# forward voltage



## mercrazy (Sep 21, 2017)

i must use a diode to protect a driver circuit from reverse polarity. i'm already using a schottky but i need less voltage drop across it. is there a way to do this?
thanks


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## DIWdiver (Sep 21, 2017)

Yes, you can use an FET for almost zero voltage loss. What's the max voltage and current? I'll pick a good one for you.

Google 'FET reverse voltage protection" and you'll see the circuits that are used. If your max voltage is less than 10-15V, all you need is the FET.


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## mercrazy (Sep 21, 2017)

it's a 12volt supply but it is used on an outboard engine charging system. some of these use a stator and regulators to produce DC. they can produce some vicious voltage spikes, even AC, when the regulators screw up. the battery catches most of these but some get through. i'm using a transient suppression diode but i still worry about damage. i'm driving 3 LEDs(3.6Vf each) in series using a pam2861 which also uses a ss14 in the driver circuit. my current circuit requires 11.8 volts with ss14 diodes to run at 400mA. i need to get required voltage down to 11.5 or less, the lower the better. if it makes a difference, i plan to drive 2 strings of 3 LEDs in parallel at combined current of 800mA. i haven't had any failures with the single strings using 2 driver chips and can't afford any either so i must have a foolproof circuit. i'm not opposed to changing driver circuit completely if you know a better pwm circuit. i looked at some of the mosfet info but most of it is over my head. any help is appreciated but it will have to be down to my level. by the way, you are my hero. thanks a million.


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## DIWdiver (Sep 22, 2017)

I would use the circuit shown in figure 3 here: http://sound.whsites.net/appnotes/an013.htm. I would avoid the Vishay brand transistors if possible, as DigiKey shows them to have substantially higher resistance than others. I didn't dig into the data sheets to see if that's real or just specsmanship.

Since N-channel FETs are generally lower resistance than P-channel ones (all else being the same), I'd use the circuit on the left if possible, but do NOT ground the electronics. What's shown there as ground (that little 3-line symbol) should all be tied together, but only gets grounded through the FET. If you can't use that for some reason (like something IS grounded), the one on the right will work too.

Those transistors are overkill for your application, but they should both be cheap and readily available. If you'd prefer some smaller and/or cheaper ones, let us know.

Make sure you put the protection diode between the LED circuits (including the reverse protection) and the rest of the world. If it's not wired correctly it may not do much good.


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## mercrazy (Sep 22, 2017)

would there be any voltage drop at all? i hate to do all this for .25 volt.
the fet costs over 50cents. guess i'll go back to stcs1 driver. it's not pwm but requires much less overhead voltage. i hate to limit supply voltage to lower range and i prefer pwm. many customers want to use 24volt supply and it says 20v spike will cause damage. stcs1 can't use 24v but it has over voltage/heat protection.
don't know what to do. thanks.


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## DIWdiver (Sep 23, 2017)

The zener diode in that circuit protects the FET from damaging voltages at the gate (that's the 20V limit). It's really extremely effective if it's kept wired close to the FET. The FETs in the diagrams can handle 100V across drain-source with no issues, so the transient protection should protect them quite well.

The voltage drop across the FET protection circuit is pretty easy to calculate. FET specs always specify the ON resistance, Rds-on. Multiply your load current times the Rds and you have the voltage drop. Rds varies with temperature, load current, and gate voltage. You are probably operating at moderate temps, low current, and around 12V on the gate. These all work in your favor, and you can probably count on the specified Rds-on, unless you use a FET that's quite a bit smaller than the ones in the diagram.

So what do the numbers say? Many of the IRF540 specs say Rds-on is 77 milliohms. You have 800 mA, so the drop is 61,500 microvolts, or 0.0615 volts.

If you want something lower this part has an Rds-on of 17 milliohms; TSM170N06CH C5G. It would work quite well in that circuit up to several amps. At 800 mA it would have a drop of 0.0136V

Also keep in mind that if you implement an external reverse-voltage protection, you can eliminate the diode at the VCC pin of the STCS1.

There are more options if you don't like these. Let us know what you are thinking and we'll try to help.


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## mercrazy (Sep 24, 2017)

.06 volts would be great. i would really like to stay with pwm chip. the linear driver STCS1 is ok with just a schottky. the pam2861 requires more overhead voltage. i wonder if my circuit is operating at peak efficiency. i'm using 100mH inductor with 4.7uF cap. LED vf is 3.6x3=10.8 + schottky vf=.3x2=.6 = total of 11.4. but it needs 11.8 min to run at full power. does the driver chip use .4v? one schottky is used at driver chip and other for reverse polarity protection. i guess you count the one at the chip for voltage drop??? most users have 12volt supply but some use 24v. even though 11.8v should be ok, many people have poor connections through fuse clips, wire connections, poor switches and relays. i want the extra overhead to compensate for this. like i said before, i'm far from being an EE. i just use the diagrams provided by the chip makers. i don't understand "put the protection diode between the LED circuits". should i put a resistor in series with the TVS diode so if they do hook up backwards it doesn't blow the tvs? if so, what value? i have the schematic drawn in ExpressSCH. i could email my circuit if you could check it. i use surface mount components and i need the fet in as small a package as possible. i must have protection from massive load dumps. i used a zener before that shorted from load dumps. cost me a lot of money. is the zener at fet to limit voltage to 12volts or less on gate? is it safe from load dumps? i like the relay idea but they cost too much. thanks.


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## DIWdiver (Sep 24, 2017)

It sounds like you have two schottky diodes in your circuit. One (let's call it A) is suggested by the PAM2961 datasheet between LX and Vin, the other (let's call it B) is added by you for reverse polarity protection.

Diode A is required for the circuit to operate properly. Its forward voltage does NOT affect the input voltage requirements. It DOES impact the circuit efficiency. It would be difficult to replace with better circuits.

Diode B is only for reverse protection. It's forward voltage DOES directly increase the input voltage requirement, and it DOES impact efficiency of the total circuit. Replacing it with a FET circuit for increased efficiency and lower input voltage requirement is simple.

The transient protection diode must survive anything the vehicle can throw at it, since it is the first line of defense against overvoltage.


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## mercrazy (Sep 25, 2017)

if diode A doesn't count, then chip requires even more overhead.

is this right?
gate- to supply voltage through resistor.
source- to ground side of electronics.
drain- to ground.

will these work?
https://www.fairchildsemi.com/datasheets/FD/FDN359AN.pdf
http://www.vishay.com/docs/72322/72322.pdf

should i ground the tvs diode through the fet or go straight to ground? i've always grounded through a schottky diode before and never had a problem. if i don't go through the fet and they hook up reversed, won't it blow the tvs? i'm using unidirectional tvs. if i used bidirectional, would it be safe to hook straight to ground?
thanks


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## DIWdiver (Sep 25, 2017)

You do have the connections correct, but you didn't list the zener diode between gate and source, cathode toward the gate. It should be between 10 and 18V. I'd probably go for 12V if I had to buy one. Since the max voltage between gate and source is typically 20V, you'll soon see why the zener is so important.

Many transients ring both positive and negative. The schottky would try to block the negative part of the transient. As it isn't designed for this, it may eventually fail. If it fails open, then you've lost your protection. I would use the bidirectional TVS, so it will survive the reverse voltage connection as well as the negative part of the transients. 

It sounds like you are making a circuit board, so the TVS goes at the edge of the board, right where power and ground come in. The traces that power everything else come right off the pads where the TVS is. Nothing else is grounded, except to the TVS. If not making a circuit board, the same concept applies - all power and ground connections go back to the TVS, and from the TVS it connects to the vehicle.

Now for the TVS voltage. You mentioned 'massive load dumps'. When I was designing a fuel pump controller to meet the requirements of a major manufacturer of construction equipment, their spec said load dump is up to 86V for 12V systems, 172V for 24V systems. Most modern 12V automotive alternators clamp their output voltage at 40-odd volts. A battery in good condition should keep it lower than that (worst case load dump is alternator maxed out and battery disconnected). Pick your poison. The TVS has to survive that. Transients contain very little energy, but load dumps can really pack in the Joules. You don't want to try to clamp one with a little surface mount diode.

Then the FET has to survive the max clamping voltage of the TVS. The LED driver that I sell for automotive applications has an 80V TVS and a 100V FET. Most of them go to off-roaders, so they'll certainly see lots of winch-related load dumps that will overmax the alternator. Nobody's complained about frying one, but they are all 12V systems. 

Also remember that the driver circuit needs to survive the load dump voltage. The reverse voltage protection and TVS are intended to let that voltage straight through. That's why I use a 100V FET. It's part of the driver; there is no reverse protection, and the TVS is unidirectional. The one I'm working on now has bidirectional TVS and reverse voltage protection.


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## mercrazy (Sep 26, 2017)

1) yes, i have the zener at the fet. i thought the zener voltage would have to match the maximum fet gate voltage or be a little lower? is that right?

yes i design my circuit boards but board space is limited due to heat sink housing i can't change. that's why i need smallest possible part, but i must have parts that won't fail. if they won't fit, i'lll have to do something else.
the biggest load dump from a 12volt system is from engine starter. smaller dumps come from small 1 amp pump motors. 24 volt dumps come from trolling motors which can pull 50amps. i haven't had a problem so far using a 26volt tvs that clamps at 40volts. i used this one because the driver chip is rated for up to 40volts. it's in a SMA package.
only problem i've ever had is when i used a zener to turn off the driver if voltage went to high. voltage spikes would blow the zener which shorted and turned the driver chip off permanently. this cutoff idea was designed by a very experienced EE so i didn't question it. this cost me $1000s so i won't take a chance again.

2)could you recommend a surface mount fet and zener? i don't understand all the specs for fets. what about one of these?
http://www.mouser.com/ds/2/427/sij470dp-348559.pdf
http://www.mouser.com/ds/2/427/si4058dy-1018942.pdf
[url]http://www.mouser.com/ds/2/427/sis892dn-1145258.pdf
[/URL]http://www.mouser.com/ds/2/427/sir870dp-1145240.pdf

3) if one of those will work, is 12volt zener ok?
thanks


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## DIWdiver (Sep 27, 2017)

If the driver can only handle 40V, then the TVS should clamp no higher than 40V. Then there's no reason to go much higher than 40V on the FET. Sure, a 100V fet will work, but you'll pay more and/or get a higher ON resistance. Any of the ones you posted would work, but wouldn't be my first choice.

In this application the fet won't dissipate notable power, you are only going to turn the FET on or off once in a while, and you have plenty of voltage to do so, so the things you need to focus on are max gate voltage, max drain-source voltage, and ON resitance. Oh, package and price.

The zener voltage should be high enough to make sure the gate is really on hard and low enough to protect the gate from overvoltage. Most of the FETs you are looking at have +/-20V max gate voltage, so I'd say no higher than 18V for the zener. Most of them are pretty well on at 10V. I'd aim for 12, maybe 15V.

Bummer about the fried zeners. How did that happen if they were protected by the TVS? Or weren't they protected?


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## mercrazy (Sep 27, 2017)

i didn't use a tvs at that time.
i already ordered some of the ones i listed. i'll see how they work.
is the 40volt for the drain/source breakdown rating?
thanks


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## DIWdiver (Sep 28, 2017)

Yes, Vds,br.​


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## mercrazy (Sep 28, 2017)

ordered 2 different boards and 1 screen from your pcb shop.
have you ever bought a screen there? do they come just as rolled up sheet?
thanks


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## DIWdiver (Sep 29, 2017)

Never bought one, but they should come flat. They're a PITA to use if they are warped or curved.


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## mercrazy (Oct 4, 2017)

well got a board built with mosfet. i intentionally left off the resistor to the gate just to make sure the fet was off. well, the light came on anyway. after checking the fet on the board, it had 8k continuity from source to drain with no power connection to gate. youtube video said there should be a high value resistor between gate and source. i didn't have a resistor but i do have a zener there. should i add a resistor too? i added the resistor with same result. i checked unused fets and they tested correct? could my board be shorted?
thanks


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## DIWdiver (Oct 5, 2017)

Unlike the base of a bipolar transistor, the gate of an FET is 'floating' which means it can hold a voltage with nothing attached to it, like a capacitor. Just handling it you could easily put several volts on the gate. You have to actively turn it off in order to make sure it's off.


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## mercrazy (Oct 5, 2017)

so it needs resistor connected to ground to drain gate? wouldn't it also need a diode in series with it so if they hooked up backwards it wouldn't charge the gate?
but, wouldn't this second resistor act like a voltage divider so the values would have to be set so gate voltage is high enough when input is high?
does it matter which way the fet is connected, source to load and drain to ground or vice versa?
most schematics i've seen online have source to ground. i have drain to ground.
i've powered up with 12volts both forward and reversed several times and it still works right with only .01 voltage drop across fet and hasn't seemed to damage the driver or LEDs. i guess i'm just lucky because i read 12 volts on source with polarity reversed. doesn't this mean it's conducting through drain? that's only place power is touching.
remember as you can tell by now, i'm an electronics idiot so please tell me what i'm not understanding.
thanks


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## DIWdiver (Oct 6, 2017)

In this application the second resistor is not needed. The one you already have will turn the FET off when you apply reverse input voltage. When you disconnect that resistor, there's nothing to turn the gate on or off. So you tested an incomplete circuit and asked why it didn't work.

I doubt very much that you had 12V on the source. That would indicate that the reverse protection did not work, and you'd most likely have fried the driver.


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