Choose proper rectifier diode

[giacomo]

Hi, if i have to build a bridge rectifier how can i choose the proper diode in relation with the maximum current ?

For instance if i have a 100VA 12V transformer and i want to build a linear power supply that can deliver around 4.5A. I watch the datasheet for the 1N4007 diode and i find to paramether "Non-repetitive Peak Forward Surge Current" 30A, can this diode do the job ?

[Vtile]

No it can not. You are not after peak current in your implementation, but an average forward current.

For 1N4007 it seems to be 1A.

Ps. Gyros response is better than mine, while the content is pretty much the same. Here is good basics about the terminology you see on the diode datasheets. Unfortunately I don't know any in italian language.
https://www.allaboutcircuits.com/textbook/semiconductors/chpt-3/diode-ratings/

[Gyro]

Hi, welcome.

The hint is in the words "Non-repetitive" and "Surge". They mean what they say (not every AC cycle). The figure you need to be looking at is the Ifavg (average forward current) to start with. In the case of the iN4007 it is 1A, clearly not suitable.

You should be looking for diodes with at least 8A Ifavg, probably more for safety. The other important figure is the repetitive surge voltage (the maximum surge every cycle), which is determined by the resistance of the secondary winding of the transformer, capacitor size etc.

[Seekonk]

When a supply is first turned on the capacitors will not be charged and look like a dead short for a few cycles. Only transformer resistance will limit this.  This condition also exists in many snubber applications so it is a useful number.

[digsys]

And FAR more important than that is heat dissipation required. A standard axial diode, even 5-8A will NOT get rid of the heat, so you'll need a 10-20A bridge,
mounted to a heatsink !! 2x diodes conducting x 0.7V (say) at 4A = 5.4W min. The peak capacitor currents will add to that.

[Rbastler]

Get a standard rectifier bridge with your required voltage and current. I suggest getting one that you can mount on a heatsink. A 20V 8A bridge ist certanly fine.

[mariush]

A 100VA transformer will output 100/12 = 8.33 A of current (AC)
At DC current, it will output approximately 0.62 x 8.33A = 5.2A of current.

In a bridge rectifier, there are always two diodes conducting electricity at any time. So if you want to be super safe, you'd want each diode to be capable of carrying 2.25A of current continuously. In real world, since we're dealing with low frequency (50Hz or 60 Hz), that means about half the time there's no current flow in 2 of the 4 diodes so there is some time for the diodes to cool down a bit before current flows again through them.
So if you're really cheap, in theory you can use 4 diodes rated for a bit lower current than the maximum current you're going to have.

In practice however, you also have to be aware of how much each diode will heat up (which varies with the current and percentage of time the diode will be on) and potentially derate the diode with the temperature.

For example, let's look at a datasheet for 1n4007 : https://www.diodes.com/assets/Datasheets/ds28002.pdf

You can see there:

Average Rectified Output Current (Note 1) @ TA =+75c : 1 A   

That means if you make a rectifier out of 4 diodes like these, at any point you have a guaranteed 2A continuous current but since they're not continuously on, but rather about half the time, they could probably handle a bit more.. though i wouldn't think of as much as 2.25A per diode 

Forward Voltage @ IF = 1.0A : 1v 

This tells you that at 1A of current flowing through the diode, there will be 1v drop on the diode, which means the diode will produce 1V x 1A = 1 watt of heat. This 1 w of heat will have to go somewhere, it will go in the air around the diode raising the ambient temperature and it will also go through the leads of the diode warming out the copper traces on the circuit board. If you leave the leads a bit long, they may act as a very tiny heatsink helping with the diode temperature.

Typical Thermal Resistance Junction to Ambient RJA  : 100 K/W

This tells you that  1w of dissipated power in the diode will increase its temperature by 100 degrees Kelvin over the ambient temperature. .. so if the ambient temperature is 25c and the diode has 1A of current continuously, the diode will eventually get hot, up to around 125c (which is still below the maximum 150c allowed temperature)
See figure 1 on second page ... you can see that as the ambient temperature raises, the diode must be derated to lower currents otherwise the diode temperature will go above 150c and be broken.
It's important to understand this because as they function, they'll heat the air around so the ambient temperature will increase.  In some circuits like power supplies, designers sometimes get away with it because there's some active fan blowing air over the diodes moving the air around.

Anyway... point is it's often smart to use bigger diodes or bigger bridge rectifiers than needed, because the forward voltage of each diode will be lower so the bridge rectifier will produce less heat. Also, ready made bridge rectifiers (like GBU packages) can easily be helped to stay cool with a heatsink

In your application, I would go with a 6-10A bridge rectifier like these ones : http://uk.farnell.com/webapp/wcs/stores/servlet/Search?catalogId=15001&langId=44&storeId=10151&categoryId=700000004655&showResults=true&pf=110130892,110130899,110152425,110162956,110162957,110162958,110195763

You'll probably have to derate them by about 20% if you don't use a heatsink, so even a 6A bridge rectifier will work with 5A continuously.

[giacomo]

Ok, first of all, thanks to all

I was looking in my components collection and i have found a bridge rectifier KBPC1010 with a avarage output current of 10A

I have another question inherent another paramether, the "forward voltage (per element)", now, i know that i need to improve my english but "per element" i intended that i need to consider two times that value when i calculate the voltage drop in the bridge diode, right ?

No it can not. You are not after peak current in your implementation, but an average forward current.

For 1N4007 it seems to be 1A.

Ps. Gyros response is better than mine, while the content is pretty much the same. Here is good basics about the terminology you see on the diode datasheets. Unfortunately I don't know any in italian language.
https://www.allaboutcircuits.com/textbook/semiconductors/chpt-3/diode-ratings/

Thank, i just start to read

Thank mariush, very itemized

[Gyro]

If you want some help visualising things then Duncan Amps free PSUD2 software is fun to play with: http://www.duncanamps.com/psud2/

Obviously it's no substitute for doing the calculations yourself, but it's good for showing you the shape of current and voltage waveforms on the various parts, both steady state and start up (as long as you put accurate parameters in of course).


P.S.

I have now i have another question inherent another paramether, the "forward voltage (per element)", now, i know that i need to improve my english but "per element" i intended that i need to consider two times that value when i calculate the voltage drop in the bridge diode, right ?

Yes, that's right.

[Codebird]

You can reduce heat dissipation and wasted power a bit if you use Schottky diodes. They can take more current too. But, as a previous poster pointed out, the best way is often to get a bridge - simply because they can be easily attached to a heatsink.

[Ian.M]

A 100VA transformer will output 100/12 = 8.33 A of current (AC)
At DC current, it will output approximately 0.62 x 8.33A = 5.2A of current.

The average DC current you can draw without exceeding the transformer's RMS secondary current rating depends on the rectifier topology, and whether or not it directly feeds a reservoir capacitor.  See Hammond (transformer division) Design Guide for Rectifier Use.

In a bridge rectifier, there are always two diodes conducting electricity at any time. So if you want to be super safe, you'd want each diode to be capable of carrying 2.25A of current continuously. In real world, since we're dealing with low frequency (50Hz or 60 Hz), that means about half the time there's no current flow in 2 of the 4 diodes so there is some time for the diodes to cool down a bit before current flows again through them.
So if you're really cheap, in theory you can use 4 diodes rated for a bit lower current than the maximum current you're going to have.

<snip>

That means if you make a rectifier out of 4 diodes like these [1N4007], at any point you have a guaranteed 2A continuous current but since they're not continuously on, but rather about half the time, they could probably handle a bit more.. though i wouldn't think of as much as 2.25A per diode 

There's a great amount of confusion here - YES the average current in ONE diode of the bridge is half the bridge output current. This is NOT because the load current is split between two diodes at any instant.  The two conducting diodes are effectively in a series circuit with the transformer secondary and the load and they *ALL* pass the same average current in any half-cycle (assuming steady state operation).   However as there are four diodes in the bridge and the conducting pair and non-conducting pair swap at the end of every half cycle, each diode is off half the time so the average diode current is half the load current.    You cant have two bites of the same cherry - if you are rating the diode according to half the load current, you cant also increase the rating because "there is some time for the diodes to cool down a bit before current flows again through them".

In practice, the diode datasheet will specify the conditions under which the parameters are specified.  e.g:

Maximum Ratings and Electrical Characteristics (@TA = +25°C unless otherwise specified.)
Single phase, half wave, 60Hz, resistive or inductive load. For capacitive load, derate current by 20%.
.....
Average Rectified Output Current (Note 1) @ TA =+75°C I_O 1.0 A
.....
Notes: 1. Leads maintained at ambient temperature at a distance of 9.5mm from the case.

That tells you that the datasheet has already used the cooldown time in the rating as its for half-wave operation.  It also specifies a derating factor of 20% (i.e. *0.8 ) for a capacitive load,  However as each diode in the bridge only supplies half the time-averaged load current, you do get that benefit so the max load current for a 1N400x bridge is 0.8*(2*1A)=1.6A. 

This is a MAXIMUM rating, so you then need to apply an additional derating factor for ambient temperature and not to be right on the bleeding edge.  With good cooling (i.e. forced airflow large area copper tracks and short leads) you could push it to 1.5A if you didn't care about reliability, but derating to 1A would be safer for the average project.  If you want to be really safe, or at elevated ambient (equipment interior) temperatures, you'd need to do a proper thermal analysis.

Single package bridge rectifiers tend to be rated for average output current, For 4.5A out, intermittant operation with good cooling, you can probably get away with a 5A bridge, but you will need to check the bridge datasheet for the thermal derating factor and to see if it can handle the startup surge current charging your intended reservoir capacitance. For continuous operation with non-optimal cooling, you'd save time to start by looking at 10A bridge datasheets.