SparkFun Forums

[reklipz]

OK, well, I finally took time to find a TO-3 triac (mouser works wonders...).

I ordered one of them, and am working on a heatsink next. The 15W mark still stands, FYI.

Thanks so much for your help! I probably would have made my first pcb go up in flames had it not been for your help (well, all hope is not lost yet, the flames may still come, we'll just have to wait and see! ).

-Nate

--EDIT--
Any Idea where I can find the quick connect connectors for the triac?

--EDIT--

I think I'm going to buy this guy:
http://www.aavidthermalloy.com/cgi-bin/ ... 303b00000g

~$4.50 from Mouser.

I'm going to try to mount a fan onto it, so I think this'll be a good solution.

[reklipz]

OK, well, heres my "finished" board. Oh, just realized I have to add the USB power transistor, so, nearly finished.


(display board has ground plane, but I didn't hit ratsnest before export)


(50 mil grid)

The POWER terminal is: pin 1: neutral - pin 2: hot
The TRIAC terminal is: pin 1: load return path (to use fuse) - pin 2: triac gate trigger

The triac is the TO-3 type you suggested, mounted on the heatsink I posted.

Hows it look?

[bigglez]

OK, well, heres my "finished" board. Oh, just realized I have to add the USB power transistor, so, nearly finished.

Greetings Nate,

Your design looks very good! I'm new to the "display board"
project, are they related?

I don't have your schematic, not sure if you're willing to
share the EAGLE files, so what are you using for the
temp interface? (What is IC3?).

My suggestions are:

(1) Move C14 away from the hot AC side of the transformer,
perhaps you can fit it near the 16 pin connector?

(2) Route the low voltage side of OK1 away from the hot AC
side of the transformer. Perhaps you can run the pin2
trace under the trace to R4 to compact the design?

(3) Move R4 nearer to either the 16 pin connector or
the uC, to keep its pads away from the hot AC side of
the transformer.

(4) For peace of mind you should keep a 6mm (400mil)
guard band around all hot AC traces. This is a VDE
requirement, and a tough one at that. The process of
thinking about a 6mm creepage will get you a much
better hot AC intergrity even if you can't do 6mm for
100% of the hot AC traces.

If you are familiar with the "Mark" command in EAGLE
use it to measure line-of-sight from pads and traces to
each other on the hot AC circuits.

(5) I don't see any test points. When it comes to SMT
layouts I find that sprinkling a few pads on ground,
power rails, and at least one unused uC port (if available)
will allow a scope probe or DMM to be used later.

My preferred testpoint is a 100mil diameter pad with
40mil drill. It can be used as-is or fitted with aKeystone THM test point.

(6) Your screw terminal blocks look familiar, but
I'm not sure of their size. Do you really want to run 13.5A
through these? I use two sizes, the'blue'ones have single
hole mounting and can take a 14 - 22AWG wire, I use
these up to five amps (they are UL approved for 15A).
For heavier loads I use the 'Black' ones rated to
10A but with two electrical lugs each.

Why not run heavy wiring in the project to the heaters,
Triac, and a chassis mounted fuse (15A)? Run a second
AC circuit after the 15A fuse to the PCB and use a smaller
fuse (5A or less)? This will give the required protection
and reduce the copper needed on the PCB.

Comments Welcome!

[reklipz]

Your design looks very good!

Wow, thanks! It's my first one, so your input has done nothing but help me improve the design.

I'm new to the "display board" project, are they related?

Indeed they are. The 16 pin connector is just a ribbon cable that connects the two boards. The display board (in hindsight, something like user I/O board would have made more sense) consists of 4 silicon push buttons, each with a colored LED, and a Nokia LCD.

I don't have your schematic, not sure if you're willing to share the EAGLE files, so what are you using for the temp interface? (What is IC3?).

I don't have a problem sharing the EAGLE files, but for the record, IC3 is a MAX6675 K-Type Thermocouple to Digital Converter from Maxim.

(1) Move C14 away from the hot AC side of the transformer, perhaps you can fit it near the 16 pin connector?

Perhaps I could. It would also shorten the traces a bit.

(2) Route the low voltage side of OK1 away from the hot AC side of the transformer. Perhaps you can run the pin2 trace under the trace to R4 to compact the design?

I'll give this a shot, although it may not be relevant after I remove the large AC polyfills (suggestion 6/7).

(3) Move R4 nearer to either the 16 pin connector or the uC, to keep its pads away from the hot AC side of the transformer.

Done.

(4) For peace of mind you should keep a 6mm (400mil) guard band around all hot AC traces. This is a VDE requirement, and a tough one at that. The process of thinking about a 6mm creepage will get you a much better hot AC intergrity even if you can't do 6mm for 100% of the hot AC traces.

I'll keep it in mind when I make these changes, thanks for the suggestion. I think you mentioned it earlier as well, but in my haste to get something up here I neglected it.

If you are familiar with the "Mark" command in EAGLE use it to measure line-of-sight from pads and traces to each other on the hot AC circuits.

I've accidentally stumbled upon this one before (I think by pressing the stoplight looking thing?) It never really seemed to do anything so I've never looked into it. It seems as though I may want to, though.

(5) I don't see any test points. When it comes to SMT layouts I find that sprinkling a few pads on ground, power rails, and at least one unused uC port (if available) will allow a scope probe or DMM to be used later.

Yet another excellent idea. Also reminds me that It wouldn't be a bad idea to add an ICSP port...

(6) Your screw terminal blocks look familiar, but I'm not sure of their size. Do you really want to run 13.5A through these? I use two sizes, the'blue'ones have single hole mounting and can take a 14 - 22AWG wire, I use these up to five amps (they are UL approved for 15A). For heavier loads I use the 'Black' ones rated to 10A but with two electrical lugs each.

Why not run heavy wiring in the project to the heaters, Triac, and a chassis mounted fuse (15A)? Run a second AC circuit after the 15A fuse to the PCB and use a smaller fuse (5A or less)? This will give the required protection and reduce the copper needed on the PCB.

Moot point with the connectors, as the latter idea solves this issue (although I believe the connectors I was planning on ordering are rated to 13.5A). I've not yet looked at the chassis mounted fuse, but is there some sort of standard chassis fuse holder that I should be looking for? (And even more so, any particular type of fuse?)

On another note, I wasn't sure which type of connector to use, so I just placed one of the Wago Screw type connectors, it looks to have the same footprint, so I wasn't too worried. I was looking for the type that SFE uses/offers on their boards/site, so I made a guess. Do they look right?

Thanks again for all of the suggestions, I'm working on the revisions as we speak.

-Nate

*Note; I just setup the ftp server, so let me know if it works for you or not!

[bigglez]

I don't have a problem sharing the EAGLE files, but for the record, IC3 is a MAX6675 K-Type Thermocouple to Digital Converter from Maxim.

*Note; I just setup the ftp server, so let me know if it works for you or not!

Nate,

I got a file ending in BZ2. My Wintel machine has no idea what
to do with it. Some form of UNIX/Linux compression?

How about *.sch and *.brd or even *.zip?

Thanks In Advance!

Peter

[reklipz]

A .tar.bz2 is a BZip2 compressed tar file (tar is just a lump of the files), and indeed, widely used on UNIX/Linux.

I've uploaded a .zip for you, no problem.

I made some custom libraries, so that's why it's archived.

-Nate

[reklipz]

Silly question.

I would like to drive the PNP from the 12V source on the board, as follows:


Where VCC is 5V, and the load connects between the two open nets.

Basically all of the PNP transistor datasheets I've seen specify a max Vebo (Emitter-Base Voltage) as 5.0Volts (or -5.0Volts, depends on your perspective). Is this a problem (12V - 5V = 7V which is > 5V), or am I thinking this specification is something else?

Silly question I know, I may have known the answer at one point in time, when I first read my dad's electronics books, but I tend to forget stuff when you don't use it for a while.

-Nate

[bigglez]

Silly question.

I would like to drive the PNP from the 12V source on the board, as follows:
.....
Where VCC is 5V, and the load connects between the two open nets.

Greetings Nate,

So when the PNP transistor is conducting (on) the load is connected
between Vcc (+5V) and ground?

The output of the PNP would be the collector, the input would
be the Emitter tied to the Vcc rail, and the base of the PNP
would be either Vcc (off) or Vcc - Vbe (on).

Your control signal needs to be lower than Vcc - Vbe (4.3V) to
activate the transistor and could return to the 12V rail. Or Vcc
rail when off, such as a 5V CMOS logic signal or a uC port.

If your signal rests at 12V you'll need a protection diode on the
PNP base and emitter to reduce the reverse voltage on
the base (as you noted in your question).

What current in the load? What transistor type? (Need to
know Hfe at Ic = load current).

Also, is the input signal 0V to 12V? Or 0V to 5V? Or ??

Comments Welcome!

[reklipz]

Hmm, after reading your post, and reviewing my post, I'm not quite sure why I decided to upload that image. I had another one laid out as you described, but the one in the picture is how it would fit into my schematic (Vcc rail is the output, input would be between the break). Sorry about the confusion.

The control signal swings between 12V and 0V. This would mean I would need to drop it by at least 2V in order to meet the spec. I could just use 3 diodes, but that seems like a bad idea for some reason. If it is a bad idea, I wonder if this may be better; The 12V source feeds a switching regulator, which outputs 3.3V and 6.8V. Would using the 6.8V source as the control signal be a bad idea? (I'm wondering about whether or not this is ok because it is a switching regulator, and takes a small amount of time to settle)

Sorry for the delay, and that I can't seem to get my facts straight before posting, which only provides confusion.

-Nate

[bigglez]

The control signal swings between 12V and 0V. This would mean I would need to drop it by at least 2V in order to meet the spec.

Nate,

The spec of 7V is a max rating for the transistor under
reverse bias. It would be unwise to stress the device to
the spec limit.



This is how you might switch a 5V load from a 12V control
signal. The transistor conducts when the control signal is
near ground, and when the control signal is above 5V (i.e.
at 12V) the diode protects the transistor. The transistor
inverts the signal. If you want the output to be at 5V when
the control is at 12V, a second invertion is needed.

Comments Welcome!

[reklipz]

Thanks for the drawing.

I feel as though I'm just a hassle now, as I always seem to find something negative to come back with, unfortunately.

I tried simulating the design in multisim, and cant get the transistor to ever turn off with the 1K resistor. When I increased the value to 10K (this effectively lowers the base current, allowing the transistor to stay in its conducting state, correct?)

I've also got another question for you. When I first viewed the circuit, I thought to myself, would the diode and resistor not raise the 5V rail to 12V? As it turns out, it doesn't. Is this because the transistor sinks all of the current when the input goes high, thus not affecting the 5V rail?

If you feel that I'm not fully understanding how this works, and wouldn't mind giving a bit more detail, I would truly appreciate it. I find it a bit, embarrassing, that I can put together that whole schematic / board presented earlier, yet trivial things such as this completely stump me (well, not completely, as I was on the right track, but hit a road block).

Thanks very much for the solution! I think it's the last thing I need, and it's off to fabrication (after a full review of course!).

-Nate

[winston]

I'm a hobbyist, not an engineer, so take this for what it's worth, but to switch on a 5v rail from a 12v signal, I'd just use a pair of mosfets - small signal N-channel for the signal sense which in turn switches an appropriately rated P-channel device to switch the 5v rail. Most small signal mosfets have a Vgs(max) of 20 volts and so are easily man enough to take a 12 volt signal. (Actually, when I'm switching a load from a signal, I try and arrange it so that the gnd line is the one that needs switching, then all you need is a single N-channel mosfet). I like mosfets - for the things I do they almost always result in a simpler circuit with higher impedance inputs.

[bigglez]

I tried simulating the design in multisim, and cant get the transistor to ever turn off with the 1K resistor.

Greetings Nate,

There are several different ways to do what you want,
Winston suggested a P-Ch FET. My circuit can be analyzed
intuitively, or with pSpice if you'd rather have a full simulation.
It works!

I've also got another question for you. When I first viewed the circuit, I thought to myself, would the diode and resistor not raise the 5V rail to 12V? As it turns out, it doesn't. Is this because the transistor sinks all of the current when the input goes high, thus not affecting the 5V rail?

Remember that current takes the path of least resistance,
so the current from the 12V signal defined by R1 does flow
into the 5V rail via the diode and R2. The 5V rail is very
low impedance, say one ohm, so the 5V rail will be raised
by 1/3900 or 0.025% - it would be hard to measure with a
good DMM. Under these conditions the transistor is off, the
base-emitter is reverse-biased, and only device leakage
current flows into the base.

When I increased the value to 10K (this effectively lowers the base current, allowing the transistor to stay in its conducting state, correct?)

No. The value of R3 shunts the current from emitter to
base. In a PNP device the base current flows out of the
base when the device is conducting. Increasing the value
or R3 increases the current in the base, bacause less is
shunted around the device. The function of R3 is to remove
the base charge when the input goes positive (to 12V in this
case) so that the transistor switches off quickly, and leakage
current from the transistor doesn't turn it back on (unlikely
in this case). R3 could be removed, many circuits do not
place a resistor from emitter to base, but doing so costs
one resistor and defines that the circuit will work reliably.

I'm sure there are a lot of people lurking on this thread
and learning from your mistakes. This is an open forum
for discussion, anything here can be questioned, others
will likely have different opinions or ways to approach the
problems presented.

Comments Welcome!

[reklipz]

OK, I think I've found what I did wrong.

In my simulation, I placed a switch inline with the 12V source, and added a 10K resistor from the base of the transistor to ground. The reason I added a switch instead of just setting the voltage to 0V is because the only pathway back to ground in the actual circuit would be the transformer, the smoothing cap, or the switching regulator. I assume I can count out the switching regulator as a pathway; relying on the capacitor in reverse bias is just a bad idea; and the transformer would be really high impedance, so that pathway is out too, right? I lowered the value of the resistor to 1K, and it seems to work;



Is my reasoning above correct, or can I do away with the resistor?

[bigglez]

In my simulation, I placed a switch inline with the 12V source, and added a 10K resistor from the base of the transistor to ground.

Greetings Nate,

That's a fancy simulator...

For the simulation to be accurate the switch should be in
parallel with V1, so that when the switchis open the input
is 12V and when the switch is closed the input is zero volts.

That's a bit rough on the V1 source, so a better method is
to change the value of V1 and run the simulation twice.

The reason I added a switch instead of just setting the voltage to 0V is because the only pathway back to ground in the actual circuit would be the transformer, the smoothing cap, or the switching regulator.

Actually, the real world path impedance is very low
(only a few ohms).

Is my reasoning above correct, or can I do away with the resistor?

Please run your simulation with R4 removed and V1= 12V,
then again with V1 =0V. That is what I did (scroll up and
check the table data).

Voltage sources should be treated as 'ideal' with no series
impedance. We only need the DC stimulus for this circuit,
so capacitors are open and inductors shorted (if any are
used).

Does your simulator spit out numerical data?

Comments Welcome!