Maker Pro
Maker Pro

Mosfets

Lord_grezington

May 3, 2013
124
Joined
May 3, 2013
Messages
124
I have one final question (hopefully)

Gate charge = 10nC
Voltage @ gate charge is 0 up to 4.5V

So C = Q/V = 10/4.5 = 2.22nF gate capacitance

So, If i were to run on 10V, Q = 2.22nF x 10V = 22.2nC (gate charge increases)

Ok, now we limit the current to say 0.5A (Mosfet driver limit)

t = Q/I = 22.2/0.5 = 44.4ns transition time @ 10V
t = 10/0.5 = 20ns transition time @ 5V

So increasing the gate voltage increases switching losses???

There needs to be some sort of other bias here as well because I also noticed that the data sheet states the switching time using Vgs = 10V and state the Qg @ 4.5V. This means that as Infineon want to make their stuff look good they would give the faster switching times @ 10V, so this contradicts the calculations.

Please tell me I have done something wrong here?

Thanks
 

BobK

Jan 5, 2010
7,682
Joined
Jan 5, 2010
Messages
7,682
If the MOSFET is fully on (or nearly so) at 5V and you charge the gate at 10V it may take it twice as long to fully charge the gate, but the MOSFET will still be turned on in the same time as it would at 5V, since the gate will have reached 5V in that shorter time.

Bob
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
So there is no benefit in switching @ 10V or higher?

Yes, there is. The Rds(on) will decrease with increasing gate voltage.

Also, if there are any surges in current, a higher gate voltage will help prevent the mosfet starting to act as a constant current source, resulting it far higher excursions of power dissipation.

The only drawback is that the turn off time will be delayed slightly (not slowed) as the gate has further to discharge before the drain current is limited, and finally shut off.

Check the curves for drain current vs gate voltage. You need to ensure that the gate gets above the voltage required to sustain the required drain current. This device is not so generously rated in terms of power dissipation, so taking the gate voltage higher has a benefit of keeping the power dissipation lower in the device.

Note that if you take the gate to 10V, it will reach 5V much faster than it will if you drive it only from 5V (all other things being equal.)
 

Lord_grezington

May 3, 2013
124
Joined
May 3, 2013
Messages
124
So @ 5V, you have higher Rds On which increases standard on power, but you get a lower charge which decreases switching losses.

I have allowed a gate current of 0.5A, and I have done the calculations again. 100Khz, 10A, 48V.

@ 5V I get a total dissipation of 3.27W (gate charge of 8nC and 25mOhm Rdson from curve. This also gave me a rise/fall time of 16ns each)

@ 10V I get a total dissipation of 3.34W (gate charge of 16nC and 18mOhm Rdson from curve . This also gave me a rise/fall time of 32ns each).

70mW is not much of a difference, however the gate resistor I got a dissipation of 2.5W @ 5V (5/0.5 to get 10Ohm resistance, 0.5x0.5x10 for power), and 5W dissipation for the 10V (20Ohm gate resistor).

So if I gate these into account the 5V actually saves me 2.43W.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
So @ 5V, you have higher Rds On which increases standard on power, but you get a lower charge which decreases switching losses.

not quite.

Imagine the mosfet is a valve.

Vgs represents the amount the valve is open and the gate charge is the amount of times you need to turn the handle on the valve.

If the valve is insufficiently open (Vgs is too low) then the flow is restricted. If you open it sufficiently, the appropriate amount of water can flow.

Now you can open the valve more. (Higher Vgs). This requires the transfer of more charge (more turns of the handle) and does marginally reduce the resistance to flow. (Rds).

When you turn it off again, you have to turn a bit before the flow begins to get restricted, (so there is some apparently useless winding which occurs before the valve begins to show its effect.

The turn off is delayed, but is just as fast once it starts.

However turn ON is likely to be faster. (think of it being opened by a spring which is wound up just enough, or far more than enough. On the first case the opening will start quickly but slow toward the end. In the latter case it doesn't start slowing appreciably until after the valve is sufficiently open.

@ 5V I get a total dissipation of 3.27W (gate charge of 8nC and 25mOhm Rdson from curve. This also gave me a rise/fall time of 16ns each)

@ 10V I get a total dissipation of 3.34W (gate charge of 16nC and 18mOhm Rdson from curve . This also gave me a rise/fall time of 32ns each).

Here is the problem. The calculations are quite simplistic and assume that the process is linear.

70mW is not much of a difference, however the gate resistor I got a dissipation of 2.5W @ 5V (5/0.5 to get 10Ohm resistance, 0.5x0.5x10 for power), and 5W dissipation for the 10V (20Ohm gate resistor).

That's peak dissipation, averaged out over a second it will be significantly lower. (32ns * 200,000 is only 0.0064 of a sec). The average dissipation is 5 * 0.0065 or 0.032W

So if I gate these into account the 5V actually saves me 2.43W.

Not in practice :)

the major issue will be the delay in the OFF time, so on time will be extended for some few ns.
 

Lord_grezington

May 3, 2013
124
Joined
May 3, 2013
Messages
124
OK Steve, what you have said makes sense

That's peak dissipation, averaged out over a second it will be significantly lower. (32ns * 200,000 is only 0.0064 of a sec). The average dissipation is 5 * 0.0065 or 0.032W

I have a question about this, the "200,000", is this the frequency? My frequency is only 100Khz, so will this go down to only 0.016W?

Also, Surely the duty cycle needs to be taken onto account as well as this is not switching loss. So should I not multiply this with the duty cycle?

Also the Power Dissipated by the Rdson, should this not be multiplied by the duty cycle? (Ie 0.6 for 60%)
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
200,000 because 100kHz involves 200,000 switchings of the mosfet every second.

So if it takes 32ns to switch, at 100kHz it is switching for 200,000 x 32 ns

duty cycle does not have an impact unless the on time is less than the switching time.

Duty cycle has an impact on the static dissipation (I^2R losses in the mosfet). I have assumes nearly 100% duty cycle because that it the worst case. Yes you could multiply it by your duty cycle.

However, the calculations err on the side of caution. This helps allow for things like less than perfect mounting of the device on the heatsink, slightly elevated temperatures, dust on the heatsink. You don't want to be running on a razor's edge.

If I went on with the heatsink design, I would be recommending that you design for a heatsink maybe twice as good as you calculations suggest. Certainly with any forced air solution this is sensible because dust will build up quite quickly and begin to reduce the efficiency of the heatsink.

If you were to go ito production, you might be able to take practical experience of your prototype and tweak the design a little to reduce costs. For a one off project, you're far better being safe.
 

Lord_grezington

May 3, 2013
124
Joined
May 3, 2013
Messages
124
Thanks Steve

I have made a mosfet calculator in a spread sheet from the discussions on this thread. I will post it on here once I have completed the thermal calculations. As you mentioned its not exact, but it would have been a good tool and saved loads of time.

Can I attach a spread sheet on here or will I need to ZIP it first?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
You will probably have to zip it.

AND... It's a great idea.
 

Lord_grezington

May 3, 2013
124
Joined
May 3, 2013
Messages
124
Mosfet Calculator attached - Thermal Calculations not yet completed though.

I posted this before it was ready, I have already spotted a few issues with gate resistor
 

Attachments

  • Mosfet.zip
    5.4 KB · Views: 80
Last edited:

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
I don't see any formulae in that spreadsheet. Was that intended?
 

Lord_grezington

May 3, 2013
124
Joined
May 3, 2013
Messages
124
Sorry, Calculations should be in there now. Since I copied it from another spreadsheet with all my other calculations on I did a past special by accident.

Calculations now added!!!
 

Attachments

  • Mosfet.zip
    6 KB · Views: 79

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
That's fine but there are a few things you might want to change (or not)

1) the minimum rise time is often specified in the datasheet (as is the minimum fall time). If may be useful to allow these to be entered and to make these the minimum rise and fall times regardless of the calculation based on Ig and Qg

2) it might be useful to allow the gate voltage to be specified at which the mosfet will allow the desired drain current to pass. That will allow the use of a higher gate voltage to reveal a lower power dissipation (rather than the higher one that it does now).

If I have time I'll try to knock something up.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
OK, I've made a few modifications. What do you think of this?
 

Attachments

  • Mosfet (2).zip
    3.4 KB · Views: 65

Lord_grezington

May 3, 2013
124
Joined
May 3, 2013
Messages
124
This Is Good, thanks Steve. I will need to take a better look at it later

Hopefully others will find this useful as well.
 

Lord_grezington

May 3, 2013
124
Joined
May 3, 2013
Messages
124
Wow, I'm impressed. You went into much more detail.

1) Min rise and fall times is something I thought about adding in, but this was original intended to be a mosfet selection aid for specifying units for my project and this was going to add too much complexity, but you have done a great job.
2) No comment on this, never even crossed my mind, essential feature though to show the difference in gate voltages. I was going to start a new thread where I was going to try and use a Mosfet to limit the current (up to 60V 60A) using a gate voltage (finding resettable fuses in this current range is impossible to find, and when I do probably expensive).

Little confused about your gate resistor though, Surly you use the gate resistor to limit the current going into the gate? Will this not make driving parallel mosfets easier and also ensure that you don't take too much current from the Mosfet driver (if its not internally limited).

Your Thermal Calculations are great, I will use these.
 
Last edited:

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
The gate resistor is not typically used to limit the gate current. It is more often used to isolate the voltage at the gate from the device that is driving the gate. This allows the device (it may be an op-amp) to swing its output normally.

If connected directly to the mosfet the voltage will follow the graph shown in the datasheet and that may be inappropriate. It also allows you to place mosfets in parallel as it allows the gate voltages to be slightly different (which is important if one starts switching at a lower voltage than the rest).

Another factor is that the coupling of capacitances between the channel and the gate can cause a voltage to be induced at the gate. Tn a similar manner to the above, a gate resistor helps prevent this from affecting other parts of the circuit.

In that spreadsheet I opted to recommend the gate resistor be small compared to the effective series resistance of the driver to ensure that the gate resistor did not act to limit the current to the gate. This is important as your gate voltage approaches the Vgs required to switch the device fully on. A large gate resistance will cause the final stage of the charging of the gate capacitance to proceed slowly and will affect the simple t = Q/i equation used here.

The effect of specifying the Vgs required to turn the mosfet sufficiently ON is that it determines the time required to do most of the switching on or off and removes the misleading "feature" that a higher gate voltage causes slower switching (when it actually allows faster switching).
 
Top