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I am reviewing the power stage of a board which is to be connected to a wide range of cars and motorbikes, with internal combustion engines, old, new, 2 stroke, 4 stroke, etc.
Current schematic looks like this:

schematic

simulate this circuit – Schematic created using CircuitLab

D1 serves as reverse polarity protection, D2 is a TVS, and F1 limits overcurrent if the board breaks and gets shorted. Power consumption is low, peak current is around 1.1 A, and considering D1 voltage drop of about 0.5 V, power losses due to D1 are in the range of 0.6 W. (Note: the input capacitors for the SMPS are not shown)

Power consumption of the next revision is set to increase, with peak current of about 2 A, bringing losses up to 1 W, and heat to match: my suggestion is to swap D1 with a P-MOS, to make use of a more efficient reverse polarity protection method.

Mr. R (as in Retired electronics designer) made the following observation: placement of D1 is wrong, it should be placed before the caps, because right now the caps are just parallel with the whole vehicle; we should move the caps after D1 so they will serve as current tanks for our board only. We should forgo the P-MOS solution because it would conduct both ways and not solve this problem.

Assuming we can draw infinite energy from the vehicle, what should I take into account when choosing a solution over the other?

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4 Answers 4

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D1 serves as reverse polarity protection, D2 is a TVS, and F1 limits overcurrent if the board breaks and gets shorted.

D2 will short the supply and blow the fuse if reverse voltage is applied hence, no need for D1 hence, heat dissipation problem solved and, diode forward volt-drop problem solved.

Assuming we can draw infinite energy from the vehicle, what should I take into account when choosing a solution over the other?

I'd be concerned that the TVS is not adequately rated for well-known automotive surges like load dumps. That concern also applies to whatever you connect to the output and the voltage ratings of the capacitors in your image.

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    \$\begingroup\$ Right... so perhaps i should also move D2 after D1, so no current and no blown fuse? \$\endgroup\$
    – Jack
    Feb 29 at 10:46
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    \$\begingroup\$ @Jack or, leave it as it is and use a resettable fuse like a polyfuse. Like these: ttieurope.com/content/dam/tti-europe/manufacturers/sgx/… maybe? \$\endgroup\$
    – Andy aka
    Feb 29 at 11:11
  • \$\begingroup\$ These are probably better ones: 4donline.ihs.com/images/VipMasterIC/IC/LFSI/LFSI-S-A0007909423/… \$\endgroup\$
    – Andy aka
    Feb 29 at 11:19
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    \$\begingroup\$ You need to look up automotive surge requirements. The automotive environment is one of the most challenging to design for. If you are using a resettable fuse what was your first comment all about? @Jack ||| if you can choose to use a diode or a MOSFET you can also choose to use neither. \$\endgroup\$
    – Andy aka
    Feb 29 at 11:39
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    \$\begingroup\$ @jpa - Who are you addressing? It always helps to use @replies (see details here) when multiple people have commented on a post already. This avoids confusion and ensures a notification goes to that specific user to alert them of your comment, so they can reply (if they want to). \$\endgroup\$
    – SamGibson
    Mar 1 at 10:27
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The most trivial reverse voltage protection implementations using a MOSFET requires just one MOSFET and nothing else, replacing the diode altogether:

schematic

simulate this circuit – Schematic created using CircuitLab

The body diode won't conduct the wrong way, so I disagree with Mr. R here. This is \$V_{OUT}\$ in response to a sweep of source voltage going from -20V to +20V:

enter image description here

Considering that your only other concern is power dissipated in the block/pass element, this solution wins hands-down. The right MOSFET could dissipate less than 100mW with a 2A load current. Jelly-bean IRF9xxx models would dissipate well under 1W.

A slightly better solution would protect the gate from horrible spikes on the supply:

schematic

simulate this circuit

I agree with Mr. R regarding capacitor placement, but supply decoupling closer to the downstream electronics is of far greater importance, and will naturally have to come after the transistor.

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The main tradeoff that I see is cost vs dissipated power, plus something about robustness of the design. Usually transistors are more expensive than diodes, but once they will turn on they dissipate so much less. They are more delicate though.

Keep in mind that just a PMOS will likely not cut it, you need some circuitry to protect its gate since the battery voltage on a car can show huge spikes; you need to make sure your TVS can effectively protect the FET.

I agree that caps should be moved after the reverse polarity protection - to be honest I have no idea why you have four modules with those values; unless those are placeholders of course. There is little to no reason to have 100 pF or even 100 nF in that stage of the circuit, this is a bit fishy to me - but not detrimental, just a bit of added cost.

I am not sure as to why your colleague said the PMOS conducts both ways, as this is clearly not true.

Finally, I am unsure why you need this at all. If this is to be plugged in the "cigarette lighter" plug, then polarity is guaranteed; if this is to be plugged somehow with screw terminals or whatever, odds are you have a metal enclosure that you can refer to the chassis and provide a screw terminal for the positive only, which makes polarity reverse impossible.

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    \$\begingroup\$ I said P-MOS for short, but would actually include Zener to clamp gate voltage and all bells and whistles. When P-MOS is polarized it's just an Rds(on), it does not limit flow of current to a single way, so if the supply drops, current could flow from my capacitors to the rest of the vehicle. This is needed because sometimes people would call at 3 a.m. asking "which is plus, red or black?" but sometimes they don't and just connect backwards :') \$\endgroup\$
    – Jack
    Feb 29 at 10:36
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schematic

simulate this circuit – Schematic created using CircuitLab

Mr R is right that the caps should be placed after D1. Here I positioned D1 so that if reverse current appears, it will bypass the SMPS and other attached devices. Eventually blowing the fuse if this current persists. This set up won;t consume anything.

P and N MOSFETS conduct both way because of their internal reverse diode. This internal diode is there... to protect the FET against reverse current. So you better get just the diode instead.

100uF will offer more stability than 10uF. Other caps may not be necessary except for the 100pF one who is there for a reason, I imagine. Intermediary caps are not needed. Other 10uF or 1uF caps should be placed after the buck down regulator, on PCB or right before the device if there is none. These devices and the SMPS should have their own cap already if they were properly designed.

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  • \$\begingroup\$ I don't think this schematic addresses the point of Mr. R, about the energy flowing back to the power source in case it drops due to a temporary fluctuation; also, i think the body diode is present because there is a P-N junction intrinsic in the construction of the MOS itself \$\endgroup\$
    – Jack
    Feb 29 at 16:26
  • \$\begingroup\$ No, there is really a diode added to the MOSFET inside the chip. Yes, it doesn't address the problem of current flowing back to the source. For that, you will have to put the diode in series, but as said, it will consume energy. It may be necessary if these drops in power are long (several seconds). If they are short (just a few millisecond), the 100uF will address this problem. \$\endgroup\$
    – Fredled
    Feb 29 at 18:01
  • \$\begingroup\$ The buck down regulator will also add latency in case of drop in power. It will auto-regulate immediately according to the fluctuations. Capacitors after the regulator are important because they will maintain voltage the time the regulator reacts (a millisecond or so) or if for a very short time it can't adjust enough. You can use 1000uF before and 100uF after the regulator. But I think that 100uF and 10uF should be enough. You see. \$\endgroup\$
    – Fredled
    Feb 29 at 18:05
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    \$\begingroup\$ @Fredled: While some MOSFETs might include a diode separate from the PN junction, the body diode of a typical MOSFET exists because a MOSFET is inherently a four-terminal device which includes diodes between the source and substrate, and between the drain and substrate. Shorting out one of those diodes means that the top surface of the silicon only needs two different non-substrate connections rather than three. \$\endgroup\$
    – supercat
    Feb 29 at 18:12

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