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You can find two such circuits using a diode that protects against reverse polarity, i.e. against wrong power connection.

I wonder what is the point of using the D2 diode, since the current flowing through it will damage it - at least that's what the simulation shows. Diode D1 will not allow current to flow through it if we do not exceed the maximum voltage value. Am I right? Please explain.

enter image description here

JYelton
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Jerzy Przezdziecki
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    I like D1 circuit if you can live with V1 losing one diode drop. D2 only works well if there is a fuse! In SPICE there probably is no limit to the current of V2. – ronsimpson May 01 '23 at 13:04
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    The answers here are pretty good. Both of these circuits have downsides. The D1 circuit is inefficient since you have to drive current through the diode when working. The D2 circuit needs a fuse (and even then, it will not self-recover). An alternative arrangement is to put a FET in the supply (P-ch) or return (N-ch) line. These will act as an ideal diode (like D1, but with lower voltage drop). The body diode is used until the voltage available at the load is high enough to turn on the FET. FET selection and bias circuit are dependent on expected voltages and currents. – Troutdog May 01 '23 at 14:06
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    @Troutdog if V_in is fixed, and you're using a 7805 as shown, it just becomes a matter of *where* the power is wasted - unless you run out of headroom for the linear regulator and make it misbehave. This could even help if the regulator is running hot! – Chris H May 02 '23 at 10:01
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    The solution proposed by @Troutdog, with a MOSFET, is usually the best. What is needed though, for general MOSFET gate safety, is however a resistor/Zener protection of the gate: both for inputs close to VgsMAX (that usually ranges from 20V down to 8-10V, depending on MOSFET type) and for nasty ESD/spikes. With that protection on you can additionally provide some more gate capacitance, and also limit inrush current, if your point of load causes that need. – LuC May 02 '23 at 11:53
  • Moreover, if your power line is noisy and/or connected manually/through an electromechanical switch, consider also having at the input a bidirectional TVS or a varistor - evaluate at least something like a SMA24CA onward, for the voltage/energy depending on your application – LuC May 02 '23 at 11:59
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    @ChrisH This is true. The diode is simple, which is a plus. The choice boils down to the circuit requirements. In this case, with a 1A load, a diode (D1) is a good solution assuming there is enough voltage headroom at the input. If, however, the load was say 4A, I would definitely explore using a FET there. Your point about distributing the thermal load is a good one. – Troutdog May 03 '23 at 15:20

3 Answers3

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Every circuit might need a different form of protection. In your particular circuit (a linear 5 volt regulator) D1 is the preferable choice because it also helps reduce the power dissipation in the linear regulator by dropping voltage. Clearly if you have a strong voltage source, the D2 implementation needs some type of fuse but, you wouldn't use a BAT54.

In different regulator circuits where the input voltage runs much closer to the drop-out voltage of the linear regulator, D1 would prove problematic and the D2 circuit (with a fuse) would be used.

Andy aka
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  • Thanks for clarification. I wonder if adding a 100k resistor in series with a D2 will make it more efficient? This will lower the possible current to 180uA and won't damage the D2? Or it will just work as a fuse as you have suggested? – Jerzy Przezdziecki May 01 '23 at 13:12
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    D2 protects the circuit by shorting out the supply, so no you can't put a resistor in series with it. Try it in your simulation. Remember that in reality the supplies will not supply infinite current into a short. – Drew May 01 '23 at 13:14
  • You can put a resistor in series with the whole circuit. That would waste quite a bit of power though and is only appropiate for low power applications. – Christian May 01 '23 at 21:39
  • @Christian you might be missing the point here; you would only use D2 if the supply to the regulator was close to its drop-out voltage hence, adding a resistor in series with the power input isn't going to be useful or desirable. – Andy aka May 02 '23 at 13:57
  • @Andyaka you're right, I was thinking about overvoltage protection with a zener diode – Christian May 02 '23 at 14:18
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There are many circuits (battery powered primarily), where it's not ideal to drop the input voltage by having a series diode. If you've got 2 AA's, that's about 2V when they're near dead. It's difficult to power a circuit from that, even without wasting a few hundred mV on the diode.

Circuit 2 is called a crowbar configuration. It shorts out the supply to protect the circuit in a reverse condition. In practice it would be used with either a fuse, a power supply with overcurrent protection, or batteries which cannot supply too much current. It's preferable in many cases because it's simple, cheap, and does not drop any voltage. The downside is that a reverse connection will destroy the batteries or maybe a blow a fuse.

Drew
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  • (+1) Specifically also because you mentioned what a *crowbar* protection circuit is. – LorenzoDonati4Ukraine-OnStrike May 02 '23 at 14:51
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    I'd add that sometimes that crowbar configuration is indeed used in battery-powered devices with no fuses *as a last ditch* protection. That is, it's there in case the user reverses the battery momentarily (think for example of an AA cell) in the battery holder before realizing it won't fit in. The diode is not meant to survive a sustained polarity reversal, but on average will keep the device safe for the most common blunder a user could do, allowing those 2-3 seconds it takes to realize "OOPS! I tried to put a cell in reverse". – LorenzoDonati4Ukraine-OnStrike May 02 '23 at 14:58
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From the use of BAT54 we could infer that less 200mA are required from the power supply. If you compare the datasheets for this diode and a fuse (see links below), you could see that a 200mA fuse would drop 1.7V while the diode would drop around 0.7V at this (maximum) current.

It depends on the actual current specs (including short bursts), but it is clear that the fuse could drop more voltage than the diode.

A fuse from this manufacturer (chosen randomly as an example) would have to be rated to at least 630mA to provide less voltage drop than this specific diode.

enter image description here

diode fuse

devnull
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