EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: Vindhyachal.takniki on March 19, 2016, 07:55:09 am
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1. I have a application in which I have schottky connected at input. At input a solar panel is connected & at output battery is connected along with charge controller. Schottky serves two purposes; one is protect the circuit in case reverse connection of panel are connected & second prevents reverse flow of current from battery to solar panel known as dark current.
2. Max current from panel is 10A, so I have connected two parallel schottky PDS1040. Datasheet of PDs1040 state max Vf drop of diode around 0.55V at 10A(wrost case Tj=-65) & 0.52V at 5A.
Max panel voltage goes to around 21V.
3. These two diodes gets heat up very quickly & I want to replace them with Mosfet. Now I was looking for how I can replace it with Mosfet.
4. First option is to use ideal diode. I have checked linear LTC4358, which is for 5A. Operating voltage range is 9V-26.5V. I am thinking I will connect these two in parallel. One problem is its lower voltage range is 9V so from datasheet I didn't understand what will happen when voltage falls below 9V, will it turn off, or it may partially tun on which can increase the voltage across mosfet & heat it up. Second it very costly around $5.38 unit price.
5. Other option I found here ( https://www.google.co.in/search?q=ideal+diode&espv=2&biw=1366&bih=639&source=lnms&tbm=isch&sa=X&ved=0ahUKEwj6zu3ioczLAhXQwI4KHeDWAFcQ_AUIBigB#imgrc=CPkV_TNPueyWpM%3A (https://www.google.co.in/search?q=ideal+diode&espv=2&biw=1366&bih=639&source=lnms&tbm=isch&sa=X&ved=0ahUKEwj6zu3ioczLAhXQwI4KHeDWAFcQ_AUIBigB#imgrc=CPkV_TNPueyWpM%3A) )
It uses p channel mosfet & BCM856DS. Its an doubled matched transistor.
6. Any other circuit which I can use?
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Seems to me the current mirror circuit can only withstand 2 x Vebo, ~ 12 V,
before the two PNPs avalanche when under reverse connection of the panels
and self destruct.
Regards, Dana.
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I want reverse polarity protection fro solar panel only & reverse current going into panel.
Currently I have decided that I will place three PDS1040 in parallel & large coper pads with vias on it on PCB for 10A current & will see the result.
Since ideal diode from linear is around $5.
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Place pads for series R for each schottky. Reason is they will not current share equally
unless you do. Since datasheet does not predict worst case min Vfwd you cannot guarantee
the sharing.
This approach sucks to some extent because you wind up throwing some power away. But try
a spice simulation with some variation in Vth and see what you get.
Regards, Dana.
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I'm using a simple circuit shown in the attached schematic in several of my projects as a replacement for diodes and so far had no problems with them.
The Zener Diode is in there to limit the Source to Gate-Voltage and might be redundant in your application.
The resistor-values can be chosen more or less arbitrarily as long as the relative resistance remains the same. - Experiments on the breadboard and simulations in LTSpice have shown that lower resistors equal faster switching-times, but also increase current consumption.
In your case you can probably get away with very high resistances to keep the supply-current for the control-circuit low.
With the values shown in the schematic, the circuit works pretty well up to 2MHz.
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1. Can I connect 4 diodes in parallel. Max total current is 10A & volatge is 35V I am using PDS1040.
2. However I was reading on internet, then diode Vf decreases as its get more heat up & this may cause current hogging. One solution suggested that add series resistor for each.
3. Can I add 0.010 resistor in front of each diode?
So that max dissipation across in worst case is 10*10*0.010 = 1W.
4. Will 10 milliohm is ok for this application? ( http://www.digikey.com/product-detail/en/panasonic-electronic-components/ERJ-8BWJR010V/P.010AVCT-ND/1711707 (http://www.digikey.com/product-detail/en/panasonic-electronic-components/ERJ-8BWJR010V/P.010AVCT-ND/1711707) )
Ckt is attached.
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How about the SM74611 (http://www.ti.com/lit/ds/symlink/sm74611.pdf)?
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I have made the PCB. Will test it.
But I want to know whether 0.010 ohm resistor is ok to balance the current in case of mismatch
or should I use higher value resistor.
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Spice simulation would be a good way to answer that question.
Also depends on the degree of matching you are trying to achieve.
Regards, Dana.
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how to select which resistor value is suitable. I have attached the pic of If Vs Vf for PDS1040.
In my application maxx current is 10A total with 35V max.
Pic shows max Vf mentioned in red line. So how to calculate most optimal resistor value.
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I'm using a simple circuit shown in the attached schematic in several of my projects as a replacement for diodes and so far had no problems with them.
The Zener Diode is in there to limit the Source to Gate-Voltage and might be redundant in your application.
The resistor-values can be chosen more or less arbitrarily as long as the relative resistance remains the same. - Experiments on the breadboard and simulations in LTSpice have shown that lower resistors equal faster switching-times, but also increase current consumption.
In your case you can probably get away with very high resistances to keep the supply-current for the control-circuit low.
With the values shown in the schematic, the circuit works pretty well up to 2MHz.
The maximum reverse blocking voltage for that circuit is no more than the sum of the emitter-base breakdown voltage of Q2 & Q3 or around about 10V to 14V. Diodes should be connected in series with the emitters to provide reverse polarity protection for higher voltages.
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how to select which resistor value is suitable. I have attached the pic of If Vs Vf for PDS1040.
In my application maxx current is 10A total with 35V max.
Pic shows max Vf mentioned in red line. So how to calculate most optimal resistor value.
Looking at the diode datasheet (http://www.diodes.com/pdfs/ds30470.pdf), the relevant figure is Fig. 5. It depends a lot on the layout of your circuit. Assuming the minimum layout (see Note 2), your diodes will be in serious trouble conducting more 3.75 amps at 65ºC. With a good layout (Note 4) the problems will start conducting 6.25 amps at 100ºC (start of high slope runaway).
So the first lesson is that, more than your ballasting resistor or the diode characteristic curves, you must layout your diodes very carefully, with special attention to heat dissipation.
Ok, let's assume I do a really sloppy job with the layout and use the minimum required. I know that my diodes will be in the dangerous, out of spec zone, conducting more than 3.75 amps at 65ºC. Normally, your 10A would be evenly divided among the diodes, 2.5A for each. Note that with this poor layout that would mean trouble at 100ºC, even with perfect balance of currents. The layout is that important.
According to the datasheet, at 25ºC and 2.5A of current, the diode drop is about 0.4V. That means a power dissipation of 1W. The datasheet specifies that the thermal resistance from junction to ambient air for the minimum layout is 120 ºC/W, which means a delta temp of 120ºC at 1W power. Even conducting 2.5 amps in balanced state, the power dissipation will be too much for your diodes without a significant disspation provided from your pads: there's 2 ºC / Watt thermal resistance from junction to pad, according to the datasheet, so you need pads that will guarantee that your diodes will not fry dissipating 1W or more.
So for your application the minimum layout is not good at all. Let's assume you use the pads suggested in Note 4, 18.8mm x 14.4mm for the anode, 5.6mm x 3.0mm for the cathode, 2oz copper. In that case, your diodes can conduct 2.5 amps at a max of 135ºC, which is good enough. As said, trouble will begin conducting 6.25 amps at 100ºC.
So with a good layout, you need resistors that guarantee that no diode will conduct 6.25 amps at 100ºC. The datasheet predicts about 0.35V drop in these conditions. If we assume that the other diodes remain at 25ºC (worst case), each would need to conduct about 2A to compensate. The drop is about 0.4V @ 2A 25ºC. With equal resistor R ballasting all the diodes, the drop in the runaway diode is 0.35 + 10R, while in the cool diodes it is 0.4 + 2R. Making them equal: 0.35 + 10R = 0.4 + 2R, thence: 8R = 0.05, so R = 0.05/8 = 0.00625 Ohm, which is dangerously close to your 0.01 Ohm.
I would say that your 0.01 Ohm could manage with very good thermal layout. But also that the layout itself is much more important than the ballasting resistors.
Anyway, beware my back of the envelope calculations, and watch for an expert's answer, if any is forthcoming.
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Thanks @orolo.