# inadequate overhead voltage?



## mercrazy (May 6, 2018)

i have 2 strings of 3 LEDs with switches to select some or all LEDs.
i must use diodes to accomplish this.
i want to swap LED1(2.2Vf) and LED4(3.3Vf) because of different colors.
i have them arranged originally like schematic because i'm limited to 12volt supply.
i tried this with schottkys at D3 and D5 but didn't have enough voltage.
they were connected like the Pfets are now and everything worked correctly with current arrangement but not if i swapped LED1 and 4.
can i replace these D3 and D5 with mosfets with no voltage drop?
how do i hook them up?
or is there a better way?
how do i post schematic? tried 2 ways and can't show either.
thanks


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## DIWdiver (May 7, 2018)

The way to post an image so it shows up in the post is to place it in a file sharing service like Dropbox, Picassa, Google Drive, etc, then paste a link to the image in the box that pops up when you click the "insert an image" button.  Remember the maximum image size of 800 x 800.

Alternately, you can just paste a link to the image in your text.


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## DIWdiver (May 8, 2018)

I've copied the picture you e-mailed me and posted it here so everyone else can see it.  I hope you don't mind.  If you do, I'll happily remove the post.









Discussion follows in another post.


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## mercrazy (May 8, 2018)

no problem, thanks for posting.
anxiously awaiting discussion.
thanks again.


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## DIWdiver (May 8, 2018)

It took me a while to figure out what I was looking at.

Talking for now only about string 1, the core of the circuit is C1, C2, D1, R1, L1, LEDs 1-3 and the 2861 (which it took me a while to figure out was a PAM2861 LED driver chip).  This core is a constant current LED driver circuit using buck-mode switching topology that allows high efficiency with large difference between input and output voltages.

D4, D6-8 are transient suppression, and Q3, Z2, and R4 are reverse polarity protection.  The other components appear to be there to turn LED1 on and off without changing LED2 and LED3.

Let's look at these step by step.  

The basic driver circuit looks fine.  As you haven't put values in, I can't say whether you've made good choices.  Maybe you haven't made any yet.  I'll assume you can pick proper values there.

Unless K1 and K2 are wired at some distance from the rest of the circuit, I would wire the anode of D4 to P1 and eliminate D6 and D8.  D4 and D7 could be combined in a single bidirectional TVS.

Most power mosfets can handle +/- 20V on the gate, so with a 12V supply, Z2 is probably unnecessary, especially if your transient suppression is good.  That said, it's not doing any harm, so you may want to leave it in.

Now about switching LED1 on and off.  Turning on Q1 would short LED1, thus effectively turning it off.  The driver will compensate for the lower voltage drop in the string.  This part of the design seems over-engineered to me.  Since K1 cannot be used for anything other than turning LED1 on and off, why not just put K1 across LED1 and be done?

The circuit as originally designed (not as shown here) could be useful if K1 were in fact a 3-position switch with an off position.  You could then use the single switch to allow 3 states: off; all on; LED2 and LED3 on.

As drawn here, Q4 cannot be turned on.  In order to turn on an N-channel FET, the gate needs to be driven positive with respect to the source.  This can't happen because the gate is tied to the lowest voltage in the circuit (ignoring drop across Q3, which should be very small).

A P-channel FET is turned on by driving the gate negative WRT the source.  I'm guessing in the original design Q4 was a schottky diode and Q1 was a Pfet.  In this case, when the circuit is powered through C2, the diode would be reverse biased and R3 would pull the gate of Q1 low, turning it on and shorting LED1.  If the circuit were powered through C3, the diode would be forward biased and the gate of Q1 would be pulled high, turning it off and allowing LED1 to light.

In the case of low overhead voltage, the voltage drop of the schottky would be detrimental, and a FET would nearly eliminate that effect.  Unfortunately, you cannot use an Nfet because you don't have any supply to drive the gate higher than the input voltage.  Using a Pfet for Q4 would be the right idea, but I can't see how to wire a Pfet to work here either.

This post seems long enough now.  If I'm on the right track, let me know and I'll share some ideas on how to improve the circuit.


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## mercrazy (May 8, 2018)

you are correct on all counts.
the TVS diodes will be bidirectional but couldn't find one is this software.
the switch does have center off.
K1 is main power switch with choice of LED1 on or off.
K2 controls string 2 with choice of OFF or ON with LED4 ON or OFF.
P1 is actually a wire that brings in power to switch.
the circuit works properly as is with schottkys in place of Q4 and Q5 but i run out of voltage.
now i understand driving the gate high on Nfets.
i thought it just had to see a 10 volt difference.
so Nfets has to drive gate high and Pfets has to be driven low?
u got any idea how to alleviate my problem?
thanks


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## DIWdiver (May 11, 2018)

Well done!

Having done some research, you e-mailed me this, with permission to post it:






This is pretty much where I was going.  I think this will work as shown, though you might have to increase Z1/Z3 from 12V to 15V.  With 12-13V input voltage, Z1/Z3 may leak enough to turn on Q6/Q7.  There are several things in this circuit that could produce unexpected results.

As I said before, I think with decent transient protection at P1, the zener diodes are all unnecessary, but not harmful.

The scheme I came up with is essentially the same thing, with a few less parts, more direct action, and less unknowns.  I've drawn only part of the circuit, but it should be pretty obvious how it fits in with your design.





With S1 in the upper position, all three transistors are off and power is applied directly to the current regulator core (R1 and R2 in your diagram, R4 in mine). In this state the circuit should work well with input voltage down to just above the Vf of L1+L2+L3.

With S1 in the lower position, all three transistors are on, and power is applied to the regulator core through Q1.  Q2 is shorting L1, keeping it off.  In this state the circuit should work well with input voltage down to just above the Vf of L2+L3.

I hope this is helpful.  If not, keep asking questions until it is.


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## mercrazy (May 12, 2018)

i do like the idea of fewer parts/unexpected results.
wouldn't some of the LEDs be on all the time unless i add a 3rd switch to cut 12V?
i don't have that option in my light housing.  i'm limited to 2 switches, SPDT, center OFF.
i want to turn on the following LED combinations with 2 switches:
2,3
1,2,3
2,3 -   5,6
1,2,3 -   5,6
2,3 -   4,5,6
1,2,3 -  4,5,6
and i want the first switch to kill all power to both strings.
is that possible with your circuit and 2 switches?
remember, i'm very electronically challenged.
it takes me 2 days to understand the simplest circuit.

for the zeners, i want to be able to power with 12 or 24 volts, so i guess they're necessary.
i'll be using a SMAJ26 bidirectional TVS diodes immediately across both rails and ground.
your comment about 12v zeners might leak enough to turn on transistors makes me think 15v zeners will leak at 24v.  is that right?
if so, maybe i/we have another problem to solve?
thanks


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## DIWdiver (May 12, 2018)

Leakage is the conduction of a small amount of current under conditions where ideally there would be no current. A 15V zener and 10K resistor with 12V across them would ideally not conduct any current.  In reality there will be a small leakage current.  With 24V, we would expect current to flow.  In fact, we would expect the zener to clamp at around 15V, with the rest, 9V, across the resistor.  9V across 10K would give about 0.9 mA.  We would not consider this leakage, as it is expected even with ideal components.

In your circuit, with K1 set at C2, as the supply voltage approaches the zener voltage, the Z1 will begin to conduct; leakage at first, then normal conduction.  This current will flow through R3.  When enough voltage develops across R3, Q6 will turn on when it is supposed to be off.

My circuit won't have this problem.  However, for supply voltage greater than the zener voltage, you will want to add a resistor between the collector of Q3 and the gates of Q1 and Q2.  As far as turning off, with the switch in the center position, all the transistors are off, and the only current that will flow is a small leakage through Q1.  This will not be enough to light any of the LEDs.

If you want to have the first switch kill all power, just connect the common of the second switch to the drain of Q1 (where it connects to R4).  With S1 in the center position, this node has no power and all the LEDs are off.


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## mercrazy (May 12, 2018)

DIWdiver said:


> In your circuit, with K1 set at C2, as the supply voltage approaches the zener voltage, the Z1 will begin to conduct; leakage at first, then normal conduction.  This current will flow through R3.  When enough voltage develops across R3, Q6 will turn on when it is supposed to be off.
> My circuit won't have this problem.  However, for supply voltage greater than the zener voltage, you will want to add a resistor between the collector of Q3 and the gates of Q1 and Q2.  As far as turning off, with the switch in the center position, all the transistors are off, and the only current that will flow is a small leakage through Q1.  This will not be enough to light any of the LEDs.
> If you want to have the first switch kill all power, just connect the common of the second switch to the drain of Q1 (where it connects to R4).  With S1 in the center position, this node has no power and all the LEDs are off.



as you've probably guessed by now, i've had other help to get me this  far. I always questioned how Z1 affects the circuit but they said it  wouldn't have any affect. They may have assumed 12 volt supply only. Is  24 volt supply the only time it would be a problem?
All this is still foggy to me but i plan to try and do another schematic your way.
In my circuit, the common of switch 2 is connected to the always hot lower rail of string1. I don't understand this switch 2 connection.
Is it possible to do all the different LED combinations with your circuit?
thanks


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## DIWdiver (May 12, 2018)

I agree, with a true 12V supply Z1 (and Z4) should never do anything.  If you were talking about a "12V" automotive bus it would be a disaster, because the actual voltage is typically around 13.8V while the vehicle is running.

Imagine just Z1 and R3 connected across 13.8V.  Z1 would clamp at around 12V, and the remaining 1.8 would be across R3.  This voltage also appears at R7, and is plenty to turn on Q6, whether you like it or not.  

I don't see a way to 'fix' this circuit for 24V.  With different choice of zeners, it could work up to 16V or so, but not much beyond that.  Above this it looses the ability to turn off LEDs 1 and 4.

My recommendation doesn't have this problem.  Connect your switches exactly as you have them in your circuit and you can get all the combinations you mentioned.  The switch connection that I recommended is exactly what you've done.  I just didn't notice that you had already done it.


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## mercrazy (May 12, 2018)

it could be a marine outboard type bus with as much as 14.5 volts so i'm forced to used 15V zeners.
some engines have stator based charging systems where the voltage regulators go bad and put AC or monster spikes on the line.
other things like engine starters have massive load dumps.
or some may want to use up to 26 volts that supply other noise generating devices such as motors.
with tvs diodes, will it be safe?
i've had problems with blown zeners before and i don't want to risk that.
so i guess i'm forced to use your design which i like better any way.
only thing i don't like is having to connect an additional wire direct from power source to circuit board around the switch.
this is a worry since the circuit board will be hot all the time.
any way around this?
please check my schematic for mistakes.
what value resistors should i use?
do Q1 and Q2 need zener protection?
someone said to use 60volt fets and i wouldn't need zeners?  is that right?
i could only find a 40Vgs but i guess that would work if his theory is correct?  would this one work?
i'm using TVS diodes that clamp at 42volts.
https://www.diodes.com/assets/Datasheets/ZVP4525E6.pdf
i think i found the problem with 40v fets.  they have higher ON resistance.  doesn't that defeat the purpose?  
thanks
https://www.dropbox.com/preview/2halfr9.PNG?role=personal


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## DIWdiver (May 14, 2018)

In that application I would use a big TVS and as low current a fuse as possible.  Then I would make sure that everything else would handle the max clamping voltage of the TVS.

I've seen blown zeners before too, but that's in applications where there is sufficient power available to overheat them.  The zeners in this design would not have that problem, and thus I would not consider them a reliability issue.

As I have said before, there is no reason to have "an additional wire direct from power source to circuit board around the switch." You can wire the switches exactly as you did in your circuits.

When someone says "use a 60V FET" they are referring to VDS​, not VGS​.  It's uncommon for the max VGS​ spec to be above 20V, and extremely rare for it to be above 30V.  You can usually clamp the gate voltage with a zener and protect it quite easily.  You usually cannot do this with VDS​, so you are forced to use higher voltage specs there.

Now lets talk about voltages in automotive and similar systems.  I'm aware of several things that can cause overvoltage events in these syetems:
1.  Transients.  Caused by fast switching of loads and parasitic inductances (including wires), these can reach to thousands of volts.  Fortunately there is very little energy in these, and in a properly designed system even a small TVS can protect against these.
2.  Double voltage jump.  Most tow trucks and other 'aid' vehicles are 24V systems.  When they give you a jump start, they can put 20-30V into your vehicle.
3.  Reverse voltage.  OOPS!  Anything more need to be said?
4.  Load dump.  An alternator is actually a current output device.  The voltage regulator tries to adjust the output current of the alternator to maintain a steady 13.8V or so on the battery.  When you have a high load like a starter, the battery voltage drops and the voltage regulator asks for lots of current to compensate.  When the load is removed, or 'dumped', the regulator asks for less to compensate.  But this takes some time, and with the low load and high output current, the system voltage rises.  Normally the 

I would normally choose the TVS first, so it would survive whatever I was going to throw at it.  The I would choose a FET with VDS​ at least a few volts higher than the max clamping voltage of the TVS.  Then I would protect the gate with a zener a few volts below it's max VGS​ rating.

Lastly, I know how you can post your images directly into the thread.  Click the "Insert Image" icon and paste your dropbox link into the dialog box that comes up.  Then edit the link to change the "www.dropbox.com" part of the link to "dl.dropboxusercontent.com".  Click OK and the image should appear directly in your posting!  Just make sure your image size is no more than 800x800 pixels, or the moderators will be on you to correct it.


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## mercrazy (May 14, 2018)

now i'm really confused.
i had figured out how it works with the extra hot wire but without it i don't understand.
when switch is up, Q4 is off, Q6 gate is NOT low so it's off so power can't pass through. is that right?
how does power get to driver chip?
same for string 2.
maybe i have it drawn wrong?

i need help selecting tvs.
this is what i've been using and haven't had a problem.
i chose 26 working volts so i could power from 24 volt supply.  it clamps at 42.1 which is a little above voltage max for my Pfet of 40v.
are these ok?
1. https://www.digikey.com/product-detail/en/bourns-inc/SMAJ26CA/SMAJ26CABCT-ND/2254170

this is Pfet.  should i choose 1 with higher Vds?
2. https://www.mouser.com/datasheet/2/115/DMP4065S-537770.pdf

is this fuse ok?  max current draw for light is 900mA.
3. https://www.digikey.com/product-detail/en/bel-fuse-inc/0ZCF0185FF2C/507-1747-2-ND/4156147

4. what value for R9 and R10?  are the other resistor values ok?

https://www.dropbox.com/s/l07grfwvp8e4p5s/2Hr9.PNG?dl=0


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## ssanasisredna (May 14, 2018)

mercrazy said:


> now i'm really confused.
> 
> i chose 26 working volts so i could power from 24 volt supply.  it clamps at 42.1 which is a little above voltage max for my Pfet of 40v.
> are these ok?
> ...



- SMAJ26CA can start clamping as low as 28.9V, which is low. 24V systems while charging can hit 30V in regular operation. I would go a bit higher.
- Go to a 60V FET, so you have some tolerance if designing for a 24V system.
- Fuse is rated too low of voltage, 33V. If the fuse ever pops, it will by definition be in an over-voltage situation and once it opens, it will likely exceed the rated voltage. It would be okay for 12V only. The fault current at 10msec (approx rated for SMAJ) is likely going to be too high to protect during a large voltage transient, so TVS may pop with 24V double battery.


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## mercrazy (May 14, 2018)

i'm forced to use tvs that clamps close to 40volts because of driver chip.
the 24volt system used here doesn't use alternator charging system.
these batteries are charged with remote charger when light is usually off.
i will take your advice on the FETs and Fuse.
thanks SS
i'm waiting on DIW because i'm confused on the circuit now.


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## ssanasisredna (May 15, 2018)

mercrazy said:


> i'm forced to use tvs that clamps close to 40volts because of driver chip.
> the 24volt system used here doesn't use alternator charging system.
> these batteries are charged with remote charger when light is usually off.
> i will take your advice on the FETs and Fuse.
> ...



It does not clamp at 42.1V, that is just the "400W" point. It's the point where the voltage * current at that voltage will be <400W, the rated power for the part. It will conduct and clamp (and heat up) earlier.

Lead-Acid chargers will typically charge at 14.8V, but if temperature compensated they may go higher and some employ higher voltage pulses as part of a conditioning phase. Just something to be aware of. This would scale to 24V battery systems. It's generally a given that someone will connect the charger while the light is turned on ... at least once. The worst case occurs when the charger is connected but you lose the connection to the battery. Some chargers will put out quite a bit higher voltages for a period of time.


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## mercrazy (May 15, 2018)

thanks for the good information.
1. i've always wanted a way to show the light was subjected to excessive voltage or AC.
is there a cheap way to do this?

 2. will this 48V fuse work? 
https://www.mouser.com/ProductDetai...%2bBXi4wRUHZtKAyWcVPAKDrjEiPOyCLtM4F41FU0dw==
i just want a way to prevent a dead short or fire.  i guess that requires a current limiting fuse? customers are warned to use fuses on supply line but some don't.  i've seen melted wires in boats and they always look for a device to blame for the lawyers.

3. sounds like 24volt supply is more trouble than it's worth.
this is the best smaller package 60V Pfet i could find. https://www.mouser.com/ProductDetail/621-DMP6023LE-13
i've been using smaj26 with no problems so guess i have to stay with 12V only.
i can't afford to not protect the driver chip.

4. do you have any thoughts on the circuit?  DIWdiver must be busy and i need to order circuit boards right away.
thanks again.


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## DIWdiver (May 15, 2018)

Hi,

Thought I posted this last night, but I don't see it here.  

R4 pin 1 should connect to K1-C3, not C2.  Likewise R7-2 should be at K2-C3.  Hopefully the circuit makes a lot more sense that way.


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## mercrazy (May 15, 2018)

i'm embarrassed.
i should have been able to figure that out.
makes perfect sense now.
1. are 10k resistors OK for R9 and R10?
2. the fuse has some resistance.  does it reduce effectiveness of TVS diodes?
and if fuse opens, then TVS has no effect.  what's the proper technique here?
3. i noticed some 15V zeners start acting down to 13 volts.  should i go up to 16V zener?
4. does Q1 and Q2 need separate zener protection?
thanks a million.
https://www.dropbox.com/s/qh129q6uce17fq2/2halfr9.PNG?dl=0


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## DIWdiver (May 15, 2018)

10K is fine.
Fuse is in the wrong place. It should be between P1 and K1. Then it will not disturb the TVS operation. Sorry, I didn't notice that before.
The zeners are only to protect the FETs from excessive VGS​. You could use anything from 10V to 18V.
Yes, Q1 and Q2 need separate zeners.


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## mercrazy (May 15, 2018)

thanks guys, we're getting close.
60V Pfets are a problem.
they're either too large physically, too expensive, too high ON resistance, or too fine pitch.
1. if i stay with 12V supply only, can i stay with 40V Pfet?
2. putting the fuse at P1 requires off board wiring which i don't like. Can i connect the TVS to a .004" wide, 1oz. copper trace to use as proof of excessive voltage? Do you think it would burn through if the TVS shorted? All wiring is 20 gauge or larger. Switch is 10A.
3. please check my zener/resistor placement at Q1 and Q2.
https://www.dropbox.com/s/x56ou12g14ofedj/2Hr91.PNG?dl=0
thanks again


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## DIWdiver (May 15, 2018)

The circuit really should be fused as close to the source as possible, to protect from the possibility of failed insulation on the wire from P1 to K1, or a short to ground within K1. In those cases, a fuse on the board won't help you.

If you are only worried about about what happens on your board, how about fusing the ground? It's pretty unconventional because of the reason mentioned above, and if the fuse blows the entire circuit is considered 'hot', but in this case it might make sense.

Otherwise, with two separate power inputs, you need two separate TVSs and two separate fuses.

Yes, with 12V only, you can go to a lower TVS voltage and 40V FETs. If you do that, you might as well switch the Nfets too.

I don't know how fast a circuit trace like that would open. I've seen them literally explode, but that's with massive overload. I think a commercially made fuse is much more predictable.

Keep in mind that with just a TVS and fuse, there is some range of input voltages where the TVS will fail before the fuse. Say you have the light switched to a low power mode, so it's drawing only 500mA. If you have a 1A fuse, there is a voltage that will put 500mA in your light, 500 mA in the TVS, and never blow the fuse. If this happens to be 28V, while you are getting a double voltage jump, the TVS would be dissipating 14W. It won't last long at that level!

The trick is to design it so the dangerous range is one you are never likely to encounter. If you can be confident you will never encounter double voltage (either in a jump or for some other reason), then maybe this design is fine!

The zener/resistor placements look fine.


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## mercrazy (May 15, 2018)

i'll just tell them fuses are mandatory on their side.
if the TVS blows, i just hope it blows open.
now to layout the board.
thanks again guys.


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## ssanasisredna (May 16, 2018)

mercrazy said:


> i'll just tell them fuses are mandatory on their side.
> if the TVS blows, i just hope it blows open.
> now to layout the board.
> thanks again guys.



If the TVS does not blow open, then that is what the fuse is there for.


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## ssanasisredna (May 16, 2018)

It's late and I need sleep, but I am a little concerned about potentially strange effects from R11 and R3 biasing Q4. DIW thoughts?


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## ssanasisredna (May 16, 2018)

Trace Fuse --- You can make them as fast as a regular fuse, just takes careful design. If the unit is going to be potted or coated though, throw that out the window.


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## ssanasisredna (May 16, 2018)

mercrazy said:


> thanks guys, we're getting close.
> 60V Pfets are a problem.
> they're either too large physically, too expensive, too high ON resistance, or too fine pitch.
> 1. if i stay with 12V supply only, can i stay with 40V Pfet?
> ...



Don't get hung up on super low RDSon. The IC is only capable of 1A output. If you have a 100mohm FET that is only 100mV drop. The self heating will be low. RDSon does go up with temp, so for safety assume at least 1.5x RDSon, if not 2x depending on how hot this gets.


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## DIWdiver (May 16, 2018)

ssanasisredna said:


> It's late and I need sleep, but I am a little concerned about potentially strange effects from R11 and R3 biasing Q4. DIW thoughts?



I don't see a relationship between R11, R3, and Q4. Perhaps we're looking at different schematic revisions?


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## mercrazy (May 16, 2018)

ssanasisredna said:


> Trace Fuse --- You can make them as fast as a regular fuse, just takes careful design. If the unit is going to be potted or coated though, throw that out the window.


boards get humiseal so that's out. blowing open is ok because i have proof of electrical damage and light may still work. blowing shorted causes blown fuses and light is unusable. wish i knew which way it blows. i may layout the board with reset fuse only to TVS. 
i was concerned about fuse resistance reducing effectiveness of TVS. i could place jumper instead of fuse if need be.
i've been using smaj24, 26, and 33 for years without a fuse and haven't seen but one damaged board. this guy had major stator/voltage regulator failure and fire which no TVS could stop. so maybe i'm worrying too much.
for RDSon, i'm trying to save every last bit of voltage i can. i'm pulling 800mA and a schottky was dropping 400mV. this only gave me about 200mV extra overhead depending on LED Vf variance. i'm trying to have close to 1 volt extra overhead to compensate for bad connections, switches, etc.
you guys are up too late and too early like me. i'm old is my excuse. 
thanks again


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## ssanasisredna (May 16, 2018)

-----


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## mercrazy (May 16, 2018)

i try to stay below 200mOhms on inductors.
i don't have a choice on LED bins, that's why i'm going to all this trouble.


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## mercrazy (May 23, 2018)

what about this idea?
thanks
https://www.dropbox.com/s/ylcs1eucc0nxxw7/g4.PNG?dl=0


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## DIWdiver (May 23, 2018)

I don't see any reason that shouldn't work. You lose some efficiency and overhead voltage switching the FET reverse polarity protection to diode. In this application I think that's not a problem, so schottky diodes are probably better than the FET solution; cheaper, smaller, and work a bit better, though the latter is probably insignificant.


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