# Diode with low voltage drop?

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#### Clarkdale44

##### Member
Hello

I need a diode with less voltage drop like 10SQ050 that is 50V 10A...

But i need something that is rated at least 30A or 40A with 50V or more...

Is there such a diode available?

Also do any of you know what kind of diode is that in image below (in red circle)?

Regards!!

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#### JLNY

##### Active Member
Most kinds of Schottky diodes will have a lower voltage drop similar to the part you mention. Yes, there should be many kinds of power schottky diodes available with those specs. Most will probably be in high-power packages that need to be mounted to a heatsink.

That diode in the picture looks to be some kind of DO-4 or DO-5 package diode. no way of telling its characteristics without some kind of part number.

#### Clarkdale44

##### Member
OK so these are called stud diodes.. used in high power applications...

If i buy one of these say 40A rated.. how big of a heat sink i am gonna need..?

I will be drawing nearly 25Amps...

#### Rich D.

##### Active Member
That diode in the image appears to be a normal stud diode that is mounted in an aluminum block. The large block is likely not a part of the diode itself.

Note In the category of nothing is ever always that easy:

A diode (Schottky I assume you are looking for) has a voltage drop rating and a current rating. Often those are not possible at the same instant.
A voltage rating is made for a a particular current only at a particular temperature.
Lower currents will produce a lower voltage drop.
Higher currents will produce a higher voltage drop.

As an example I'll grab a random diode, the SB320S Schottky diode from Vishay has a maximum specified rating of 3Amps average current (in a particular thermal condition), and a forward voltage drop of 500mV.

At 25 degrees C, when conducting 2 Amps, it's voltage is 400mV, but when conducting 200mA it's voltage is more typically 250mV.
For non repetitive current peaks, it can easily handle 20 Amps, but the voltage drop at that current is 800mV.
And then of course at different temperatures it acts a bit differently but let's not go there.

Just be aware that for best results you should verify that the diode's voltage drop depends on how much current it is conducting at the time.

#### JLNY

##### Active Member
OK so these are called stud diodes.. used in high power applications...

If i buy one of these say 40A rated.. how big of a heat sink i am gonna need..?

I will be drawing nearly 25Amps...
Calculating heat sink sizes correctly is a bit of a science, and will depend on things like airflow and the heat sink material. I'm sure there are guides online that can give a better explanation than I can, but as a first step you will at least want to calculate the power dissipation we are dealing with here:

As an example, I will use the part MBR6080. (a 60A forward current and 80V reverse voltage-rated schottky diode in a DO-5 package. You might also consider other diodes or package types if your application does not specifically require a stud-mounted diode.)

According to the trace curves, at 25A the forward voltage will be slightly under 0.7V, resulting in a power dissipation of 17.5 Watts-- which is quite a lot for just a single diode. If the application has a lower duty cycle-- e.g. if the device will only be powered for brief periods with enough time to cool down in between, or pulsed on and off so that it only conducts part of the time-- a smaller heatsink may be possible, but if this is going to be running a continuous 25A, you may need a reasonably-size heat sink with fins and possibly a bit of airflow.

Just out of curiosity, what is the application here?

#### Clarkdale44

##### Member
Calculating heat sink sizes correctly is a bit of a science, and will depend on things like airflow and the heat sink material. I'm sure there are guides online that can give a better explanation than I can, but as a first step you will at least want to calculate the power dissipation we are dealing with here:

As an example, I will use the part MBR6080. (a 60A forward current and 80V reverse voltage-rated schottky diode in a DO-5 package. You might also consider other diodes or package types if your application does not specifically require a stud-mounted diode.)

According to the trace curves, at 25A the forward voltage will be slightly under 0.7V, resulting in a power dissipation of 17.5 Watts-- which is quite a lot for just a single diode. If the application has a lower duty cycle-- e.g. if the device will only be powered for brief periods with enough time to cool down in between, or pulsed on and off so that it only conducts part of the time-- a smaller heatsink may be possible, but if this is going to be running a continuous 25A, you may need a reasonably-size heat sink with fins and possibly a bit of airflow.

Just out of curiosity, what is the application here?

Inverter drawing from battery 20 to 25amps and due to my unusual requirement... i need to use a diode.

#### dr pepper

##### Well-Known Member
The diodes dissipation and running temp defines the size of heatsink.
There are 2 main factors amongst others for working out dissipation, one is forward voltage and the other is switching losses which is mainly affected by switching time.
Trying to get the sink spot on from the start on a prototype probably isnt a god idea, if you work out dissipation based on forward volt drop then add 50% or so and try it, you can work out from the actual sink temp how much energy is going into it, then you can if the sink is the wrong size calc a new sink based on the dissipation and required max sink temp.
There are probably going to be several arguments for this process, however it works for me as I have a box of salvage parts.

#### Rich D.

##### Active Member
Might I recommend a STTH61W04SW made by ST Microelectronics. It is in a TO-247 case. It ain't cheap, but it's the cheapest one I could find that would surely be rugged enough for your requirements. I found it at Mouser for $4.02 each, 10 for$3.42 each.

http://www.mouser.com/ProductDetail...=sGAEpiMZZMtoHjESLttvkvpio6NcApNgnGcYmKygf5A=

The only way I know of to reduce the cost would be to relax the voltage and/or current requirements.

It's rated for up to 60Amps average but of course that's if it is bolted to a heatsink with plenty of surface area or good airflow, 400V, and a Vf rating of 930mV.
At 25 amps DC it will disipate about 25 watts which would get it pretty warm. Expect about 1.1 volts drop across it.
If your power inverter will not be conducting full time (and I assume it won't by using a switching converter) the power in the diode will be less, but that can't be predicted here. Best to plan for worst case.
Don't forget to use a thin layer of heatsink compound!

#### Rich D.

##### Active Member
I was looking at and finding plenty of diodes that are packaged in pairs. Typically they can take 20A each, but since there are two of them, running them in parallel can give you 40 Amps, right?

Well...maybe. If there are slight variations in Vf then they will not match together very well. The one with the slightly lower voltage will run more current, and if the balance isn't good enough then it will eventually burn out, leaving the other one to take the full power for a while until it too blows. I don't know that they can guarantee exactly matching Vf for each diode with the manufacturing process variations. And if you could guarantee or measure the difference, how accurately can you calculate the current imbalance and how can you trim it out? Probably pretty iffy, engineering-wise.

A common method to balance two diodes in parallel is to use individual low-value resistors in series with each diode. Problem there is that those big-ass resistors will drop some more voltage, and might very well be no better than a bigger diode with a higher Vf to begin with. More complicated and not necessarily cheaper, so I would go with the KISS method and ante up for the more costly diode to begin with.

#### dr pepper

##### Well-Known Member
Reading back over the posts I agree with Rich, if your putting this diode inline with the supply to a inverter the voltage across the diode will not reverse so switching losses will be really small, so 25w dissipation is probably close.
I'd suggest a sink of about 4 degrees c per watt if this is an automotive application.

#### crutschow

##### Well-Known Member
Inverter drawing from battery 20 to 25amps and due to my unusual requirement... i need to use a diode.
So what function will the diode be performing?
If it's for reverse polarity protection, a power MOSFET would be a better choice.

#### JLNY

##### Active Member
Voltage drop is only 0.3v confirmed by seller, rest i don't know... Although it is for solar application but i think it should work just fine for mine as well...
I would ask for a part number from the seller and then look up the datasheet before buying.

Even if the low-current Vf is 0.3V, it will be higher at 25A. Unfortunately, this is a high-power enough application that a diode can no longer be considered a constant voltage drop at all current levels. you will need to go through the datasheet and look at the Vf vs current plots.

#### Tony Stewart

##### Well-Known Member
Yes Schottky diodes in this range of current and Vf are about -25V not -1000V, so it must be Silicon >0.7V

#### Clarkdale44

##### Member
Part number is 60HF100

#### JLNY

##### Active Member
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