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I have a circuit that does not work as expected. It uses a MOC3023 phototriac to directly control a 12V AC pump.

Since the motor is very low power I assumed I wouldn't need an external triac. This may very well be my mistake, but I still want to use the situation to learn from my mistakes and directly understand the problem.

I have an AC LED module (with rectification) that I use to test. It can be controlled using my circuitry without problem. When I connect the pump instead though, once the phototriac is triggered it latches and won't turn off again.

I then replaced the MOC3023 with a MOC3043, wich is the same plus zero crossing detection, since I thought the inductive load may cause some issues there. With absolutely no luck, the result is the same.

My two questions are is the external triac absolutely mandatory and if so, why? The second one is why does the phototriac latch up when a inductive load is present?

schematic

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – Voltage Spike Apr 8 at 14:57
  • \$\begingroup\$ Thanks, no further discussion took place, now the valid information is gone, great work! \$\endgroup\$ – Julian F. Weinert Apr 8 at 14:58
  • \$\begingroup\$ "Comments are disposable: unlike posts, there's no public revision history, and they can be deleted without warning by their authors, by moderators, and in response to flags." meta.stackexchange.com/questions/19756/how-do-comments-work In addition the comments are not deleted, they are moved to chat, where anyone can view them. SE does not want comments detracting from answers. If there is information in the comments that is necessary for understanding edit the question or answer \$\endgroup\$ – Voltage Spike Apr 8 at 15:00
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Thought experiment

Some small water pumps use a lot of current then with motion they cutout and return with a spring like an oscillating solenoid and use AC with a diode to pump at the line frequency on a half cycle. The contacts have hysteresis so you have a relaxation oscillator that is very noisy unless submerged. So if it is a small aquatic pump measure the DC resistance in both polarities and confirm my hunch.

The Tiny uC might be able to drive the emitter with 10mA at 2.1V from 3.3V dropping 400mV @ 10mA from Figure 22-21 in the atTiny85P spec meaning it has a “typical” Ron of 40 Ohms at Vdd=3V so consider that as a max for 3.3V. Thus a 2.9Vout to 2.1Vf = 800mV/10mA means you need a 80 ohm series instead of 680.

From fig 3 of the Triac spec, it has an output resistance slope of a nominal value = 3.125 Ohms plus 2 diode drops. So once you fix the trigger current, now you have the pulse current to worry about burning up the Triac from I^2*DCR resistance of the pump coil. This current reduces with velocity of BEMF but with a 250mW power limit in open air @ 25’C, you may be out of luck. 5W average could be as much as 20W peak for 25% df (guesstimate).

  • this is all based on Ohm’s Law and linear behaviors in conduction mode. and the nonlinear current of aquatic pumps like all solenoids and motors alike with V/DCR surge currents.
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  • \$\begingroup\$ The pump measures 27.6Ω both polarities and is a 2W version. The part with the resistances went a bit over my head. The 3043 / 3025 have a maximum trigger current of 5mA, assuming 3.3V - ~1.3V gives me 400Ω at 5mA or with the 680Ω would be 3mA. \$\endgroup\$ – Julian F. Weinert Apr 7 at 3:54
  • \$\begingroup\$ I misread the data sheet and you're right, I should drive more current, at least 5mA for my variant. Now what did you mean with 40Ω and 2.9V out? Can you please clarify? Thanks \$\endgroup\$ – Julian F. Weinert Apr 7 at 3:56
  • \$\begingroup\$ Read again. See Fig.22-21 and compute RdsOn , the slope of V/I of 5V typ.logic (+/-25%)= 40 ohms thus 400mV drop from 3V or 3.3V @ 10mA. 3.3-400mV=2.9V \$\endgroup\$ – Tony Stewart EE75 Apr 7 at 11:34
  • \$\begingroup\$ If you “Linearize” the model of the Vo , Ron, for CMOS then you can use Vdd otherwise use the typical plot. and same for LED, Vth+I*R=Vf (@ T) you can compute a much more accurate current or Rs value for a desired nominal or worst case If min. Using KVL. Especially with low voltage drops, so 680R is way off \$\endgroup\$ – Tony Stewart EE75 Apr 7 at 11:49
  • \$\begingroup\$ (3.3-2.1) (Nom @ 25’C) / 680R = 1.6 mA.. mind you Vf has a NTC while RdsOn has a PTC which tends towards cancel almost. (2.3cool - 2.1 hot= -200mV)... so -0.2V/10 mA= 20 Ohms equivalent change from Shockley effects on diode NTC, you can confirm with DMM across 80 Ohms to get 800 mV or not. Given there are tolerances on Rled and RdsOn \$\endgroup\$ – Tony Stewart EE75 Apr 7 at 11:55

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