0
\$\begingroup\$

I am trying to build speed control for 500W synchronous, single-phase AC induction motor. It is for a fan in my greenhouse.

Here is current design: TRRIAC control

I am trying to verify it on breadboard with resistive load (100W lightbulb). The zero crossing detection circuit works well, but TRIAC trigerring does not work.

Notes:

  • R5 is set such that there is ~12mA current going through the LED in the optotriac. It is above maximum triggering threshold for MOC3052 and well below its current rating (60mA)

  • R8 is set such that the maximum current through the optotriac is always below 1A limit (on 230VAC mains)

  • I have tried with BT139-800E TRIAC (on schematic, with sensitive gate) as well as with BTA140-800

Where have I gone wrong? Could you recommend robust/reliable triggerring circuit?

Thank you,

\$\endgroup\$
  • \$\begingroup\$ Remove U3 and put a regular LED in place of the opto-LED. Confirm that this blinks. \$\endgroup\$ – Transistor Aug 22 '16 at 20:23
  • \$\begingroup\$ Yes it does. Looking at it with oscilloscope the timing is also correct. \$\endgroup\$ – student Aug 22 '16 at 20:42
  • \$\begingroup\$ Time for some photos of the setup then. Schematic looks OK. Also, give us some info on the timing of your trigger pulses. \$\endgroup\$ – Transistor Aug 22 '16 at 20:53
  • \$\begingroup\$ The timing: wait for 1 ms after zero crossing and create 0.1ms trigger pulse. I have verified this with osc. \$\endgroup\$ – student Aug 22 '16 at 21:38
  • \$\begingroup\$ what is ZCS pulse width ? This should be as short as possible so your delay is gauranteed to be after Zero crossing or use the trailing edge with no delay. \$\endgroup\$ – Sunnyskyguy EE75 Aug 22 '16 at 21:56
1
\$\begingroup\$

Just some notes: \$I_F=\dfrac{V_{CC}-1.2V}{R}\$, you get \$ I_F=\dfrac{(5V-1.2V)}{330}\approx11.5mA\$ The reccomandation is to use a current between 10mA and 60mA, so your at the minimum. Consider that browning will reduce the CTR for 50% after years of use.
Using MOSFET may complicate things, since there is a need of only max. 60mA, a small signal BJT NPN transistor should be just fine, but it's up to you if you prefer MOSFETs....
Make sure that MT1 and MT2 aren't swapped, short the transitor Q1 so that LED will turn on - in such way TRIAC can't remain turned off. Replace the MOC if it doesn't work.
At desired angle turn the LED on and turn it off at an angle close to zero cross, for example at 150 degrees. Add a bleeding resistor accross LED (1.1k) to reduce turn-off time, recalulate the series resitor.

\$\endgroup\$
  • \$\begingroup\$ It was not working due to swapped MT1/MT2, thanks \$\endgroup\$ – student Aug 23 '16 at 9:40
1
\$\begingroup\$

Is it firing all the time? or not at all? Short out U3 output to test the triac for full turn on. Are you firing with a pulse or delayed DC after ZCS pulse.

To minimise noise pickup, keep gate connection length to a minimum. Take the return directly to MT1 (or cathode). If hard wired, use twisted pair or shielded cable. Fit a resistor of 1kΩ or less between gate and MT1. Fit a bypass capacitor in conjunction with a series resistor to the gate.

If 60mA is the worst case trigger current at low voltage with R8=470 this would be a trigger voltage of 28V should fire in all 4 quadrants.

\$\endgroup\$
  • \$\begingroup\$ It does not fire at all. When I short the gate to MT1 then the main TRIAC latches. \$\endgroup\$ – student Aug 22 '16 at 20:46
  • \$\begingroup\$ How did you calculate the trigger voltage? \$\endgroup\$ – student Aug 22 '16 at 20:49
  • \$\begingroup\$ Is MT1 is the emitter so when shorting Gate to MT1 should disable it. The shunt R helps decay leakage current. I meant shorting the opto output switch so the 470 ohm resistor achieves up to 60mA at I*R=28V for low duty cycle. It should trigger before this and you can monitor the gate current and voltage above 0.7 to determine when it latches. \$\endgroup\$ – Sunnyskyguy EE75 Aug 22 '16 at 21:35
  • \$\begingroup\$ I might be mistaken here, but I would be feeding the gate with up to 0.75A during one half period since peak voltage is 350V and the gate behaves as PN junction ... this might stress the gate above its power rating. \$\endgroup\$ – student Aug 22 '16 at 21:48
  • \$\begingroup\$ Sorry, it wouldn't -> 0.7V drop on the gate and 0.75A going through gives is about 0.5W peak power. \$\endgroup\$ – student Aug 22 '16 at 21:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.