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Author Topic: 8x8x8 multiplexed LED cube with an Arduino Mega 2560  (Read 29477 times)
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Hofstadter's Law: It always takes longer than you expect, even when you take into account Hofstadter's Law.
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I bought a sheet of adapters like
http://www.ebay.com/itm/SMD-CONVERTER-ADAPTER-PCB-SOT-TO-MSOP-SIP-DIP-14-/270832318388?pt=LH_DefaultDomain_0&hash=item3f0edd37b4
Probably had like 20 each of the different adapters on it.

That looks excellent, I'll order one.
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Unfortunately, P-channel MOSFETs and PNP transistors seem to be pricey compared to N-Channel & NPNs.
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Designing & building electrical circuits for over 25 years. Check out the ATMega1284P based Bobuino and other '328P & '1284P creations & offerings at  www.crossroadsfencing.com/BobuinoRev17.
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Cluj-Napoca, Romania
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From my point of view the biggest problem is not that they are pricey, it's that they are rare. Otherwise I could buy them here at home or on eBay smiley I just talked to a friend and he says that Farnell only ships to Romania for companies, not for individuals. I'll have to check that out.
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Okay, so can they ship to Un4Seen, Incorporated?
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Arduino for Teens available at Amazon.com.

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Hofstadter's Law: It always takes longer than you expect, even when you take into account Hofstadter's Law.
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I just searched eBay for P-channel mosfets and got 156 results. The SMD ones are inexpensive, the TO220 ones cost a lot more and are mostly not logic level.
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Might be worth trying RS as well, http://ro.rsdelivers.com/.
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OK, I have figured it out smiley As I wrote, one NDP6020P costs about 2.5$ at Farnell. The transport to Romania would be 5 Euro, regardless of what is in the package. So, assuming that I'd buy 15 pieces of NDP6020P (8 for the 8x8x8 cube, 5 for the 5x5x5 cube and 2 for spares), the total cost would be around 27E or 35$. That's more than the 1000 LEDs and about 33% of the total project cost smiley I there's no other choice, I'll go for it, but if there's some more widely available, less expensive MOSFET, I'd prefer that smiley

All the suitable but less expensive p-channel mosfets I found were in SMD packages - and they are MUCH less expensive. So I think your alternative is to download Eagle and design a pcb to hold a 74HC595 and eight SMD mosfets. Then get Itead to make 10 of them for $9.90. Add mosfets such as http://www.ebay.co.uk/itm/20PCS-AO3401-SOT-23-P-Channel-MOSFET-TRANSISTORS-/150653948679?pt=UK_BOI_Electrical_Components_Supplies_ET&hash=item2313acd307.
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I ordered some of these from thaishine.com
N
Manufacturer Part No: IRF3205

Specification
Mosfet Type N-channel
Current Rating 110A
Rds (On) 8mΩ
Voltage Rated 55V


P
Manufacturer Part No: IRF9540

Description
MOSFET
Transistor Type:MOSFET
Transistor Polarity:P Channel
Drain Source Voltage, Vds:-100V
Continuous Drain Current, Id:-23A
On Resistance, Rds(on):117mohm
Rds(on) Test Voltage, Vgs:-10V


They were under $1 each in packs of 10.

I also found these (below), after I had ordered the above fets.

Mosfet Type: P-channel
Current Rating: 2.6A
Rds (On): 40 mOhm
Voltage Rated: 12V
Package: SSOT-3

Mosfet Type: N-channel
Current Rating: 4A
Rds (On): 100 mOhm
Voltage Rated: 60V
Package: SOT-223

They are smaller package, but probably suitable for my 1 amp needs. they may be suitable for 1.3 amp, you have to check.

Im not sure which one I got it from, thaishine also sells through ebay.
www.thaishine.com
www.taydaelectronics.com

I am way behind on those cubes, and I havnt had a chance to use the parts, but I assume they will be good for cubes/multiplexing.
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dc42, are those tiny sot23 parts going to be able to handle 1.28 amps? they seem pretty small for that. I looked, but I didnt see any datasheet.

Wow, 4 amps from such a tiny package. I wish I had gotten those now.
« Last Edit: December 29, 2012, 12:46:24 pm by Hippynerd » Logged


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Yes, see http://www.aosmd.com/pdfs/datasheet/AO3401.pdf. They are rated at 4A continuous (although 3.7A would be a more realistic limit when driven from 5V, since that is the current at which Rds(on) is measured for Vgs=4.5V) and 1.4W power dissipation (although I wouldn't want to dissipate more than a few hundred mW in one).
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Cluj-Napoca, Romania
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Guys, did I understand correctly, that the IRF9540 would be able to do the job and I could use it instead of the NDP6020P?
I've found them on eBay and for a good price too: http://www.ebay.com/itm/10pcs-IRF9540-P-Channel-Power-MOSFET-23A-100V-TO-220-IR-Free-Shipping-/180816152092?pt=LH_DefaultDomain_0&hash=item2a197b861c

It seems to be a fast switching P-channel MOSFET in TO-220 packaging, rated at 23A and 100V. What do you think?

Thanks,
Andras
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Seems to be a logic level part
http://www.irf.com/product-info/datasheets/data/irf9540n.pdf

Will have at least 1V drop across it, see Figure 2.

TPIC6B595 will have 3-4 ohms of resistance, see Figure 7, at 20mA = another 0.07V
http://www.ti.com/lit/ds/symlink/tpic6b595.pdf

So as long as your LEDs have Vf <(5-1-.07) = 3.93V you should be okay.
Adust your current limit resistors accordingly.
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Thank you, CrossRoads!

1-2 days ago we used a different calculation to obtain the value of the current limiting resistors. You wrote back then:
"Resistors = inexpensive resistors, 1/8W, carbon composition. Value will depend on the LED color & the current you want to put thru them.
If have a 5V source, and the anode transistor has 0.45V across it, and the cathode shif register has 0.25V across it, that leaves the remaining voltage across the LED and the resistor.
For and LED with Vf of 3.2V way, and using 20mA as the current, then:
(5V - .45 - .2v - 3.2)/.02 = 55 ohm.  56 ohm is a standard value."

If I understand correctly, now instead of the .45 voltage drop across the anode transistor we have the (at least) 1V drop across the IRF9540. In your earlier explanation you wrote that the cathode shift register drops 0.25V, but I think your current calculation is the correct one and it drops only about 0.07V. I use standard blue LEDs with a voltage drop of 3.3V. So, basically, if your old calculation was wrong and the new one is right, it means that the remaining voltage that needs to be handled is 5V - 1V - 0.07V - 3.3V = 0.63V. Our target current is 20 mA, so R = V/I = 1.18/0.02 = 31.5. In other words 33 Ohm resistors are fine.

Is my above calculation correct? If yes, I'll go for the IRF9540 MOSFETS. That's great news! smiley
« Last Edit: December 29, 2012, 06:24:41 pm by Un4Seen » Logged


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Hmm, are your sure you interpreted figure 2 (http://www.irf.com/product-info/datasheets/data/irf9540n.pdf) correctly? It's the first time in my life that I try to read a diagram like this, but it seems to me that the IRF9540 drops 1V at 2A. In the case of the LED cube one IRF9540 handles 64 LEDs, which means that the current going through it ranges from 0.02A (1 LED lit up) to 1.28A (64 LEDs lit up). The way I read this figure 2, at 0.02A the voltage drop is 0 but I can't figure out from the diagram how much the voltage drop is at 1.28A... I don't really understand it, but can I really use 1V as voltage drop across the IRF9540 regardless how much current (0.02A to 1.28A) is passing through it?

Thanks,
Andras
« Last Edit: December 29, 2012, 05:47:39 pm by Un4Seen » Logged


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After looking some more at the graphs, I reached the conclusion that we should try to read figure 1, not figure 2, because figure 1 is at 25 degrees Celsius, while figure 2 is at 175 degrees Celsius. so, looking at figure 1, the second graph line from the bottom is the one for 5V. I don't really understand why the grid lines on the graph are not distributed evenly, but assuming that the distance between the 1A and 10A values on the vertical axis is distributed linearly, it seems that the voltage drop at 1.28 is aroun 0.6-0.7V and the voltage drop at 0.02A is not even represented on the drawing but it should be less than 0.3V, possibly even 0? Am I making any sense here? smiley
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