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In my application, my load current is 1.4 amps, and for reasons of cost, availability, and recovery time, I want to use this particular 1 amp diode. Like many diodes, this one has a negative temperature coefficient; the hotter they get, the lower their voltage drop for a fixed current. This means that if you parallel two diodes, they are not guaranteed to share current well; one device will take more current than the other, heating it up more, causing it to take even more current.

How can I estimate how many 1-amp diodes are needed in parallel to carry 1.4 amps?

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  • \$\begingroup\$ What type of circuit is it being used in? \$\endgroup\$ – Andy aka Oct 13 '15 at 16:09
  • \$\begingroup\$ @Andyaka For the purposes of the question, let's assume I only want DC blocking. My real-world application is different, obviously, or I wouldn't be talking about a fast-recovery diode! \$\endgroup\$ – Stephen Collings Oct 13 '15 at 17:15
  • \$\begingroup\$ Won't a capacitor do that? The problem here is that you are giving an answer to your own question and not spilling the beans on the application. If someone else had made this answer (given the limited amount of info in the question) I'd be pulling them up on the naivety of their answer because it doesn't account for wider issues i.e. there are other things to consider. \$\endgroup\$ – Andy aka Oct 13 '15 at 17:22
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A diode junction can be approximated as a voltage drop in series with a resistance. These two values are occasionally stated explicitly on a data sheet, for they can always be extrapolated from the V-I curve for the device: pick two points in the linear region of the curve, and make a line between those two points. The V-axis intercept is the voltage drop, and the V/I slope is the resistance. Since the V-I curve changes with temperature, so will these values.

Putting two of the specified BYG23T diodes in parallel, assuming one to be at the 150C max temperature and the other to be at the 25C "cold" temperature, we get this approximation, with V2/R2 as the hotter of the two diodes:

schematic

simulate this circuit – Schematic created using CircuitLab

Obviously this is a worst-case scenario; it's almost unimaginable that one diode will be at 150C while the other is at 25C. But if we survive this scenario, we'll also survive any more realistic ones.

Let us also assume there is exactly one diode at 150C, and an arbitrary number n of diodes at 25C. The single diode at 150C will limit the entire parallel combination from taking more current, and all the other diodes being at 25C will minimize their contribution towards the overall current rating of the paralleled set.

We know that the total current through the paralleled set is the sum of the currents through each branch.

\$I_{total}=nI_1+I_2 \$

We also know that the voltage across each branch is identical.

\$V_1+I_1R_1=V_2+I_2R_2\$

\$V_1\$, \$V_2\$, \$R_1\$, \$R_2\$, and \$I_2\$ can be obtained from the datasheets. \$I_{total}\$ is application-dependent. That leaves us with two variables (\$I_1\$ and \$n\$) and two equations. Solving for n and leaving out a few steps:

\$n=\lceil\frac{R_1(I_{total}-I_{2})}{V_2-V_1+I_2R_2}\rceil\$

Alternately, if we know \$n\$ we can solve for the actual current that will be seen by the hottest diode.

\$I_2=\frac{I_{total}R_1+n(V_1-V_2)}{R_1+nR_2}\$

In this particular case, with the values from this particular diode and a 1.4 amp load current, three 25C diodes are needed in parallel to guarantee that the 150C diode will not see more than 1A. Actual current through the hot diode in that situation comes to just under one amp.

That gives us a total of four paralleled 1A diodes to carry a 1.4 amp load. This is almost certainly overkill. Our calculations could be improved if we could account for the effect of temperature on the V-I curve of the device. But the equations get much more complex, and we would have to have some idea on just how much variation in temperature response we could expect from device to device. That's not a spec I've ever seen on any datasheet.

From the equations I've run for this particular set of diodes, 25% imbalance would cause one diode to take all the current at half an amp; 31% at 1 amp. Since the data sheet doesn't list variation from component to component, by paralleling diodes we are always taking a risk that the imbalance might be enough to ruin the whole scheme.

Supposing 10% imbalance, I get that two diodes aren't quite enough to carry 1.4 amps. At 20% imbalance, three aren't quite enough.

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  • \$\begingroup\$ I really think you need to consider the AC implications of this as well as the DC. The diode in your question is a fairly fast device in terms of reverse recovery time but it also has a forward recovery time (i.e. time to begin conducting after being reverse biased) that is 620 ns typical (as per the data sheet). Now my understanding is that this will vary between devices and, on a fast application could easily kill the device that recovers more quickly. Duty cycle is important. Peak current is specified at 18A and this may indeed be a get-out-of-jail spec for using the device on its own. \$\endgroup\$ – Andy aka Oct 13 '15 at 17:09
  • \$\begingroup\$ What about thermally linking two with a drop of epoxy or silicone? \$\endgroup\$ – rdtsc Oct 13 '15 at 17:10
  • \$\begingroup\$ @Andyaka Your point is quite valid, switching characteristics and recovery losses will make everything worse. I started off considering using this as the output of a SEPIC converter, decided not to, but was intrigued by the question anyway. Since I'd done the math, I decided to share it. \$\endgroup\$ – Stephen Collings Oct 13 '15 at 17:20
  • \$\begingroup\$ @rdtsc I'm sure that would help, by some amount that is very difficult to quantify. \$\endgroup\$ – Stephen Collings Oct 13 '15 at 17:21
  • \$\begingroup\$ Interesting, here is a small twin-schottky heatsink. \$\endgroup\$ – rdtsc Oct 13 '15 at 18:42

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