Simple phase angle controller circuit

Thread Starter

E_enthusiast

Joined Aug 20, 2020
31
Hi everyone, i am trying to make a simple phase angle controller circuit to control the speed of my Ac motor.. my question is can a triac control directly the motor? is there any improvement i can make to the circuit? Thanks for the help in advance.
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Ian0

Joined Aug 7, 2020
3,238
Hi everyone, i am trying to make a simple phase angle controller circuit to control the speed of my Ac motor.. my question is can a triac control directly the motor?
Yes, provided that it is a universal motor. It won’t work quite so well on an induction motor, and not at all on a synchronous motor.
is there any improvement i can make to the circuit?
1. a snubber across the triac (say 100Ω and 100nF class-X)
2. you can probably leave out R7.

Note: a common mistake is not to give the triac long enough to switch off. Turn off the opto current at 9ms after zero-crossing (for a 50Hz mains)
 

Thread Starter

E_enthusiast

Joined Aug 20, 2020
31
Yes, provided that it is a universal motor. It won’t work quite so well on an induction motor, and not at all on a synchronous motor.

1. a snubber across the triac (say 100Ω and 100nF class-X)
2. you can probably leave out R7.

Note: a common mistake is not to give the triac long enough to switch off. Turn off the opto current at 9ms after zero-crossing (for a 50Hz mains)
thanks for this .. i really appreciate it.. can you please tell me why should i leave out R7?
 

Ian0

Joined Aug 7, 2020
3,238
thanks for this .. i really appreciate it.. can you please tell me why should i leave out R7?
You need 25mA to trigger the triac. There will be enough current to trigger the triac through 100Ω provided that the instantaneous mains voltage is greater than 2.5V.
if you switch at mains peak the poor old MOC3051 has to take 3.25A for the microsecond or so while the triac thinks about switching on. With two resistors, it’s 6.5A. Admittedly, more current will hurry it up a bit.
There‘s no spec in the datasheet for peak currents that short, but let’s not tempt fate.
 

Thread Starter

E_enthusiast

Joined Aug 20, 2020
31
You need 25mA to trigger the triac. There will be enough current to trigger the triac through 100Ω provided that the instantaneous mains voltage is greater than 2.5V.
if you switch at mains peak the poor old MOC3051 has to take 3.25A for the microsecond or so while the triac thinks about switching on. With two resistors, it’s 6.5A. Admittedly, more current will hurry it up a bit.
There‘s no spec in the datasheet for peak currents that short, but let’s not tempt fate.
if i may ask, how can you calculate the resistor for the triac gate? apparently i have done it wrong because it gave me a resistance value of 50ohm, that's why i put two 100ohm resistor in parallel. Also can you kindly tell me why a triac can't drive an asynchronous motor?
**EDIT**
 

Ian0

Joined Aug 7, 2020
3,238
Also can you kindly tell me why a triac can't drive an asynchronous motor?
presumably you mean a synchronous motor?

It can, but it can only switch it on and off. You can’t vary the speed. The clue is in the name -it’s a synchronous motor, because the rotation of the rotor is synchronised to the rotation of the magnetic field in the stator.
The magnetic field in the stator rotates at the mains frequency divided by the number of pole pairs.
if you then vary the current (by phase-firing the triac), the torque varies with the current.
Either:
there is enough torque to rotate the shaft load, and it runs,
or:
there isn’t, and it stalls.

If you want to vary the speed of a synchronous motor, you need a variable frequency drive.

Induction motors run at synchronous speed minus the “slip”, which is generally about 5%, so you often see speeds of around 2800rpm for a 2-pole motor. The slip varies with the torque, so by controlling the current you can change the speed by varying the slip. It works better of some types of load (e.g. fans) but not well on others. The VFD is a better choice.
 

Thread Starter

E_enthusiast

Joined Aug 20, 2020
31
presumably you mean a synchronous motor?

It can, but it can only switch it on and off. You can’t vary the speed. The clue is in the name -it’s a synchronous motor, because the rotation of the rotor is synchronised to the rotation of the magnetic field in the stator.
The magnetic field in the stator rotates at the mains frequency divided by the number of pole pairs.
if you then vary the current (by phase-firing the triac), the torque varies with the current.
Either:
there is enough torque to rotate the shaft load, and it runs,
or:
there isn’t, and it stalls.

If you want to vary the speed of a synchronous motor, you need a variable frequency drive.

Induction motors run at synchronous speed minus the “slip”, which is generally about 5%, so you often see speeds of around 2800rpm for a 2-pole motor. The slip varies with the torque, so by controlling the current you can change the speed by varying the slip. It works better of some types of load (e.g. fans) but not well on others. The VFD is a better choice.
oh yeah i got it now.. i totally forgot about the sync speed and all that stuff, thanks for clearing things up. About the resistor i still don't get it.. the gate voltage of the triac is 1.3V and the maximum current is 35mA ( but i am gonna limit it to 30mA) so i need a 43ohm resistor --> 50ohm.. why should i put only 1 100ohm resistor instead of 2 in parallel to get 50ohm? Thanks in advance
 

Ian0

Joined Aug 7, 2020
3,238
if i may ask, how can you calculate the resistor for the triac gate?
The available voltage divided by the current required.
The minimum current is specified in the datasheet as 35mA (I thought it was 25mA in my previous reply), the maximum is 4A.
The available voltage is simply the instantaneous mains voltage which varies from zero at either end of the cycle to (325V-1.3V) at mains peak.
With an opto triac, you can‘t trigger at zero volts, because there is no current to trigger it, so you have to assume a minimum usable phase angle. It’s a compromise between the maximum trigger current at mains peak, which might overload the opto triac, and the minimum usable phase angle.
I would take 1A as a cautious limit, and use R=330Ω.
That would mean that it won’t trigger below V=13V, which is a phase angle of 2.3°
 

Thread Starter

E_enthusiast

Joined Aug 20, 2020
31
The available voltage divided by the current required.
The minimum current is specified in the datasheet as 35mA (I thought it was 25mA in my previous reply), the maximum is 4A.
The available voltage is simply the instantaneous mains voltage which varies from zero at either end of the cycle to (325V-1.3V) at mains peak.
With an opto triac, you can‘t trigger at zero volts, because there is no current to trigger it, so you have to assume a minimum usable phase angle. It’s a compromise between the maximum trigger current at mains peak, which might overload the opto triac, and the minimum usable phase angle.
I would take 1A as a cautious limit, and use R=330Ω.
That would mean that it won’t trigger below V=13V, which is a phase angle of 2.3°
what about the power of the resistor? a 1W resistor is enough?
 

Ian0

Joined Aug 7, 2020
3,238
what about the power of the resistor? a 1W resistor is enough?
Peak power = peak voltage x current = 325V * 1A
but it only lasts for a couple of microseconds (the time taken for the triac to latch) every 10ms.

so 1 * 325 * (2*10^-6/10*10^-3) = 65mW

but watch out for the voltage rating.

You will end up with a 1W resistor or two ¼W resistors in series just to get enough voltage rating.
 

Thread Starter

E_enthusiast

Joined Aug 20, 2020
31
Peak power = peak voltage x current = 325V * 1A
but it only lasts for a couple of microseconds (the time taken for the triac to latch) every 10ms.

so 1 * 325 * (2*10^-6/10*10^-3) = 65mW

but watch out for the voltage rating.

You will end up with a 1W resistor or two ¼W resistors in series just to get enough voltage rating.
thank you very much, i really appreciate this.. it helped a lot
 
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