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Looking at this 1N4148 datasheet, it seems that the forward continuous current is 0.3A, thus 300 mA. So if I add two in series, I will get 600 mA.

I want to use diodes to reduce a 5V 1.8W fan speed. The current through the fan is P = V * I <=> I = P / V = 1.8 / 5 = 360 mA.

Can I use two 1N4148 diodes in series, resulting in a voltage of 5 - 0.7 * 2 = 3.6 V (having thus about 72% fan speed), because 360 mA < 600 mA?

(note: I know a 1N4001 is better, having a 1A limit, but I don't have them yet).

Or did I make a mistake in my calculation or is there another effect why two 1N4148's cannot be used?

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    \$\begingroup\$ The diodes are in series so have the same current through them, it's not split in two. Draw the circuit and see where the current loop is... Also, your fan may be 1.8 W at 5 V, but may well be different at 3V6. Better to use PWM anyway. \$\endgroup\$
    – awjlogan
    Oct 4, 2018 at 14:38
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    \$\begingroup\$ How do you feel about a 555? \$\endgroup\$
    – awjlogan
    Oct 4, 2018 at 14:46
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    \$\begingroup\$ Nah, use a 741 to drive the fan \$\endgroup\$
    – Finbarr
    Oct 4, 2018 at 14:47
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    \$\begingroup\$ Fans use pretty efficient SMPS to commutate the phase current so their average rate VI load characteristics look like a variable resistor down towards stall speed only when SPINNING. So choose your % RPM above start Voltage or to decide on then choose V drop with power diodes... and by every means possible leave an optimal gap from fixed grill or heatfins to moving blade to reduce noise and then seal the gaps so air does not leak to keep it quieter from Eddy Current winds. Or the if want a thermistor to FET controlled fan speed, make one. With some BJT in between for gain and offset bias. \$\endgroup\$ Oct 4, 2018 at 15:56
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    \$\begingroup\$ You can gang a bunch of 1N4148’s together in xSyP arrays or get the right part or use BJT super bias \$\endgroup\$ Oct 4, 2018 at 16:16

3 Answers 3

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The 300 mA rating is in the Absolute Maximum Ratings table - you don't normally want to go near those ratings.

The 1N4148 is intended as a small signal diode. If you look at the Electrical Characteristics table, you will see that most specifications are given with a 10 mA test current, so you should only use a 1N4148 with currents in that range.

As others have said, for diodes are in series, EACH diode carries the full current.

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    \$\begingroup\$ +1 for mentioning the problem with designing using the 'Absolute Maximum Ratings' table. An important detail novice's frequently misunderstand. \$\endgroup\$ Oct 4, 2018 at 15:27
  • \$\begingroup\$ Yes indeed, I also was not so aware of this. For the time being, I think I will find another diode (or just use full power if that's it), and in future project I know what to use better, thanks. \$\endgroup\$ Oct 4, 2018 at 15:39
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Components in series have the same current through them — it doesn't divide. All of the current goes through all of the devices.

If one diode has a limit of 300 mA, then two of them in series still have a limit of 300 mA.

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  • \$\begingroup\$ Ok clear, than I need to check if I can user a better suitable diode. \$\endgroup\$ Oct 4, 2018 at 14:43
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If you place two components in series, current flowing through first diode and second diode is equal. It means that through both of your diodes will flow 0.6 A (actually 0.5 A -> you have to use new 3.6 V if you want to compute current to fan) -> they are going to be burned. You should choose other diode with bigger forward current.

Also, when you want to compute current to fan, you should use 3.6 V instead of 5 V, because of voltage drops across these diodes.

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  • \$\begingroup\$ Btw, I could use maybe 3 or 4 diodes until I get under 300 mA right? \$\endgroup\$ Oct 4, 2018 at 14:45
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    \$\begingroup\$ Actually, if you look on that formula I = P/U, with lower voltage u get higher current. I would prefer using some kind of PWM modulation - 555 as seen above \$\endgroup\$
    – Tomomet
    Oct 4, 2018 at 14:52
  • \$\begingroup\$ Yes indeed, I only could use resistors, but they need to be of high power. \$\endgroup\$ Oct 4, 2018 at 14:55
  • \$\begingroup\$ Power loss is unavoidable in linea mode and 50% at V/2. In fact 40 yrs ago the Japanese would expect a hot R and Thermistor to be cooled by fan to detect Fan Fail and cut off fan power \$\endgroup\$ Oct 4, 2018 at 16:17

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