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transistor power in inverter

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malik_zohaib

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hi guys.
i am confuse about the power and number of power transistors use in single phase inverter.
first>> let i am making 150w inverter and using 2n3055 then how many transistors i required???how to calculate the power of single transistor for inverter??? is the formula Pd=Vce*Ic is valid???
second>> if i use mosfet (like stp55nf06l or less power) then how many transistors i required???? and the power is like Pd=RDon*Id?????
 

no i just wanna know how to calculate the required numbers of transistors. i don't have schematic. one thing more as we use two branches of transistors for +ve and -ve cycle; does it means the half of the power is delivered by one branch?? like if we have 1000w inverter then each branch should be capable to handle 500w ???? am i right???
 

You should use MOSFETs even at 150W because of the increase in efficiency and decrease in losses over BJTs.

Let's say you're making a 12V 150W inverter. So current should be 12.5A. But you have to consider the losses in the inverter. Let's assume the efficiency of the converter is 75%. Also, you have to remember that the battery voltage will be lower than 12V for a large amount of time, depending on the load and capacity of the battery.

Let's assume the battery voltage drops all the way to a 10V minimum (although it should not do so in practice). So, max average current would be (150/10)/0.75 = 20. If we're using push-pull configuration, peak current would be about 1.5*20 = 30A.

So, the MOSFET combination in the inverter should be able to take 30A current. Of course other factors such as gate driving, gate discharge, heat sinking, etc will be important.

So, we can use IRFZ44N as it can handle upto 49A. Power loss in the IRFZ44N would be about 20*20*0.0175 = 7W. You might be happy with that and provide good enough cooling. However, keep in mind that 49A rating is for 25'C. At 100'C it is 35A. So you could parallel one more IRFZ44N and pull the overall loss down to 3.5W plus increased current handling capacity of 35*2 = 70A at 100'C. So, you're completely safe. So, you'll need 4 IRFZ44N's, 2 parallel per push-pull "leg". But if you use IRF3205, you can use 1+1 and still have lower losses. So, you'll have lower losses, better efficiency and you'll need 2 less MOSFETs.

Hope this helps.
Tahmid.

---------- Post added at 12:38 ---------- Previous post was at 12:37 ----------

no i just wanna know how to calculate the required numbers of transistors. i don't have schematic. one thing more as we use two branches of transistors for +ve and -ve cycle; does it means the half of the power is delivered by one branch?? like if we have 1000w inverter then each branch should be capable to handle 500w ???? am i right???

No. Each branch should be able to handle 1000W. Since when one branch is on, all the current flows through that leg and that leg needs to withstand all the power.
 

    dselec

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    ismu

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    boOi

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Thamid good explination ,thanks
 

hi thamid

normaly in datasheet 90NF03 MOSFET power dissipation=150W.

that max condition is Vds=30v and Id=80A Rdson=0.007E

so max power is =Rdson*[Id]^2 =45W .

1) then how can it make 150W?
2) when wev are using with 90F03 mosfet is very lees heat , But with 80NF10 getting more heat at same setup, but both have almost same Rdson value. yu know what was the reason?





You should use MOSFETs even at 150W because of the increase in efficiency and decrease in losses over BJTs.

Let's say you're making a 12V 150W inverter. So current should be 12.5A. But you have to consider the losses in the inverter. Let's assume the efficiency of the converter is 75%. Also, you have to remember that the battery voltage will be lower than 12V for a large amount of time, depending on the load and capacity of the battery.

Let's assume the battery voltage drops all the way to a 10V minimum (although it should not do so in practice). So, max average current would be (150/10)/0.75 = 20. If we're using push-pull configuration, peak current would be about 1.5*20 = 30A.

So, the MOSFET combination in the inverter should be able to take 30A current. Of course other factors such as gate driving, gate discharge, heat sinking, etc will be important.

So, we can use IRFZ44N as it can handle upto 49A. Power loss in the IRFZ44N would be about 20*20*0.0175 = 7W. You might be happy with that and provide good enough cooling. However, keep in mind that 49A rating is for 25'C. At 100'C it is 35A. So you could parallel one more IRFZ44N and pull the overall loss down to 3.5W plus increased current handling capacity of 35*2 = 70A at 100'C. So, you're completely safe. So, you'll need 4 IRFZ44N's, 2 parallel per push-pull "leg". But if you use IRF3205, you can use 1+1 and still have lower losses. So, you'll have lower losses, better efficiency and you'll need 2 less MOSFETs.

Hope this helps.
Tahmid.

---------- Post added at 12:38 ---------- Previous post was at 12:37 ----------



No. Each branch should be able to handle 1000W. Since when one branch is on, all the current flows through that leg and that leg needs to withstand all the power.
 

hi thamid

normaly in datasheet 90NF03 MOSFET power dissipation=150W.

that max condition is Vds=30v and Id=80A Rdson=0.007E

so max power is =Rdson*[Id]^2 =45W .

1) then how can it make 150W?
2) when wev are using with 90F03 mosfet is very lees heat , But with 80NF10 getting more heat at same setup, but both have almost same Rdson value. yu know what was the reason?

1) You only considered conduction loss. There are also switching losses and gate losses. Besides that, remember that Rdson is not constant at the value you stated. It will be higher at higher temperatures.

2) Consider the gate charges of the two devices and calculate corresponding losses. Consult the datasheets and find the required values. What circuit are you using these MOSFETs in? An inverter?

Hope this helps.
Tahmid.
 
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    ismu

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yes for inverter , can u give me a simple calculation about all mosfet loses,like what u have done earlier for mosfet load?
 

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