Mosfet Selection - Electronic Circuits Projects Diagrams Free

jnnewton · 15th May 2008, 10:30 PM

If I am using a mosfet as a switch, I assume there is no trick to the vds needed. If i want to switch 100V, i need vds to be greater than 100V correct?

I have been trying to find a surface mount mosfet to switch 500V with a load current of 20A. This seemed easy until i actually took the Rtheta into account. It seems that all of these 500V mosfets rated at 20A will melt far before they ever get to 20A. Does this seem correct?

If both of these statements are correct, is it time for me to look at igbt's?

Thanks for any help.

mneary · 15th May 2008, 11:07 PM

Best solution usually is to put a few in parallel. It decreases the total dissipation by n, and divides the remaining power among n devices.

500V surface mount devices might have other problems, like spacing (500V on the surface of a board has to be handled carefully).

crutschow · 15th May 2008, 11:28 PM

Quote:

Originally Posted by jnnewton

is it time for me to look at igbt's?

IGBT's generally have over a 2V on voltage drop and thus would also dissipate a significant amount of power at 20A.

jnnewton · 16th May 2008, 02:43 AM

So is it safe to say that several mosfets or igbts in parallel are the way to go (is this how manufacturers make H-bridge circuits for say 5-10hp motors?

mneary · 16th May 2008, 02:52 AM

igbts in parallel don't reduce the voltage drop and you want surface mount, so for only 20A I would still go with MOSFETS.

dknguyen · 16th May 2008, 03:44 AM

IGBTs also do not balance out for current sharing and one will current hog in the same way that parallel diodes don't work because one will hog the current and burn out, and the cycle repeats with the remaining diodes until they are all fried.

jnnewton · 16th May 2008, 07:07 AM

I cannot find a mosfet that will do this at room temperature. So I am going to change the way I do this (a little). First of all, I am using the Following method to validate the mosfet for my application. Please tell me if something is wrong here.

Vbus - DC Bus Voltage (applied to soruce pin of mosfet)
Iload - Load current (also current from source pin to drain pin of mosfet)
Ids - max current (given as a function of Rds(25)
Tamb - Ambient temp inside the enclosure
Pdisp(a) - Power dissipated in the mosfet (ambient)
Tmax - Max Junction Temp
Rtheta(ja) - Temp rise per watt of dissipated power (junction to ambient)
Rds(a) - Resistance at temperature (junction) (given in graph)

So,
The equations are as follows:

Pdisp(a) = (Tmax - Tamb) / Rtheta(ja)
Iload(a) = sqrt(Pdisp(a) / Rds(a))

so for FCB20N60 from fairchild, whose spec says 600V, 20A:

Pdisp(a) = (150 - 40) / 40
Pdisp(a) = 2.75W

Iload = sqrt( 2.75W / 1.25)
Iload = 2.2 A

Which means I can get 2.2A out of a 20A mosfet Which means this one won't work.

Questions:
1. Does my math have errors? If so, please offer suggestions.
2. Is there actually a mosfet that will do what I want? What is a P/N?

Optikon · 16th May 2008, 04:57 PM

Quote:

Originally Posted by jnnewton

I cannot find a mosfet that will do this at room temperature. So I am going to change the way I do this (a little). First of all, I am using the Following method to validate the mosfet for my application. Please tell me if something is wrong here.

Vbus - DC Bus Voltage (applied to soruce pin of mosfet)
Iload - Load current (also current from source pin to drain pin of mosfet)
Ids - max current (given as a function of Rds(25)
Tamb - Ambient temp inside the enclosure
Pdisp(a) - Power dissipated in the mosfet (ambient)
Tmax - Max Junction Temp
Rtheta(ja) - Temp rise per watt of dissipated power (junction to ambient)
Rds(a) - Resistance at temperature (junction) (given in graph)

So,
The equations are as follows:

Pdisp(a) = (Tmax - Tamb) / Rtheta(ja)
Iload(a) = sqrt(Pdisp(a) / Rds(a))

so for FCB20N60 from fairchild, whose spec says 600V, 20A:

Pdisp(a) = (150 - 40) / 40
Pdisp(a) = 2.75W

Iload = sqrt( 2.75W / 1.25)
Iload = 2.2 A

Which means I can get 2.2A out of a 20A mosfet Which means this one won't work.

Questions:
1. Does my math have errors? If so, please offer suggestions.
2. Is there actually a mosfet that will do what I want? What is a P/N?

These are static power calculations... are these parts on a heatsink? I dont see that included. Also, I thought you were going to parallel power mosfets (I also recommend this method for your I & V)?

I recommend as large of a physical package as possible to help get the heat out.

Can you get ahold of any of these? problem solved.

Technology

mneary · 16th May 2008, 10:46 PM

Simply putting two of the FCB20N60 in parallel on a 10C/W heat sink will give you the 20A rating that you want.

Bob Scott · 17th May 2008, 03:35 AM

After making sure that your MOSFET is good for the required voltage and current, check the Rds(on) rating. This varies with gate voltage. Make sure that you have enough gate voltage.

P=I^2*R

Power dissepated by the MOSFET will be determined by P=(I^2)*Rds(on). If the power dissepated causes too much temperature rise, look for a MOSFET with lower Rds(on). If you cannot find a MOSFET with a low enough R, connect MOSFETs in parallel. Every time you double the number of MOSFETs in parallel, the total resistance is halved and the power dissepated as heat in each MOSFET drops to 1/4.

I use a rule of thumb. It's my own rule of thumb. The circuit requires lower resistance if the MOSFET is too hot to hold my thumb on. I don't use heat sinks. You can add more MOSFETs for the price of a heat sink.

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15th May 2008, 10:30 PM	(permalink)
jnnewton	Mosfet Selection If I am using a mosfet as a switch, I assume there is no trick to the vds needed. If i want to switch 100V, i need vds to be greater than 100V correct? I have been trying to find a surface mount mosfet to switch 500V with a load current of 20A. This seemed easy until i actually took the Rtheta into account. It seems that all of these 500V mosfets rated at 20A will melt far before they ever get to 20A. Does this seem correct? If both of these statements are correct, is it time for me to look at igbt's? Thanks for any help.

15th May 2008, 11:07 PM	(permalink)
mneary	Best solution usually is to put a few in parallel. It decreases the total dissipation by n, and divides the remaining power among n devices. 500V surface mount devices might have other problems, like spacing (500V on the surface of a board has to be handled carefully).

16th May 2008, 02:43 AM	(permalink)
jnnewton	So is it safe to say that several mosfets or igbts in parallel are the way to go (is this how manufacturers make H-bridge circuits for say 5-10hp motors?

16th May 2008, 02:52 AM	(permalink)
mneary	igbts in parallel don't reduce the voltage drop and you want surface mount, so for only 20A I would still go with MOSFETS.

16th May 2008, 03:44 AM	(permalink)
dknguyen	IGBTs also do not balance out for current sharing and one will current hog in the same way that parallel diodes don't work because one will hog the current and burn out, and the cycle repeats with the remaining diodes until they are all fried. __________________ NO, that picture isn't me so don't bother asking if we can be friends.

16th May 2008, 07:07 AM	(permalink)
jnnewton	I cannot find a mosfet that will do this at room temperature. So I am going to change the way I do this (a little). First of all, I am using the Following method to validate the mosfet for my application. Please tell me if something is wrong here. Vbus - DC Bus Voltage (applied to soruce pin of mosfet) Iload - Load current (also current from source pin to drain pin of mosfet) Ids - max current (given as a function of Rds(25) Tamb - Ambient temp inside the enclosure Pdisp(a) - Power dissipated in the mosfet (ambient) Tmax - Max Junction Temp Rtheta(ja) - Temp rise per watt of dissipated power (junction to ambient) Rds(a) - Resistance at temperature (junction) (given in graph) So, The equations are as follows: Pdisp(a) = (Tmax - Tamb) / Rtheta(ja) Iload(a) = sqrt(Pdisp(a) / Rds(a)) so for FCB20N60 from fairchild, whose spec says 600V, 20A: Pdisp(a) = (150 - 40) / 40 Pdisp(a) = 2.75W Iload = sqrt( 2.75W / 1.25) Iload = 2.2 A Which means I can get 2.2A out of a 20A mosfet Which means this one won't work. Questions: 1. Does my math have errors? If so, please offer suggestions. 2. Is there actually a mosfet that will do what I want? What is a P/N?

16th May 2008, 10:46 PM	(permalink)
mneary	Simply putting two of the FCB20N60 in parallel on a 10C/W heat sink will give you the 20A rating that you want.

17th May 2008, 03:35 AM	(permalink)
Bob Scott	After making sure that your MOSFET is good for the required voltage and current, check the Rds(on) rating. This varies with gate voltage. Make sure that you have enough gate voltage. P=I^2R Power dissepated by the MOSFET will be determined by P=(I^2)Rds(on). If the power dissepated causes too much temperature rise, look for a MOSFET with lower Rds(on). If you cannot find a MOSFET with a low enough R, connect MOSFETs in parallel. Every time you double the number of MOSFETs in parallel, the total resistance is halved and the power dissepated as heat in each MOSFET drops to 1/4. I use a rule of thumb. It's my own rule of thumb. The circuit requires lower resistance if the MOSFET is too hot to hold my thumb on. I don't use heat sinks. You can add more MOSFETs for the price of a heat sink. Last edited by Bob Scott; 17th May 2008 at 03:42 AM.