Diode and forward voltage

[nbrand86]

I am planning to use 2 schottky diodes for a reverse polarity protection circuit which has 2 separate power supplies connected to it.
I selected a diode which has specifications as below

1. Voltage - DC Reverse (Vr) (Max) = 400V
2. Current - Average Rectified (Io) = 20A
3. Voltage - Forward (Vf) (Max) @ If = 1.7V @ 20A

My application is run by 20V Li-Po battery and tested maximum current draw in the full functional operation is 15 A.
Even though the spec 1 & 2 are fulfilling my requirement, I am little bit doubted about the 3rd spec.
Please explain me how it affects and is this diode still usable or not for my application?

 

[papunblg]

The forward voltage is the voltage drop across the diode if the voltage at the anode is more positive than the voltage at the cathode (if you connect + to the anode).

You will be using this value to calculate the power dissipation of the diode and the voltage after the diode.

---------- Post added at 13:02 ---------- Previous post was at 12:59 ----------

When current flows through a diode there is a small voltage drop across the diode terminals. A normal silicon diode has a voltage drop between 0.6–1.7 volts, while a Schottky diode voltage drop is between approximately 0.15–0.45 volts. This lower voltage drop can provide higher switching speed and better system efficiency.

 

[mister_rf]

In order to minimize power losses at 15Amps try to use a high current diode, up to 30-40 Amps and you may lower the reverse voltage up to 45-60V.
See for example MBR3045, MBR3060,
http://www.datasheetcatalog.org/datasheet/on_semiconductor/MBR3045PT-D.PDF

 

[nbrand86]

Yes, it is understandable. I an still little bit confused about the 'if' word and the "@ 20A" part.
so, is it all about the power dissipation on the diode?
Will this diode be able to stay healthy in a 20v system?
will it be damaged by the high voltage?

The total power maximum dissipation through the whole system would be

P= V(max) x I (max)
P = 20 x 15
P= 300W

 

[papunblg]

I am planning to use 2 schottky diodes for a reverse polarity protection circuit which has 2 separate power supplies connected to it.
I selected a diode which has specifications as below

1. Voltage - DC Reverse (Vr) (Max) = 400V
2. Current - Average Rectified (Io) = 20A
3. Voltage - Forward (Vf) (Max) @ If = 1.7V @ 20A

My application is run by 20V Li-Po battery and tested maximum current draw in the full functional operation is 15 A.
Even though the spec 1 & 2 are fulfilling my requirement, I am little bit doubted about the 3rd spec.
Please explain me how it affects and is this diode still usable or not for my application?

Also read this Advanced Power Switching and Polarity Protection for Effects http://www.geofex.com/Article_Folders/mosswitch/mosswitch.htm

 

[mister_rf]

For that diode the forward voltage at 20Amps = 1.7V, this means a diode power dissipation Pd= 1.7*20= 34W.
In practice at 15Amps you may get around 25-30W.
But if we choose another diode, for example the MBR3045 series, see the datasheet Vf = 0.6V at If = 20Amps, so using this at 15Amps Pd = 15* 0.6V= 9W only.

 

[papunblg]

Design Guidelines for Schottky Rectifiers : **broken link removed**

 

[betwixt]

The voltage in your system isn't the issue, the power lost in the diode is what matters.
What the specification is telling you is that if you had 20V at your battery and you were drawing 20A from it through the diode, there would be 20-1.7V at the equipment side, in other words 18.3V. The voltage drop depends on the current and at 15A it would be a bit less than 1.7V, lets say 1.5V but check the data sheet for the exact amount. So you would expect about 18.5V to be available.

Now the important factor, the amount of heat the diode will produce. Devices of this rating are normally mounted by a flange which fits to a heat sink to dissipate some of the heat. The power lost in the diode is Vf x I, in other words, the voltage you are dropping multiplied by the current passing through it. At 1.7 @ 20A the amount is 34W, at your requirement and assuming 1.5V drop it would be 1.5 x 15 = 22W. That is quite a lot of power and hence heat to have to lose somewhere.

Other things to watch out for are Vr (Max) which in your case doesn't have to exceed the battery voltage and Io which should be at least 15A CONTINUOUS current, you will see specifications which mention "average rectified" and "non-continuous peak" which are only ratings for part cycle (when converting AC to DC) or short bursts.

Brian.

 

[mister_rf]

@nbrand86
I am also convinced you need to change the design of the circuit to include a MOSFET protection instead schottky diodes.
See details here:
https://www.infineon.com/dgdl/Rever...b2614&fileId=db3a304412b407950112b41887722615

 

[nbrand86]

After some more research, I changed my mind to change my design in to MOSFET instead of Diode circuit I had (Figure 1)


Figure 1 : with Diode


I replaced the diodes with N type MOSFET as displayed below.


Figure 2 : with MOSFET

I am not sure about the accuracy of this configuration. Please show me if there is any mistake. And any suggestions for a MOSFET number
so that I can purchase and get done with this circuit.
Thank you.

 

[papunblg]

After some more research, I changed my mind to change my design in to MOSFET instead of Diode circuit I had (Figure 1)

View attachment 72981
Figure 1 : with Diode

You CANNOT use a schottky diode since it's reverse leakage current is too much and may still damage your circuit in reverse polarity.

I replaced the diodes with N type MOSFET as displayed below.

View attachment 72982
Figure 2 : with MOSFET

I am not sure about the accuracy of this configuration. Please show me if there is any mistake. And any suggestions for a MOSFET number
so that I can purchase and get done with this circuit.
Thank you.

http://www.ciphersbyritter.com/RADELECT/BREADBD/BREADBD.HTM

http://hackaday.com/2011/12/06/reverse-voltage-protection-with-a-p-fet/
**broken link removed**

 

[nbrand86]

Thanks a lot, That's really helpful.
So, a P-channel FQP47P06 will do my job right?

(My requirement Vmax=20V , Imax= 20A)