600/800W 220~230v AC Inverter SMPS from 12v DC

[sam781]

Magnetic characteristics from EPCOS ETD34 datasheet -

Σl/A = 0.81 mm–1
le = 78.6 mm
Ae = 97.1 mm2
Amin= 91.6 mm2
Ve = 7630 mm3

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While learning the power calculation I want to finalize the design parameter for 12 to 325v transformer. I want to start designing for a 200w transformer. If its efficiency is 80%, it can be easily used for 160/150w load (200w*.8=160w).

Primary Voltage, Vp=12v
Secondary voltage=325v
Considering 80% duty cycle, turns ratio, N = 325/(12*0.8)=33.75=34
So, secondary voltage, Vs=Vp*N=12*34=408v

Primary current, Ip=200/12=16.67A=17A
Secondary current, Is=Ip*N=12/34=0.49=0.5A

So, I need to chose wire as per primary and secondary current rating. Which AWG table should I follow in this case? As, I do not have neither a LCR meter nor datasheet (for Al value) of the core, could you please let me know what should be the safe value of per volt turns here?

 

[rajudp]

i thing you checked the epcos profile selection. you can get datasheet of each part in that they given Aw, for ETD34 it is 1.22
since you are using 400/Sqcm you can select the coil so that 1 Sqmm of coil can handle 4A of current
ie area of the wire will be = Current/4 Sqmm
you have to consider skin effect also for the coil

 

[sam781]

So, primary wire area -> 17/4= 4.25 Sqmm. i.e, 10 AWG (5.26 Sqmm)
Secondary wire area -> 0.5/4=0.125 Sqmm. i.e, 26 AWG (0.129 Sqmm)

It (primary wire) seems a very thick (2.6mm diameter) wire. Is is possible to use in winding? :sad:

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Could you please let me know the safe value of per volt turn needed in this transformer?

 

[rajudp]

what frequency you are operating,

Skin_Depth = 7.6/SQRT(Frequency) [cm]

 

[sam781]

Operating frequency 50kHz

So, skin depth=7.6/sqrt(50e3) cm=0.034cm=0.34mm

 

[rajudp]

yes that means diameter is 0.64mm for 50khz, you can use multiple coils parallel to get the current

 

[sam781]

Required primary wire diameter is =2*SQRT(4.25/pi)=2.33mm
So, Number of 22 AGW (0.644mm dia) wire required =2.33/0.644=3.61=4
Right?

What about the turns?

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As per Goldsmith's note on transformer design, inductance calculation is as below -
Vp=12
Ip=17A
f=50kHz

10% of Ip=1.7A
inductive impedance, xL=12/1.7=7 ohm
Now, wL=7
L=7/w=7/(2*pi*f)=22.28uH
So, primary side inductance is 22.28uH

 

[rajudp]

for primary you need 4.25Sqmm wire 22 Awg is having 0.324 Sqmm so you need 4.25/.324 = 13.1 wires for the current density

 

[sam781]

Oh! sorry, I wrongly calculated wire number using diameter instead of area. :sad:

Now number of turns....

 

[rajudp]

turns primary is 4.2t

 

[sam781]

turns primary is 4.2t

So, primary turns is 5 and secondary turns 34*5=170. But how to calculate turns?

 

[rajudp]

general formula is t/v = 10e8/( 4 x freq x flux x Ae )

 

[rajudp]

sorry , it is
Skin_Depth = 6.6/SQRT(Frequency) [cm]

 

[sam781]

freq=50kHz
Flux=1600
Ae=0.971 Sqcm

t/v=10e8/(4*50000*1600*0.971)=3.21 :sad:

 

[goldsmith]

Hi Sam
In my presented way , the number of turns will be same . you have tried to consume the time ! however that way is correct too . but don't forget that a designer should try to consume the time as low as possible .
Another thing ! a gapped core will change all things ! i hope that you knew it !
Best Wishes
Goldsmith

 

[sam781]

I found mismatch between two value between rajudp's mentioned value and calculated value (using rajudp's formula).

@Goldsmith: Yes I agree. And your process is more accurate. But currently, I do not have LCR meter. Is there any alternative way?
Designer should try to consume time as low as possible but my aim is not only make it but also know detail as much I can. So, at this stage I'm giving more emphasis on learning part rather making it physically. :smile:

 

[rajudp]

freq=50kHz
Flux=1600
Ae=0.971 Sqcm

t/v=10e8/(4*50000*1600*0.971)=3.21 :sad:

some problem with your calculation
t/v=10e8/(4*50000*1600*0.971)=0.3218

sorry it is 1E8 = 100000000

 

[goldsmith]

I do not have LCR meter. Is there any alternative way?

Hi
Of course there is another way instead of LCR meter ! you can calculate the inductance of your winding and then for calculating the number of turns , use these formulas : 1000 or 100 sqrt(AL/L) 100 or 1000 depends on material of your core( ferrite or iron powder ) .
Good luck
Goldsmith

 

[sam781]

Thanks rajudp

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you can calculate the inductance of your winding and then for calculating the number of turns , use these formulas : 1000 or 100 sqrt(AL/L) 100 or 1000 depends on material of your core( ferrite or iron powder ) .
Good luck
Goldsmith

How can I calculate inductance of winding without LCR meter?

 

[goldsmith]

Sam , did you understand what i have tried to say , really ? you know how much inductance you need ( you have calculated it at top (e.g 22uH) . you just need calculate number of turns . one was my presented way with LCR meter . the other way is using those formulas (1000 or 100 sqrt(AL/L) L is your desired inductance in uH and AL will be derived from datasheet of core ! , when you have LCR meter you won't need to read dtasheet ! ) you understand ? ( i'm talking about these simple applications . of course when you are designing an HF system you should pay more attention to the datasheet but for these low frequencies , the presented tables will free you ! )
Best Wishes
Goldsmith