Using a optioisolated trac with an MCU

[bunrockter]

Using an optioisolated triac with an MCU

I am designing a project that switches ac loads using a micro controller. I have never used triacs before, but i decided the best route to go was to use one that is optically isolated. The device I am currently considering using is a VO2223A https://www.vishay.com/docs/81924/vo2223a.pdf.

It says in the document that the trigger current is 10mA. and that the hold current is 25 mA. Never having used a triac before, I am assuming that it takes 10mA of current to the LED to turn the gate on and 25 mA of current to keep the triac switched on. Is this correct?

I saw the 10mA and I was excited because I was hoping to directly drive this with a digital output from my MCU. but if the holding current needs to be 25mA and I make it 30 for good measure, then I am not sure if it is a good idea to source the LED current from my MCU.

Unfortunately my MCU datasheet doesn't actually tell me how much current it can source, it only tells me what the short circuit current is and that appears to be 86.5 mA. I don't know how to divine from that if it is a good idea to try and get my MCU to source the current into the triac.

In general is it a bad idea to directly drive LEDs and such with MCU pins?

Thanks,

Bunrockter

 

[bunrockter]

Re: Using an optioisolated triac with an MCU

After looking at the datasheet again I think the I-hold it is referring to is the minimum current that the out put load needs to sink in order to keep the triac on. Is This correct?

 

[flame0510]

In my opinion you can use a BJT to amplify the current

 

[bunrockter]

Re: Using an optioisolated triac with an MCU

I was trying to keep the component count and cost down, and I was hoping not to have to use one. Is it bad practice to connect a MCU digital output pin directly to a LED or similar device?

 

[cubanflyer]

What you need to rememer is the available current on the output, also by not directly driving a device directly, you are providing a degree of protection for the MCU.

 

[bunrockter]

What you need to rememer is the available current on the output, also by not directly driving a device directly, you are providing a degree of protection for the MCU.

Thanks for your reply. Are you saying the protection is provided by the phototriac, or by adding a bjt between the phototriac and the digital pin on the mcu?

 

[Kurenai_ryu]

Thanks for your reply. Are you saying the protection is provided by the phototriac, or by adding a bjt between the phototriac and the digital pin on the mcu?

by adding a bjt between the phototriac and the digital pin on the mcu - as @cubanflyer & @flame0510 states...
consider using a small soic or tsop transistor...

 

[bunrockter]

by adding a bjt between the phototriac and the digital pin on the mcu - as @cubanflyer & @flame0510 states...
consider using a small soic or tsop transistor...

Parden my ignorance, but how does the bjt protect the mcu, and how many mA will it take the mcu to drive the bjt? It only takes 10 mA to drive the phototriac.

Thanks

 

[Kurenai_ryu]

asumming a common BC548 (IC=100mA or so) in common emiter (for saturation purposes, let's asume a beta of 50) to get IC=10mA you need IB=10ma/50 = 200uA
to get 200uA from mcu pin,(asuming 5V mcu io, and common 0.7V Vbe drop) (5-0.7)V / 200uA = 21.5 kohm ... let's put a mere 22kOhm or a common 20Kohm (if you don't mind calculating, just do the 10K rule)

so, you will use 200uA from Mcu pin, if you drive a BC548 with a 20kOhm resistor

(yeah yeah, the most puritan electronics students will argue many things in all of this, but "hey, it works" and will drive your led in the optotriac (BUT don't forget to protect your led with a suitable resistor!!!))

BTW, it will protect your mcu in the way that: you don't draw too much current from the mcu (so it will not heat) and if something happens to the optotriac, the first one to burn will be the transistor and not your mcu.

 

[bunrockter]

What about using a logic level MOSEFT instead of a BJT?