push pull converter 12Vdc to 310 V dc drain voltages problem,im confused

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fets are driven 120 nanosec rising and 110 nanosec falling time. too slow?
No, that's okay. You could go lower with some efforts, but the gains would be minimal and the emissions would be worse.


what do you thing that a 12 volt to 310 volt dc 150 watt converter efficiency. i mean what percent of efficiency we can say that this is good for this push pull converter.
Difficult to be accurate. If you did your homework properly, it could be anywhere between 85 and 95%. It could be much worse if you didn't, but then something would heat up and burn very quickly, which doesn't seem to be the case

that would give a Rds on between 1.25 ohms and 1.6 ohms.[/FONT][/COLOR]

P = I^2 x R

P = 8.5A^2 x 1.25 ohms

P = 72.25 x 1.25 ohms = 90w minimum??? Thats crazy, for a fet that is advertized as 17.5 milli ohms.

Does anyone see what the problem is?
A number of problems there, the most obvious being the Rdson: the values shown are normalized, not absolute, and the average current between FETs is halved, but the rms value is /√2
 
No, that's okay. You could go lower with some efforts, but the gains would be minimal and the emissions would be worse.



Difficult to be accurate. If you did your homework properly, it could be anywhere between 85 and 95%. It could be much worse if you didn't, but then something would heat up and burn very quickly, which doesn't seem to be the case


A number of problems there, the most obvious being the Rdson: the values shown are normalized, not absolute, and the average current between FETs is halved, but the rms value is /√2

looking at the datasheet on figure 4 as powerbob mentioned it say a huge rds on value. what is that grapshs exactly telling to us?

RDS(on) Static Drain-to-Source On-Resistance ––– ––– 17.5 mOhm VGS = 10V, ID = 25A „ Pulse width <400μs; duty cycle < 2%. this is from datasheet.
so my power loss for rds on is like 7 watts ? assuming I=20 A. 0.017*20*20 =P loss or 0.017x10*10=1.7 watts for each fet and 3.4 watts for total ?


So do you thing this 7 watts loss on fets can make the fets hot with a 2x2x1 cm aliminum on fets for cooling? is there any calculation for temperature rise on fets? if syes how?
 
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looking at the datasheet on figure 4 as powerbob mentioned it say a huge rds on value. what is that grapshs exactly telling to us?
Just the correction factor wrt 25°C (or sometimes 300°K)

RDS(on) Static Drain-to-Source On-Resistance ––– ––– 17.5 mOhm VGS = 10V, ID = 25A „ Pulse width <400μs; duty cycle < 2%. this is from datasheet.
so my power loss for rds on is like 7 watts ? assuming I=20 A. 0.017*20*20 =P loss or 0.017x10*10=1.7 watts for each fet and 3.4 watts for total ?
No, you have to think in rms (quadratic) terms.
In this case, it is easy: your 20amp current always passes through the (corrected) 17.5 milliohm resistance of one MOS or the other. Total power is therefore 20²*0.017=6.8W, 3.4W/MOS.
That is already quite large, and it is even larger if you take into account the switching losses and the Rdson correction factor.
You will need sufficient heatsinks (more than 2x2) or use better devices, or parallel 2 or more IRFZ44.
 
Just the correction factor wrt 25°C (or sometimes 300°K)


No, you have to think in rms (quadratic) terms.
In this case, it is easy: your 20amp current always passes through the (corrected) 17.5 milliohm resistance of one MOS or the other. Total power is therefore 20²*0.017=6.8W, 3.4W/MOS.
That is already quite large, and it is even larger if you take into account the switching losses and the Rdson correction factor.
You will need sufficient heatsinks (more than 2x2) or use better devices, or parallel 2 or more IRFZ44.

Ty so there is not much of wrongs things in the curcuit. İt is just the reason of voltage up i mean from 12 to 310 . So making like two batteries in series and from 24 to 310 will end up less loss on fets since less currwnt passing throuh. İ just wanted to make suee about the heatsinjg size being so small also.
 
Just the correction factor wrt 25°C (or sometimes 300°K)


No, you have to think in rms (quadratic) terms.
In this case, it is easy: your 20amp current always passes through the (corrected) 17.5 milliohm resistance of one MOS or the other. Total power is therefore 20²*0.017=6.8W, 3.4W/MOS.
That is already quite large, and it is even larger if you take into account the switching losses and the Rdson correction factor.
You will need sufficient heatsinks (more than 2x2) or use better devices, or parallel 2 or more IRFZ44.

what should be my toroid choke inductor selection?

iron powder? ferrite? how can i determine. lets say i want a 5 mh output inductor for 310 vdc 150 watt converter
 
Both could work, although iron powder might be stretched to its limits at these frequencies. Higher perf metal powder would be preferable. Ferrite with a suitable air gap is also OK.


well the problem is i do not know where to start. im confused. i have found some formules that exlainin how to wound a toroid but that formules does not include any AL values or permaability values. even they do not include any toroid size.

so here i want 6 mH inductor will work as an output filter at 50 kHz. so what is the first steps to choose the toroid?

any formulas and document will help me.
 
ın a push pull confıguration, snubber values( capacitor and resistor values) depends on only leakage inductance of transformer?

I mean very low leakage on transformer means very low capacitor values and very low loss on snubbers?


or what else effects the snubber values ?
 
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