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I'm trying to rectify an AC current at 220V. Many written tutorials don't mention any transformer before the diode bridge and many video tutorials use transformers that provide an output of 6V or more.

In my understanding, diodes can withstand only very small voltages in forward bias since current is an exponential function of voltage. With only two diodes working at the same time during each half cycle, every diode will get no less than 3V in case of a 6V transformer. If no transformer is present, each diode will have 110V across it.

An electrician I know stated that many power adapters (such used by laptops) use a transformer after the bridge, which would imply that the diode bridge is connected to the mains ! He also added that some diodes can withstand 1000V, like the 1N4007 diode (I even read that on some websites) but when reading the datasheet for the 1N4007 diode, 1000V is the maximum reverse voltage, not the forward one, and during rectification, the diode needs to work both ways, so it will hold in reverse bias but it would normally burn in forward bias.

Am I missing something or understanding something wrongly or are all the tutorials I came across wrong? And also how much voltage can a rectifier diode withstand in forward bias, isn't it very close to 0.7V?

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  • \$\begingroup\$ "I'm trying to rectify an AC current at 220V." - Why? What are you trying to achieve? Note that mains electricity is dangerous, and I get the impression you don't have the experience to work with it safely. Please be careful. \$\endgroup\$
    – marcelm
    Jan 19 at 21:24

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Edit: As Andy pointed out, I didnt quite pick up on the fact that you are trying to feed the primary of the transformer with DC. This will in fact be very damaging to all components involved, and potentially hazardous as well. please do not do this! You will blow something up, load or no load! The only reason power supplies can even operate like this, is because it is rapidly switching on/off the incoming HV DC power towards the tranformer in order to simulate some kind of AC. A diode bridge straight into a transformer is no good. Disregard everything I said in the comments about the transformer, as this is not applicable to your situation. The answer @Andy aka provided goes into more detail about why some power supplies can be made to work while having a rectiefied input. My apologies!

Your reasoning is almost on point. Diodes can happily be connected to mains voltage depending on how you use them. With power supplies this is indeed very common as you pointed out. If you had 1000V of forward voltage on a diode, that would indeed be strange for the scenario's you are describing. The forward voltage of a diode however isn't determined by just the voltage on one if its 'legs', its measured across the device. And this voltage is developed based on the current that is flowing trough it. enter image description here https://www.farnell.com/datasheets/639187.pdf

This picture is directly taken out of the 1N4007 datasheet. Perhaps this will make it easier to understand the forward current and forward voltage relationship. The remainder of the supply (mains?) voltage is dropped by the thing the rectifier is powering.

As other people are pointing out as well, the only way to develop such a high voltage with a forward biased diode (conducting diode), is to force the current/voltage trough/across it by shorting it, bypassing the load or essentially having the diode itself as the only load. If you connect a diode across mains like that it will fail very quickly and violently.

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  • \$\begingroup\$ My confusion is more about the last statement you added. If there is a transformer after the diode bridge (like the technician said), and in the case when no load is connected to the circuit (like any charger not connected to a device), won't the primary coil of the transformer close the circuit and cause a high current through the diodes ? \$\endgroup\$ Jan 18 at 17:20
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    \$\begingroup\$ If there is no load drawn by the circuit, the primary of the transformer won't be drawing any current from the mains either apart from some losses. With no load on the secondary, the primary becomes just a huge inductor \$\endgroup\$
    – Thijs
    Jan 18 at 17:25
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    \$\begingroup\$ The current that is drawn by the primary of a transformer when there is nothing connected on the secondary (transformer output) is only a few percent of the normal maximum operating current of the transformer. With the secondary opened, a back force will be generated opposing any excessive current flow of the mains. The primary of the transformer is not just a bit of wire, that would be a problem :) This is a good read: electronics-tutorials.ws/transformer/transformer-loading.html \$\endgroup\$
    – Thijs
    Jan 18 at 17:32
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    \$\begingroup\$ If the transformer would actually 'close the circuit' (like a short between the primary winding or something?) yes that would be a huge problem and if the only thing catching fire are the diodes you are lucky. Luckily transformers don't do this when you remove the load :) \$\endgroup\$
    – Thijs
    Jan 18 at 17:35
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    \$\begingroup\$ Thank you, but I'm not planning to use a transformer after the diode bridge anyway. I just wanted to know if my understanding of diodes is correct and that they require a load to use the remaining voltage or would burn. \$\endgroup\$ Jan 18 at 22:08
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The diodes in the diode bridge only experience large forward bias if they are short circuited without a load.

Compare the two circuits below:

schematic

simulate this circuit – Schematic created using CircuitLab

In the first, the forward-biased diodes only have 0.7 V of forward bias on them, and most of the voltage is across the load where it can do useful work. In the second circuit, the full AC voltage is applied (in forward bias) across either D8/D6, or D5/D7 depending on whether we're in the positive or negative half-cycle.

Here's a link to a simulation. As you can see in the third graph at the bottom, the voltage across the diode never goes above 0.7 V or so in forward bias.

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  • \$\begingroup\$ It is a bit clearer. But if there is a transformer after the diode bridge (like the technician said), and in the case when no load is connected to the circuit, (like any charger not connected to a device), won't the primary coil of the transformer close the circuit and cause a high current through the diodes ? \$\endgroup\$ Jan 18 at 17:18
  • \$\begingroup\$ @IlyesFerchiou No, the primary coil does not close the circuit. A transformer where the secondary is drawing very little current will also draw very little current on the primary. \$\endgroup\$
    – nanofarad
    Jan 18 at 18:07
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    \$\begingroup\$ @IlyesFerchiou yes, a transformer connected to DC acts as a short circuit and your diode bridge will probably break. \$\endgroup\$
    – user253751
    Jan 20 at 2:26
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Am I missing something or understanding something wrongly or are all the tutorials I came across wrong ? And also how much voltage can a rectifier diode withstand in forward bias, isn't it very close to 0.7V ?

You are missing the fact that the output of a bridge rectifier is connected to a load. That load defines the current that flows through the diodes in the bridge rectifier. The diodes are chosen so that they can handle the peak load current. OK they may drop about a volt for each diode but, that is normal and clearly shouldn't be a cause for concern.

With only 2 diodes working at the same time during each half cycle, every diode will get no less than 3V in case of a 6V transformer. If no transformer is present, each diode will have 110V across it.

No, that is incorrect. If the incoming peak voltage is 8.5 volts (as from a 6 volt RMS transformer) then 1.4 volts will be dropped across two diodes leaving about 7.1 volts peak volt across the load.

If the supply is 110 volts RMS then the peak voltage will be about 155.6 volts and the peak voltage across the load (after rectification by the bridge) will be about 154.2 volts.

In all cases, about 1.4 volts will be nominally dropped across two diodes in the bridge rectifier.

But if there is a transformer after the diode bridge (like the technician said), and in the case when no load is connected to the circuit, (like any charger not connected to a device), won't the primary coil of the transformer close the circuit and cause a high current through the diodes?

I see your misunderstanding and please correct me if I'm wrong. The transformer is not connected directly to the output of the bridge rectifier; it connects to a switching circuit that converts the DC bridge output voltage to a high frequency AC voltage suitable for driving the transformer primary: -

enter image description here

Picture (red-lined by me) from here.

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  • \$\begingroup\$ It is a bit clearer. But if there is a transformer after the diode bridge (like the technician said), and in the case when no load is connected to the circuit, (like any charger not connected to a device), won't the primary coil of the transformer close the circuit and cause a high current through the diodes ? \$\endgroup\$ Jan 18 at 17:18
  • \$\begingroup\$ Aha I see your misunderstanding - correct me if I'm wrong @IlyesFerchiou - the transformer is not connected directly to the output of the bridge rectifier - it connects to a switching circuit that converts the DC bridge output to a high frequency AC voltage suitable for driving the transformer primary. \$\endgroup\$
    – Andy aka
    Jan 18 at 17:21
  • \$\begingroup\$ I didn't see any tutorials about a transformer after the diode bridge, it is just the technician who told me that they exist. The existence of a switch makes perfect sense to me and goes with my understanding of how everything in such a circuit works. Just to make sure one last time that my understanding is correct : If the load after the bridge rectifier is not capable of dropping the remaining voltage of the input, than the diode bridge will definitely burn, right ? \$\endgroup\$ Jan 18 at 17:39
  • \$\begingroup\$ Certainly they will burn unless there is some current limit circuit upstream of the bridge rectifier on the AC wiring side @IlyesFerchiou \$\endgroup\$
    – Andy aka
    Jan 18 at 17:46
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    \$\begingroup\$ @IlyesFerchiou you are being advised incorrect information by Thijs. Please take this into account. He is not recognizing the scenario you propose with a transformer after the bridge rectifier. \$\endgroup\$
    – Andy aka
    Jan 18 at 18:33
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This is not a full answer because you already have several of those. But...

In my understanding, diodes can withstand only very small voltages in forward bias since current is an exponential function of voltage.

..."withstand...forward voltage" is not really the right way of thinking here. A forward biased diode has a very low impedance, so in most practical cases, it makes more sense to think of the forward voltage as a function of the current, and not the other way around. You probably should be asking how much current the diode can withstand.

On the other hand, if you are planning to build a bridge to supply power to a circuit that has even lower impedance than the diode (i.e., to a dead short), then yeah, ask how much voltage it can withstand. (for a typical, silicon rectifier diode, the answer will be slightly more than 0.7V.)


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Yes, diode bridges can be connected to "high" voltages. This is actually common.

The 1N4007 is rated for 1000V maximum reverse voltage, yes. This is the non-conducting state, so essentially -1000V is across them with almost zero current flow.

And yes, when conducting, the drop across them is only a few volts.

So what is the misunderstanding?

If a 1N4007 bridge is placed after a 6VAC transformer, then it will rectify the AC to a DC peak of 1.414*6VAC = 8.49VDC pulsating, minus the voltage drop of a diode. 1.414 is an approximation for the \$\sqrt{2}\$.

If a 1N4007 bridge is placed after 220VAC mains, then 1.414*220VAC = 311V. The 1N4007's 1000V rating can easily support 220VAC. As long as the current drawn through the diodes does not reach really high levels, this works fine.

Most "switching" power supplies do work this way, to convert the incoming AC power into a higher-voltage DC. This DC is then "chopped", at a very high frequency, into a ferrite transformer. Ferrite can be used at high frequencies, and allows that transformer to be much smaller than it would otherwise be to operate on 50/60Hz AC directly.

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