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How much energy is needed to turn on NPN transistor?

MacIntoshCZ

Member
Hello everyone,
so i am facing new problem. At frequency near 10Mhz my estimated gate driving losses of mosfet are nearly 100W?!
E = 0.5 * frequency * capacitence * voltageSquared. ideally
I would like to calculate and compare how much energy i need if i choose BJT for switching purpose.

thanks for help
 

MacIntoshCZ

Member
The voltage change on a BJT between on and off is quite small so I doubt the power would be that.

What figures did you use?
No figures. Just asking. I dont mean power losses when turning load on off (Uce * Ic). Just (Ube * Ibe ).
Should i consider about parasitive capacitence, voltage drop across gate, base current, duty cycle... What else?
 

MacIntoshCZ

Member
OH sorry. It is for mosfet irfz44n.
I found in data sheet total charge capacitence around 60nC.
As max driving frequency i Have 10Mhz.
Voltage for gate 18V
E= 10Mhz * 0.5 * total gate charge * voltageSquared.
 

alec_t

Well-Known Member
Most Helpful Member
For the IRFZ44N at 10MHz, LTspice calculates the gate drive energy power as about 115W positive peak (turn on), -33W negative peak (turn off), 8.5W average, The values vary somewhat with duty cycle.
 
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MacIntoshCZ

Member
For the IRFZ44N at 10MHz, LTspice calculates the gate drive energy as about 115W positive peak (turn on), -33W negative peak (turn off), 8.5W average, The values vary somewhat with duty cycle.
Can i ask how negative and average energy Is calculated?
Aldo what it stand for please?
 

rjenkinsgb

Well-Known Member
Most Helpful Member

Diver300

Well-Known Member
Most Helpful Member
The power is 10MHz * 0.5 * total-gate-charge * voltage

Gate charge is gate-capacitance * voltage so the equation using capacitance is

10 MHz * 0.5 * gate-capacitance * voltage-squared

The data sheet says that the gate charge is 63 nC and that the gate capacitance is 1470 pF. The equations give:-

10E6 * 0.5 * 63E-9 * 18 = 5.67 W

or

10E6 * 0.5 * 1470E-12 * 18^2 = 2.38 W

Those are average figures and alec_t's calculation of peak power may be more useful.

The average power figures are similar, and far less than 100 W. The difference between the two figures is probably due to the fact that that the gate capacitance varies with voltage, and the Miller capacitor effect. It's all a bit complicated, but you certainly don't need an average power near 100 W
 

Nigel Goodwin

Super Moderator
Most Helpful Member
OH sorry. It is for mosfet irfz44n.
I found in data sheet total charge capacitence around 60nC.
As max driving frequency i Have 10Mhz.
Voltage for gate 18V
E= 10Mhz * 0.5 * total gate charge * voltageSquared.
Perhaps you'd like to clarify exactly what you've got, so far you've called it an NPN transistor, you've specified C, B and E, and then you claim it's an IRFZ44N FET instead. Not surprising you can't work anything out.
 

MacIntoshCZ

Member

MacIntoshCZ

Member
Perhaps you'd like to clarify exactly what you've got, so far you've called it an NPN transistor, you've specified C, B and E, and then you claim it's an IRFZ44N FET instead. Not surprising you can't work anything out.
Sorry there Is probably language barrier.
I only mentioned mosfet becouse i calculated that i Will need around 100w to drive it.
10Mhz * 0.5 * 63nano*18*18.
I would Like to Have equation fór npn transistor.
They just asked for more mosfet info.
 

MacIntoshCZ

Member
I was thinking about terms like difusion and junction capacitence.
Its just in picofarads. So Ube * Ibe * dutyCycle
+ Energy needed fór parasitive caps
 

crutschow

Well-Known Member
Most Helpful Member

jjw

Member
Did you read msg#9.
You have an extra V in your equation.
Power is - f x C x V x V or - f x Q x V
where C is capacitance and Q charge.
 

MacIntoshCZ

Member
Did you read msg#9.
You have an extra V in your equation.
Power is - f x C x V x V or - f x Q x V
where C is capacitance and Q charge.
I read that but only now i realized that i used Q as C.
Jesus... thanks for second warning
 

MacIntoshCZ

Member
I tried to make model od mosfet .
Values Are from data sheet to irf3205.
Vgs 10v . Vds 44v . Qgs 35nC . Qgd 54nC
...
When Vgs Is 0V Cgd Is charged to 44v. When Vgs Is 10V Its discharged to 34V. So Cgd Is 5.4nF i Guess (energy Is supplied from Vds And discharged by gate driver)
...
Vgs 10V. Qgs 35nC.
So Cgs Is 3.5nF. Its charged by driver And discharged by driver.
I need to supply
10Mhz* 0.5 * 3.5*10n*18*18 = 5.67 J
Bingo
IMG_20201121_104855.jpg
 

alec_t

Well-Known Member
Most Helpful Member
Can i ask how negative and average energy Is calculated?
I edited my previous post to replace 'energy' with 'power'. I don't know the details of LTspice's calculation methods. Here's the simulation showing how the FET power dissipation varies with time :-
FET-power.png
 
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