I am not an electronics expert. Hence little confused with the current rating of 1N4007 diode. I know it's rated for 1 Amp and 1000 V. But I want to use 4 of them in a bridge rectifier circuit which is supposed to rectify 200 Volt AC to DC. How much max current can I draw from this rectifier safely? I'll be glad, if someone can help me understand the calculation as well.


2 Answers 2


If those diodes were sold as a bridge, it would be rated at 2A average current (when rectifying AC, since only one pair conducts at a time).

However that's under some (fairly reasonable) operating conditions, and if you are outside those conditions you may need to derate the maximum current. If you want good reliability you may choose to derate the maximum current. See, for example the datasheet diode derating and conditions such as lead length and pad size:

Typically you should derate 20% or more for a capacitive load as well, since the RMS current will be higher.

enter image description here

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    \$\begingroup\$ @TonyM The average current per series pair is 1A, 50% duty cycle. Total average current is 2A. Of course if it's possible to have DC input then you can't assume 50% duty cycle. \$\endgroup\$ Commented Dec 16, 2021 at 23:51
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    \$\begingroup\$ I thought I was probably falling into an obvious pitfall when I asked :-) But the Vishay datasheet (and the Diodes Inc datasheet) specifies IF(AV) as 1 A, defined as Maximum average forward rectified current, 0.375" (9.5 mm) lead length at TA = 75 °C. I'd interpret that as giving the average: 1 A, not 2 A. In other words, haven't you taken their figure with allowance for AC average then subjected it to another allowance for AC average? \$\endgroup\$
    – TonyM
    Commented Dec 16, 2021 at 23:58
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    \$\begingroup\$ @TonyM "average forward rectified current" is obviously half-wave and is 1A. The two diodes in series have the same current, so the rating is the same (thermal considerations aside). The second pair also has average 1A so the total is the sum of the two. Maybe worth pointing out that the peak current in each diode will be much higher than 1A, about 3.1A if I did the math right. Some of the comments in this question touch on the issue. \$\endgroup\$ Commented Dec 17, 2021 at 0:08
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    \$\begingroup\$ I can see what you're driving at and I'd first looked at the typical diode current: short spikes, once per cycle per diode, topping off the reservoir capacitor across the bridge. It's the exact meaning of the terminology that I'm seeing differently to you. (Incidentally, I'll delete this comment chain later, it'll not be needed or I can summarise it better.) \$\endgroup\$
    – TonyM
    Commented Dec 17, 2021 at 0:16
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    \$\begingroup\$ This answer ignores the major design consequences at this voltage. (-1) for inrush current , ripple voltage vs RMS rating of the diode. \$\endgroup\$ Commented Dec 17, 2021 at 14:04

I want to use 4 of them in a bridge rectifier circuit which is supposed to rectify 200 Volt AC to DC. How much max current can I draw from this rectifier safely?

This is a common beginner goal but let me explain why it is not a good one.

  1. Diodes are really sturdy dynamic nonlinear switches that can rectify from microamps to 30 Amps for 1 half cycle at 60Hz absolute MAXIMUM. Yet voltage is even more dynamic and when the bulk capacitor starts at 0V and goes up to sqrt(2)=1.414 times the RMS AC voltage the diode and capacitor current is only limited by the loop resistance. In theory, the grid is 0 and the caps are 0 while the diode depends on bulk size and thus peak current becomes V/R=I which can easily burst both diode and/or capacitor. So chokes and resistance must be added for high voltage to limit the current. This causes a few cost penalties.
    1. the power current limiting resistor must waste about 10% of the load power and ends up costing a lot more than a dozen 1N400x diodes and is still boiling hot.
    1. the capacitor also has foil-dielectric interface resistance called effective series resistance (ESR) and must also dissipate heat except it is unavoidably thermally and by design, electrically insulated. So the bulk capacitor for a low ESR Cap at high voltage that can supply the 1A RMS rating of the diode in question will cost again, more than a dozen of these diodes.

This also means those >>$6 capacitors are bulky will fail faster than the diodes due to Arrhenius Law, with an MTBF that drops 50% for every 10% rise and most are rated at only 1500 hr at 85'C. You can do the math. It's just not cost-effective to use a 10 cent 1 amp diode in volume for this application.

You can of course use less than the 1 Amp rms rating to lessen the thermal and cost issues by reducing the load current, and cap size by a factor of 5 with a 1 kOhm load.


Safety, thermal, cost-effectiveness and performance all have huge trade-offs that go up-front in any "design spec" BEFORE you think of ANY parts. Learn to define all requirements early for any design, even for mechanical students.

This shows a plot during a zero-crossing powering (best case). I expect anyone to ask questions, if you want to understand further.

enter image description here

This shows a 41 amp surge current that exceeds the rating but reduces to a chosen load to illustrate why 1A rms rectified current at 200Vac.rms is a bad design even for a bulk cap with a 1A rms ripple current rating in the high-cost range with lower ESR.

It just isn't worth it to design a power supply this way for this voltage unless you only need something < 0.2 A.

A switched-mode supply is more effective but this an extreme case with high voltage. BTW, 1N4007 are obsolete by some OEM's and stocking distributors but 1N4006 is adequate at 800 V piv rating, although some lightning protection should be considered in your spec.

Depending on your surge current requirements which may limit the use, it is basically a 1A diode worst case at a 70'C local ambient but the junction will be at 150'C . The case will be cooler but your fingers will burn above 55'C and prudent engineering is to derate the max power on all parts by 50% (unless thermally cooled by design) to prevent such burning hot parts which not only protects other parts from getting too hot but reduces the failure rate. (50% every 10'C rise from Arrhenius effects). Power is P=V*I and Vf may start at 0.7V and rise to 1V at 1A = 1W and a 60 to 80 'C junction rise per 1 watt.

If you need more current get a bigger diode. They can be operated in parallel, only if they are thermally shared to some heat sink. But if operating shared at high power and one diode is slightly lower internal resistance than the others, it starts to draw more current, and get hotter and eventually burns up in a condition called thermal runaway.

This is how they 1A rating is performed with thermal isolation, but shorter leads will conduct heat better to the metal heatsink.

It can handle high current pulses but rated 1A steady and 0.5A steady is what I suggest is wise. . When used in a bridge cap circuit, if the load draws 1A, the charging will supply 10A for 10% of the time if ripple is 10% and discharge 90% of the time. You might imagine why they are also derated for bridge cap circuits even slightly below 50% steady load from the cap as heat rises with the square of the current x duty cycle.

  • \$Pd= (Vf-0.7)^2/Ri + I*0.7V\$ is around 1 Watt at 1 amp but with an internal resistance, Ri around 0.3 Ohms so Pd dissipates sharply above 1A on pulses for the forward voltage Vf which has a low current threshold around 0.7V. https://www.onsemi.com/pdf/datasheet/1n4001-d.pdf enter image description here




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