Need Help : Diac, triac and resistors.

[rodrigopires]

Hi, i´m making a project and e need a 10k 7W resistor, but the higher i can find is a 10K 5W. If i do what is in  the picture with four 10K 5 W, do i get 10K 10W? Also, i need a 1.5KE68A but can only find a 1.5KE400A. i look in this list (http://www.vishay.com/docs/88301/15ke.pdf) but, dont know what to do. Can i use the 1.5KE400A instead of the 1.5KE68A?
Also, i need a BTA40-600B but again can only find a BTA40-700B. will it work?
Sorry for the work but i dont really know much about each especific component and an opinion from some more experienced people would really help.
Thanks!

[JoeAtl30319]

For the resistors, yes, combining them in that fashion will give you the spec you want - 10k/10watts.  Two 10K resistors in series will give you 20k of resistance with the same 5 watt performance.  Two 20k (2 x 10k in series) resistors paralleled with each other will give you a total 10k value (20k*20k/(20k+20k)).  Also, because current is flowing evenly across each 20k "in series" section, you will be able to dissipate 5 watts in each leg. This brings up the total dissipation capability to 10 watts.

[Paul Moir]

Can i use the 1.5KE400A instead of the 1.5KE68A?

No.  The 1.5KE400A starts to conduct at 400v while the 1.5KE68A starts to conduct at 68 volts.  This is what TVS and diacs do, start conducting at some particular voltage.  So I'm guessing that's probably real important to your circuit.

Also, i need a BTA40-600B but again can only find a BTA40-700B. will it work?

Yes.  The 600 refers to the maximum voltage.  700 is more than 600 so that's fine. 

Now if you don't mind me asking, I'm guessing this is a real big dimmer your building?  The 10W 10K resistor suggests you might be working with a couple hundred volts?  A 40A TRIAC suggests there's little to stop it.   Just want to make sure you know how to play safe. 

[rodrigopires]

Hi, thanks for the help! I will use that resistor schematic and the BTA40-700B. In the attachments is the schematic i hope to do. If i can't use the 1.5KE400B instead of the 1.5KE68A, obviously, is there anything i can use to substitute it?
Thanks!

[eejake52]

For the resistors, in the arrangement you suggested, you can use resistors with a power rating lower than 10W.
There are two ways to determine this:
1) By inspection, each of the resistors will dissipate one quarter of the total power.
2) By analysis, if you have a total of 10W, solve for the current through each leg, then calculate the power for each resistor.

If you want a total of 10W, you could use 4 x 10k @ 2.5W. But since your original requirement was 7W, I suggest 4 x 10k @ 2W.

Jake

[Excavatoree]

A radio engineer friend used to joke that to make a 50 ohm dummy load, you use two 100 ohm DJs in parallel.   I replied that one could also use two 25 ohm DJs in series.  He said "No, they can't take the power."   I don't know if he mispoke, and meant current, or was testing me to see if I realized that it doesn't make any difference.   If both elements have the same resistance, the power will be shared equally regardless of whether the are in series or parallel.

[Paul Moir]

The TVS in the schematic is a 1.5KE68CA, which is bidirectional while the 1.5KE68A is unidirectional like a diode .  'Course you could make it out of two 1.5KE68As back-to-back (or any other TVS).

You might be able to make it out of two 68v zeners like the 1N5373 back-to-back, which is functionally near identical but they're not designed to survive current spikes like a TVS is.  But how it's used in that circuit i don't *think* it matters.

[rodrigopires]

1) By inspection, each of the resistors will dissipate one quarter of the total power.
2) By analysis, if you have a total of 10W, solve for the current through each leg, then calculate the power for each resistor.

If you want a total of 10W, you could use 4 x 10k @ 2.5W. But since your original requirement was 7W, I suggest 4 x 10k @ 2W.

Jake

So you are saying that the power dissipation is divided by the four resistors, and thus if i have 4X 10k 5W i end up with 20W dissipation and with 4X 10k 2W with 8W?

If both elements have the same resistance, the power will be shared equally regardless of whether the are in series or parallel.

I thought that if the resistors are in parallel, the resistance goes to 1/2 and the power dissipation doubles, If they are in series the resistance doubles and the power dissipation is the same, because there is only one current path. I'm wrong?

The TVS in the schematic is a 1.5KE68CA, which is bidirectional while the 1.5KE68A is unidirectional like a diode .  'Course you could make it out of two 1.5KE68As back-to-back (or any other TVS).

You might be able to make it out of two 68v zeners like the 1N5373 back-to-back, which is functionally near identical but they're not designed to survive current spikes like a TVS is.  But how it's used in that circuit i don't *think* it matters.

Thanks, i found a 6w 68V zener diode, is it enough? Also if i make 2X 2 zeners back-to-back and put them in parallel will the current be divided and thus protecting the diodes?

[max666]

Yes, 4 resistors means you can have 4 times the power dissipation (as long as it's sufficiently symmetrical). Remember P = U * I
If you put two resistors (same ohms value) in parallel the voltage across stays the same, but the current gets divided in half for each resistor, meaning the power dissipated in each resistor gets divided in half.
Same goes for two resistors (same ohms value) in series, while the current stays the same the voltage drop across each resistor gets divided in half, meaning also the power dissipated in each resistor gets divided in half.
The 2 x 2 = 4 resistors arrangement divides both U in half and I in half for each resistor meaning the power dissipation in each resistor gets quartered.

Paralleling zeners and expecting them to share the current wont work. Due to the extremely non linear characteristic of zeners and diodes in general, minute differences in zerer voltage (even caused by temperature differences) will result in hugely different currents. That's why paralleling up LEDs without series resistors won't work, the resistor because of it's linear characteristic will somewhat mitigate the nonlinear characteristics of the diodes.
Now I don't understand your circuit, but if there is a series resistor in the current path of your zener diode, you might be able to divide it also up into two, which would help share the current across two zeners. But since the current rating isn't even known, I don't know if two zeners are required or enough.

Boy that was a long post for a whole lot of nothing.

[rodrigopires]

This circuit must be able to take up to 10 amps on full power with AC mains 220-240V, but i dont know where that current passes through. Is 2X 6W 68V zeners back-to-back enough? Also R2, R3, R4 and R5 dont have a wattage value, i'm guessing 2W for each one, is that ok?
Thanks for all the help