I am using a BAT54A barrier diode, am using this diode because i need this component on another part of the circuit and i dont feel like using another dedicated component.

So the BAT54A is a dual Schottky and i need to use 1 as a barrier diode.

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I feel that it would be kind of a waste not to use the other diode, and tried to come up with a good reason to do so. Two i can come up with is an increased max forward current and a lower voltage drop. increased max forward current is not much of a gain because 1 diode is enough for my application, 2 would not be much of a gain.

A lower voltage drop though is a much more compeling. I think paralleling diodes would decrease the forward voltage on the basis of when two diode paralleled in a configuration like above they will share current and a lower current accross the diode would result in lower voltage drop. Am i correct? Are there any demerits in what im about to do ? i guess thermal runaway would be a problem, but i wont even pass a current aobve the max rating of 1 diode


3 Answers 3


Yes, it will decrease the forward voltage drop somewhat. It will also double the reverse leakage.

If we use 150mA as the forward current, and do a simulation the typical difference is 0.698V for one diode and 0.515 V for two in parallel.

At 25V and 100°C Tj the typical reverse leakage is about 55uA for one diode, and double that, 110uA for two.

Of course self-heating will be less- in the case of 150mA the thermal resistance to ambient is stated as 430K/W when mounted in the specified way- so if the single diode has 150mA the dissipation is 0.105W so 45°C rise, so if the Tj is 100°C then the ambient must be 55°C. In the dual diode situation, the temperature rise is going to be only 33°C, so the Tj will be 88°C and the actual reverse leakage will be closer to that of a single diode. It will only be double when self heating is negligible.


Take a look at the forward current-voltage characteristic of the BAT54:

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Since it's a dual diode, it should share current fairly well when you put both diodes in parallel.

However, there won't be a huge difference in forward drop due to the exponential nature of the curve.

One diode with 10 mA at 25 C looks like a forward drop of about 0.32 V. Two diodes each carrying 5 mA might drop that down to 0.31V. If your currents are higher the difference might be bigger, but it won't halve the voltage drop.

Another reason to parallel the diodes is to drop the junction temperature and increase reliability.


The reduction in voltage drop depends on If and ESR.

Effective Rs or ESR is also a function of the device Ptot = 230 mW @ 25'C

From my experience, you won't find this in textbooks;

\$ESR <= 1/P_{tot}\$

for ESR @ If continuous = 200 mA, I expect ESR<= 5 Ohms but no less than 50% of this.

Taking the plot from the datasheet and digitizing it with pixel resolution into a spreadsheet, we get ESR=3.8 Ohms @ 200 mA but it rises exponentially from the knee of the curve;

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If we also take into account the voltage drop at higher currents due to the negative temperature coefficient (Tempco or NTC) of a diode. This makes the ESR even more effectively lower at a higher temps. near rated current. Thus the current sharing is more for reliability or thermal margin improvement.

You may find that a higher current silicon diode can achieve a lower drop for the same cost as a power Schottky diode due to the ESR [Ω] and $ cost differences in making Schottky power diodes when you start using xx Amp rated diodes.

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Schottky Diodes are known to have higher leakage current.

It is important to consider all the dynamic characteristics of a diode datasheet in order to model your required performance and choose the best match with an adequate Ptot rating for the ESR you require. This may prompt you to choose a much higher current rating than required, but then a lower voltage drop will be expected.


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