Need Proper Schottky Diode of 100 Watt & operating frequency of 10KHz to 50KHz

[swapneshshinde]

Dear friends,

I need name of Schottky Diode around min. 100 Watt & operating frequency around 30 Khz, also i want Mosfet operating on same frequency.

Please help me.

Regards,

Swap.

 

[duke37]

You will need to give more information. 100W could be 100V at 1A or 1V at 100A.
Some details of the application would allow someone to help.

 

[swapneshshinde]

Yeah Voltage is 100VDC, 1A

Yeah Duke you are right, i was supposed to tell that in (i need 100V, 1Amp min. diode)current volt format. Please suggest me as soon as possible, thank you for ur help.
It's kind of step up chopper. Also, tell me MOSFET operates in same range of frequency & current 2Amp max.

 

[Harald Kapp]

Why would you need a 100 W diode? Even if your parameters call for 1 A and 100 V, for the diode it is either of two operating modes:

a) diode is reverse biased. Current is 0 A, Voltage is 100 V. Power is 0 W
b) diode is forward biased. Current is 1 A, Voltage is ~0.3 V. Power is 0.3 W

Plus losses from switching, of course.

You can select a suitable diode e.f. here: http://www.vishay.com/diodes/rectifiers/
First select a housing suitable to your project (SMD, THT...), then filter the resulting list according tor your parameters.
Same for MOSFETs: http://www.vishay.com/mosfets/

Or look up selection tables of other manufacturers.

Regards,
Harald

 

[jackorocko]

Quick search on digikey gave me 5 pages of results that fit the bill. All for high freq.

 

[swapneshshinde]

100 Watt???

I need it because when i put full load that diode must be capable to carry that much power, i need schottky because i'm operating mosfet at 30Khz which is in shunt to inductor my input is 10VDC & output is 100VDC without any inverter, u r getting now what i'm designing.

 

[BobK]

The power dissapated by a diode (or a resistor) is the voltage across it multiplied by the current through it. The voltage across the diode is not 100V it is about .3 volts as Harald has already said. If you force the voltage across any diode to 100V you will produce smoke.

Bob

 

[duke37]

You are trying to do a difficult job. The devices will have to pass 10A and resist 100V. A better circuit would be the traditional push/pull transformer. Then the two fets would need to pass 10A but resist about 30V (IRF540) and the output diodes would need to pass 1A and resist >100V (UF4007).

 

[swapneshshinde]

@ Dear Bob,
You are right, it will burn the diode, i haven't connected that diode across load in the ckt., but what i want to say is, the diode should be switched on-off fast or on-off delay should be very low, so suggest me schottky or fast recovery diode.

@Dear Duke,
My design is step up chopper, why should use push/pull transformer, can u explain please?

 

[duke37]

In what I think is your circuit, the fet and diode will have to deal with both high current and high voltage. The devices which will be optimum for high voltage will be less than optimum for high current. Thus high voltage fets will have a higher on resistance than low voltage fets.

A transformer enables the primary circuit to see high current and the secondary to see high voltage.

Also, the boost convertor works on high current pulses which need to be smoothed with large, low esr capacitors. A push/pull circuit takes a fairly constant current and delivers a failrly constant current so the capacitors have a much easier time.

If your chopper only raised the voltage by 5 times, things would be much easier.

 

[(*steve*)]

In a switchmode power supply a schottky diode is useful because it has both a low forward voltage and rapid recovery characteristics.

The former reduces the losses in the diode and directly impacts efficiency. Synchronous rectification is a technique which can further improve the efficiency.

The fast recovery, can (at least in part) be explained by the capacitance of the junction. A diode, when forward biased will take a certain time to stop conducting when reverse biased. This can be modelled as the diode having a certain capacitance. Charge can flow in the reverse direction freely until this capacitance is charged.

In buck regulators, especially where the output voltage is low, the forward voltage drop of the diode can be extremely significant in determining the overall efficiency.

As the output voltage rises, the diode forward drop becomes a smaller percentage of the output voltage and therefore has less impact.

Schottky diodes have a downside too. They tend to be very leaky in the reverse direction. This leakage gets worse with voltage and temperature. The reverse leakage (because it occurs when reverse biased, and therefore at a higher voltage) can actually contribute to more heating than the forward current!

For higher voltages, the use of non-schottky diodes may be a good option.

If you decide on a schottky diode, it would pay to look at the datasheet very carefully to determine the leakage at the voltages and temperatures it will be operating at.

 

[Harald Kapp]

You could also look for fast recovery diodes, e.g. here: http://www.vishay.com/diodes/rectifiers/fast-recovery-superectifier/

Downside is a much higer forward voltage (> 1V) compared to Schottky diodes, but less leakage current.

Harald

 

[swapneshshinde]

@ Steve: Yes u are very right, but which diode i should choose fast recovery or schottky.

 

[(*steve*)]

@ Steve: Yes u are very right, but which diode i should choose fast recovery or schottky.

I would guess you would be in the region where a fast recovery diode would be better, but that's just a gut feeling.

You need to check the forward voltage drop for each at the current you're looking at, and also the reverse leakage and use this to calculate your losses.