using the following formula
- N= (η * Vdc_mpp )/Vgrid_rms (2)
- = (0.9* 40)/220 = 0.1636
- This value is approx equal to 6. So transformer turn ratio will be 1:6. This vale can alternatively be calculated based on the current ratio of primary and secondary currents as given in page 3433 of reference given below.
- Now measure the peak primary using following formula
- Ip_max = IA* (Vgrid_peak/Vdc_min + 1/N) (3)
- = 1.286 * (311/30 +1/1/6) = 21.04 A
- Using Ip_max , we can find the value of primary inductance as follows, here the value of δmax taken as 0.5
- Lp= (Vdc_min * δmax ) / Ip_max * fs (4)
- = 30 *0.5/(21.04*200k) = 3.564 µH
- Based on the value of Lp we can find secondary inductance by using equation (5)
- Ls=Lp/(n2) (5)
- = 3.564 /(1/36) = 128.3 µ H
- During the experimentation the values of Lp and Ls needs adjustment that is in between 3-4% of the values calculated above.
- For simulation in PSIM we need to calculate the magnetizing inductance "Lm" of the transformer. It should be kept in mind that the Lm is very important in transformer design. If value of Lm is low then high peak current will flow in primary winding, Which means high current stress on primary side switch. If the value of Lm is very high then system may operate in continuous conduction mode which leads to a distorted output voltage. The expression for Lm is given below [2]
- Lm= Vdc_mpp 2 * δmax 2 / (4 Ppv * fs)
- =2.25 µ H
Remember , the good design of flyback transformer needs lowest leakage inductance
That's all folks...
Reference:
[1] Mingzhi Gao; Min Chen; Chi Zhang; Zhaoming Qian, "Analysis and Implementation of an Improved Flyback Inverter for Photovoltaic AC Module Applications," Power Electronics, IEEE Transactions on , vol.29, no.7, pp.3428,3444, July 2014
[2] Tamyurek, B.; Kirimer, B., "An Interleaved High-Power Flyback Inverter for Photovoltaic Applications," Power Electronics, IEEE Transactions on , vol.PP, no.99, pp.1,1
[1] Mingzhi Gao; Min Chen; Chi Zhang; Zhaoming Qian, "Analysis and Implementation of an Improved Flyback Inverter for Photovoltaic AC Module Applications," Power Electronics, IEEE Transactions on , vol.29, no.7, pp.3428,3444, July 2014
[2] Tamyurek, B.; Kirimer, B., "An Interleaved High-Power Flyback Inverter for Photovoltaic Applications," Power Electronics, IEEE Transactions on , vol.PP, no.99, pp.1,1
1 comments
This comment has been removed by a blog administrator.
ReplyDelete