Die Simulator Construction, your thoughts on my plan?

[jaydee]

I have some heaters Robotech sent me and I plan on using one of them for the simulator. Each one is capable of 150watts so one "should" be enough being I only want to use maybe 70watts.

What I plan to do is very similar to RoboTech's design in the picture below except I will have a heater in place of the thermoprobe in the middle and I will only be using one heater. I had planned on using all Copper on mine mainly because I have enough of it. I was thining about making it all one peice as opposed to Robo's design. I think the thermal transfer would be better but i would also have to make a new part in the event of future core revisions. Which is no big deal.

So my real question is where to stick the probe? Should I cross drill into the milled out die and epoxy it in? It seems the only logical thing to do?

Next is insulation. I thought about making a silicone sealant square around it but then started digging into silicons thermal properties and it looks like it is not going to cut it. Would the styrofoam style insulation work alright? The stuff they use on houses now days instead of the fluffy pink/yellow stuff.

I was thining about blocking it in with that and then making a plexi glass shell around it to hold it all in and tight.

Next is the power supply. Anyone know a decent place to pick up a adjustable DC power supply. I need at least 70watts and preferable 100watts, just to have some extra adjustment, of power.

I was going to use a computer supply but have failed to figure out how to make it adjustable.

Pic of what Robo is working on and similar to what I am thinking:

[pHaestus]

phenolic resin is I believe what Bill used.

Put the probe on the side offset from the die. You dont want to crossdrill a hole into the center of the die because there will be some heat shadowing that will affect the die temp above it. Again take a look at Bill's design (hardforums if they havent pruned it).

[jaydee]

Originally posted by pHaestus
phenolic resin is I believe what Bill used.

Put the probe on the side offset from the die. You dont want to crossdrill a hole into the center of the die because there will be some heat shadowing that will affect the die temp above it. Again take a look at Bill's design (hardforums if they havent pruned it).

Yeah,I am serching for that thread. I had it booked marked several times I just keep reloading Windows without saving my bookmarks....

So are you saying just attach the probe to the side of the die?

With recent discussion I am not sure this is worth doing, but I am past the point of no return. Should make some decent progress on this this Sunday. Setting up the mill at a freinds house and should be able to mill this thing out. Will have to look up that resin, I was hoping to find something in a liquid form that dries. Sounds like that might be it.

[jaydee]

Found it.....again..... http://www.hardforums.com/showthread...hreadid=266016

[pHaestus]

See the way they made a little shelf to the side for the thermal probe? That is what I mean. Not just sticking it to the outside. I dont think that phenolic resin stuff was in liquid form originally; looks cut.

[jaydee]

Originally posted by pHaestus
See the way they made a little shelf to the side for the thermal probe? That is what I mean. Not just sticking it to the outside. I dont think that phenolic resin stuff was in liquid form originally; looks cut.

Yeah, I like the little shelf idea. Yeah, that stuff looks like it came in sheets and was cut to fit. I will see if I can find some. I think i can make it work. Looks like it should support the 4 mounting holes to if I drilled them into it.

[Zymrgy]

originally posted by pHaestus
I dont think that phenolic resin stuff was in liquid form originally; looks cut.

you are correct. phenolic is very close to fiberglass...that is a resin holds sheets of fiber ( I think cloth?) together making a very strong & light material. The stuff also stinks like shit when you machine it & makes one hell of a dust cloud. you can get the stuff in sheets...I have seen it up to 3" thick before. It would make a very good insulator. the stuff can be drilled & tapped ( tho I would use helicoil inserts in the threads ) & this looks like what they have done to me...built a box around the copper core.

[jaydee]

Well I got nothing done on this this weekend. Went fishing all weekend instead. Caught a lot of fish to so it was worth it.

[pHaestus]

fishing >> block testing

I managed to fry the diode on my test CPU, so I think I may go fishing next time I decide to do some testing as well

What are those heaters exactly Jaydee? They look like something I should try to use in my die simulator instead of those minco heaters....

[jaydee]

Originally posted by pHaestus
fishing >> block testing

I managed to fry the diode on my test CPU, so I think I may go fishing next time I decide to do some testing as well

What are those heaters exactly Jaydee? They look like something I should try to use in my die simulator instead of those minco heaters....

I am not exactly sure what those little heaters are for. Robotech sent them to me so he might know what they were used for and where to get more.

[pHaestus]

Maybe I swap you a flowmeter for one of them? Will need to get specs though first

[jaydee]

Yeah, when I get home tonight I will see what the specs where. I belive they said 40VDC 150watt, but thats not 100% for sure.

[pHaestus]

Hrm may not be so easy to find a good 40VDC 4A PSU...

[jaydee]

Originally posted by pHaestus
Hrm may not be so easy to find a good 40VDC 4A PSU...

You should be able to run them at any DC voltage and amprege up to 150watts. I was going to use like 12VDC and 6amps to get 72watts. Volts x Amps = wattage right?

[bigben2k]

Originally posted by jaydee116
You should be able to run them at any DC voltage and amprege up to 150watts. I was going to use like 12VDC and 6amps to get 72watts. Volts x Amps = wattage right?

Did you test this figure?

Assuming that the resistance is linear (and it shouldn't be, because of it's nature), 150W @ 40V will draw 3.75 amps, and the resistance will be ~10.7 Ohm. If you run 12 volts through that resistance value, the current will be ~1.12 amps, for a total power of ~13 Watts.

Did you think you got to choose the amps?

[jaydee]

Originally posted by bigben2k


Did you think you got to choose the amps?

Yes, because thats what adjustable bench power supplies do right?

[pHaestus]

I am with Ben on this one There is more than one equation:



May be helpful to you in sorting out power requirements

[pHaestus]

Consider that your heater is a resistor basically Jaydee. So R is fixed. That means you can't just throw more current at it to compensate for the lowered voltage.

[bigben2k]

Well, R is *almost* fixed...

In this case, similar to a lightbulb, the more current goes through, the higher the resistance will be.

The current draw is defined by the resistance and the voltage applied: it is not affected by the type of power supply.

To get less than 150 Watts, you would lower the voltage (from 40), which would in turn lower the amps (from 3.75), and as a result, the power (P=V * I).

[jaydee]

Originally posted by pHaestus
Consider that your heater is a resistor basically Jaydee. So R is fixed. That means you can't just throw more current at it to compensate for the lowered voltage.

I understand what you guys are saying, but was under the impression these don't work that way.

Anyway I am awaiting an answer from Robo.

Also here are the specs: Watlow Firerod cartridge heaters 150watt 40VDC. Their website: http://www.watlow.co.uk/products/heaters/cartridge.htm

I can't seem to find anything relevant though.

[pHaestus]

that is the sort of heater I am looking for so I think it is not such a big deal to find a suitable supply. Here is one for $60:

http://www.mpja.com/productview.asp?product=12614+PS

Would be a lot nicer if they were AC though...

Don't forget you will need to be able to monitor current and voltage with a pretty high res... This I guess gets expensive fast (but you suspected that right?)

[bigben2k]

I had to switch over to the US site, for a bit more info, but I didn't find anything about a 40 Vdc Firerod.

http://www.watlow.com/products/heaters/ht_cart.cfm

I'm guessing it's a custom product.:shrug:

[jaydee]

Here is the reply I got back from Robo:

Yes, the heaters have a set resistance (which will change very slightly with temperature). I don't remember exactly what it is on the ones I sent you but you can put an ohm meter across the leads and measure it (~4 ohms maybe?).

No, they do not have to be powered from 40 VDC. That is the max voltage - go over that and they may overheat and burn up. Less voltage will just produce less heat - exactly what you want! You can set a specific heat generation (say 70 watts) by adjusting the voltage (the heaters will pull more and more current as the voltage increases) until you reach a point where measured voltage x measured current = 70 watts.

Say for example you adjust the applied voltage to 10.00 VDC and the heater pulls 5.0 Amps... 10 x 5 = 50 watts. Turn up the voltage a little more and maybe 12 VDC and 6 Amps... 12 x 6 = 72 watts. Hope that helps.

What I don't understand is what is making the heater want to draw the amperage. If i set the power supply to 12VDC and 6amps what is making it want to use those 6 amps?

Don't forget you will need to be able to monitor current and voltage with a pretty high res... This I guess gets expensive fast (but you suspected that right?)

Yeah, I know this and it is something I am really dreading as I havn't yet done any research on this.

[bigben2k]

Jaydee, I think you've been working with waterpumps too long

Contrary to waterflow, where the resistance (pressure drop) across a block changes with the flow rate, the electrical resistance of the heater will remain (essentially) the same, regardless of voltage. (a poor analogy, I know).

All in all, the equation V=R * I (V: voltage, I: current (amps) and R: Resistance (in Ohms)) is always true. It's a law, just like many other things in physics.

So assuming that the heater will put out 150 Watts, at a voltage of 40, using the other equation (also a law) where P=V * I, you can see that the current that the heater will draw is 3.75 amps.

Back to the first equation (V=RI), you can then calculate that the resistance of the heater is ~10.7 Ohms. This does not change (well, just a little bit, maybe up to 15%, but just because it's a heater).

So if you reduce the voltage to 12, and applying V=RI, where R is (still) ~10.7 Ohms, you would figure out that it will equal about 1.12 amps. Now knowing the amps, and using P=VI, you can then calculate the power output to be ~13 Watts.

But back to your question...

What I don't understand is what is making the heater want to draw the amperage. If i set the power supply to 12VDC and 6amps what is making it want to use those 6 amps?

Power supplies usually work in one of two ways: they allow you to control either the voltage (most common) or the current.

If you control the voltage, the resistance determines the current.
If you control the current, the resistance determines the voltage.

So your power supply may do either, but not both. Perhaps the "amp" setting is some kind of failsafe trigger, where if it's exceeded, it shuts down?

Yeah, I know this and it is something I am really dreading as I havn't yet done any research on this.

It's not so bad.

Theoretically (and someone correct me here), you simply use your voltage measurement's margin of error, in %, and add it to your current measurement error margin (also in %), then add 1 or 2% for secondary heat loss (better to quantify it yourself, but not easy), and you have a relatively accurate measurement of the power that your testbench applies, complete with error margins.

[jaydee]

Originally posted by bigben2k
Jaydee, I think you've been working with waterpumps too long

LOL, no argument there!

Contrary to waterflow, where the resistance (pressure drop) across a block changes with the flow rate, the electrical resistance of the heater will remain (essentially) the same, regardless of voltage. (a poor analogy, I know).

Yeah I understand this. Just like a resistor as pH said.

All in all, the equation V=R * I (V: voltage, I: current (amps) and R: Resistance (in Ohms)) is always true. It's a law, just like many other things in physics.

Math sucks. Let me try and put this to were I can understand it better. Resistance x Current = voltage?

So if I have 10ohms of resistance and 5 amps it equels 50 volts?

So assuming that the heater will put out 150 Watts, at a voltage of 40, using the other equation (also a law) where P=V * I, you can see that the current that the heater will draw is 3.75 amps.

Ok, so to get the amps from 150watts at 40VDC into amps I would divide 150 by 40 to get 3.75amps?

Back to the first equation (V=RI), you can then calculate that the resistance of the heater is ~10.7 Ohms. This does not change (well, just a little bit, maybe up to 15%, but just because it's a heater).

Ok.

So if you reduce the voltage to 12, and applying V=RI, where R is (still) ~10.7 Ohms, you would figure out that it will equal about 1.12 amps. Now knowing the amps, and using P=VI, you can then calculate the power output to be ~13 Watts.

Ok.
But back to your question...

Power supplies usually work in one of two ways: they allow you to control either the voltage (most common) or the current.

If you control the voltage, the resistance determines the current.
If you control the current, the resistance determines the voltage.

So your power supply may do either, but not both. Perhaps the "amp" setting is some kind of failsafe trigger, where if it's exceeded, it shuts down?

All the ones I been looking at have Amperage and Voltage adjustment. I.E.: http://www.alltronics.com/images/ac350.JPG

I belive you maybe correct though that it is just a current limit and not an actual setting. Makes since... This stuff doesn't click when you are ignorant of other important info like above as I am.

It's not so bad.

Theoretically (and someone correct me here), you simply use your voltage measurement's margin of error, in %, and add it to your current measurement error margin (also in %), then add 1 or 2% for secondary heat loss (better to quantify it yourself, but not easy), and you have a relatively accurate measurement of the power that your testbench applies, complete with error margins.

Will have to dig into this later.