bennyp
Newbie level 3
Hello
This is probably a result of overthinking, but I really want to understand this circuit.
I'm unable to post links, but the circuit is the common circuit shown in the datasheets for the MOC series, and on the SimpleIO website "TriacOut Series Gate Resistor Application Note"
(although I will be replacing the MOC isolator with a non-zero crossing one like MOC3022 or so, but the circuit on that page is almost the same as shown in the datasheet for the MOC3022)
They show calculations saying 1/4 watt would be ok. But:
When the AC current hits a zero crossing, some circuit (in my case, a microcontroller) lights LED lamp inside the isolator. The gate on the triac inside the isolator is triggered (ignoring the small phase delay to get above gate current), and current is allowed to flow through the isolator. Current is then pulled through the resistor, through the isolator and on to the external triac's gate. But what happens when the LED gets turned off?
Won't the triac inside the isolator keep conducting until the next zero crossing, being a triac and all? Therefore, the entire brunt of 120V rms would be sent through the resistor, and the power required would be (120V)^2/180ohm = 80 watts, which is certainly unrealistic.
This seems obviously wrong, but I'm not sure where. Does the gate on external triac stop drawing current after it has been successfully triggered?
Thanks
Ben P.
This is probably a result of overthinking, but I really want to understand this circuit.
I'm unable to post links, but the circuit is the common circuit shown in the datasheets for the MOC series, and on the SimpleIO website "TriacOut Series Gate Resistor Application Note"
(although I will be replacing the MOC isolator with a non-zero crossing one like MOC3022 or so, but the circuit on that page is almost the same as shown in the datasheet for the MOC3022)
They show calculations saying 1/4 watt would be ok. But:
When the AC current hits a zero crossing, some circuit (in my case, a microcontroller) lights LED lamp inside the isolator. The gate on the triac inside the isolator is triggered (ignoring the small phase delay to get above gate current), and current is allowed to flow through the isolator. Current is then pulled through the resistor, through the isolator and on to the external triac's gate. But what happens when the LED gets turned off?
Won't the triac inside the isolator keep conducting until the next zero crossing, being a triac and all? Therefore, the entire brunt of 120V rms would be sent through the resistor, and the power required would be (120V)^2/180ohm = 80 watts, which is certainly unrealistic.
This seems obviously wrong, but I'm not sure where. Does the gate on external triac stop drawing current after it has been successfully triggered?
Thanks
Ben P.
