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| Need some help building a circuit involving switchs and relays |
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| Renate:
Well, the original circuit is not bad, as long as you replace the thermistor when it fails. The good thing is that even if the thermistor fails the relay will operate correctly. I think that that 3 pin connector on the edge that has a big solder blob on it was designed to go to a light bulb on a panel to indicate "too many bulbs on". Yes, you definitely want a circuit that can differentiate: [*]No bulbs on [*]Some bulbs on (less than 10 or 20) [*]Too many bulbs on[/list] Even flashing the bulbs would take as much as 12 A. If the circuit was DC this would be easier to supply a peak current. Do you know if the models that flash use DC instead of AC? |
| Shaydzmi:
Thank you for following this long thread! --- Quote ---The good thing is that even if the thermistor fails the relay will operate correctly --- End quote --- Can you please explain to me in details what does this triac do in this circuit? Please consider that I'm not an electrician by any mean, I'm just a learner, so don't assume that I am knowing much. --- Quote ---Even flashing the bulbs would take as much as 12 A. --- End quote --- Did you get this value from where? 152x50mA= 7.6 or am I wrong? --- Quote ---Do you know if the models that flash use DC instead of AC? --- End quote --- Yes the wire to the bulbs is carrying AC, the board you know it it AC an DC, maybe, if you want, I can bring that board here also, and you can teach me how it works. |
| Renate:
Let's say that you want to build a weight scale that can weigh both a mouse and an elephant. It will have to use a very light spring to detect a mouse. But somehow you want a very strong spring that can tolerate an elephant. The triac circuit is like that. The 470 ohm resistor is the light spring. If you draw 50 mA (light bulb) or even just 25 mA (the bypass resistor) you'll be able to see the effect. (25 mA * 25 mA = 11.75 V) But what if a couple of Amps want to flow through that? That's where the triac steps in. It says "Wow, that little spring is being stretched all to heck, I better step in and start pulling." 12 Amps is all the light bulbs AND all the bypass resistors. There are ways that we can cut down all the current when too many bulbs go on, but we also want to make sure that the operator can see that is what is happening. |
| Shaydzmi:
Thank you. Ok, I see, the triac then only feeds the bulbs, it has nothing to do with operating the relay. The resistor exists to feed the transistors driving the relay? --- Quote ---12 Amps is all the light bulbs AND all the bypass resistors. --- End quote --- The bypass resistors shouldn't be considered, because they won't work as long as the bulbs are working, and we won't let all of them burn . |
| Renate:
--- Quote from: Shaydzmi on June 04, 2020, 10:36:36 pm ---The bypass resistors shouldn't be considered. --- End quote --- They must be considered or you will hurt their feelings! :P The bypass resistors are in parallel with the light bulbs. Current goes through them whether the bulb is there or not. Some of my numbers were a bit off too, so let's do it again. The light bulbs are 1.2W, 24V. Dividing, we get a current of 50 mA. The bypass resistors ar 1.2k. 24V / 1.2k = 20 mA That's a total of 70 mA. 76 feeders, 2 circuits each = 152 * 70 mA = 10.64 Amps. |
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