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iScience
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why do two diodes in parallel not share the current load?
They may, but the current sharing depends on:iScience said:why do two diodes in parallel not share the current load?
Yes, 1 device tends to hog most of the current. Diodes have a highly non=linear I-V curve, i.e. Id = Is*exp((Vd/Vt) - 1). Important to know is that "Is" is a strong function of temperature, and varies among parts. Due to mismatches in parameters, as soon as 1 diode draws more current, it gets hotter. Then Is goes up with the higher temperature, which increases diode forward current Id further. Then temp goes higher, Is increases, etc. It is a slim chance that 2 paralleled diodes share current 50%/50%. Possible, but very unlikely.nasu said:They don't? The shares may not be equal but how could they not share? You mean the current goes all through only one of them?
cabraham said:Yes, 1 device tends to hog most of the current. Diodes have a highly non=linear I-V curve, i.e. Id = Is*exp((Vd/Vt) - 1). Important to know is that "Is" is a strong function of temperature, and varies among parts. Due to mismatches in parameters, as soon as 1 diode draws more current, it gets hotter. Then Is goes up with the higher temperature, which increases diode forward current Id further. Then temp goes higher, Is increases, etc. It is a slim chance that 2 paralleled diodes share current 50%/50%. Possible, but very unlikely.
If, however, each diode is connected to a small valued series resistor, then the diode-resistor pairs are connected in parallel, the resistors force current sharing. Does this help at all?
Claude
Yes.iScience said:Definitely, thanks
and since transistors are also nonlinear this accounts for their observed behavior as well?
Two diodes in parallel do not share the current load evenly because they have different forward voltage drops. The diode with the lower forward voltage drop will have a larger portion of the current flowing through it, leaving less current for the diode with the higher forward voltage drop.
The forward voltage drop is the voltage required for a diode to start conducting electricity. The diode with the lower forward voltage drop will have a higher conduction current, resulting in the majority of the current flowing through it. This means that the diode with the higher forward voltage drop will have a lower current flowing through it, resulting in an uneven distribution of the current load.
Yes, it is possible to balance the current load between two diodes in parallel by using a resistor in series with each diode. This will create a voltage drop across the resistor, allowing for a more equal distribution of the current load between the two diodes.
One disadvantage of having two diodes in parallel is that the diodes may not share the current load evenly, as discussed earlier. This can result in one diode receiving more current than it is rated for, potentially leading to overheating and damage. Additionally, having two diodes in parallel will increase the overall cost and complexity of the circuit.
Using more than two diodes in parallel can improve current sharing to some extent, but it is not a perfect solution. The more diodes that are used, the closer the current sharing will be to an even distribution. However, this also increases the cost and complexity of the circuit, and there may still be some variation in the current load between the diodes due to manufacturing tolerances.