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Problem with simulation

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maryem

Member
Hello, i'm doing a project to control a motor with a circuit witch i send it in a PDF:
a part of alimentation 5V
a part of order is the pic16f1508
a part of interface with the network RS485
a part of isolation with the MOC3023(OK3/OK4)
a part of power with the BTA6 (T1/T3)



but when i construct it, i don't found 220V in the terminal of T1 and T3, i have only 9V. I don't know the origine of this problem. Can someone help me
 

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ericgibbs

Well-Known Member
Most Helpful Member
hi M.
What is connected to X3 1 & 2 and X4 1 and 2.?

Also, where are you measuring T1 and T3 relative too, ie: where is the voltmeter ;from and too'

E
 

maryem

Member
X3 voltage 1&2 220V, X4 1 output to motor ( sens +), X4 2 output to motor (sens-)
i put the voltmeter in the output of résistance 360Ohm
 

alec_t

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Most Helpful Member
Which version of the BTA06 are you using? Some require a gate trigger current which may be higher than your 3023 can provide.
How do you know if your PIC code is working correctly?
 

Tony Stewart

Well-Known Member
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You need to test the voltage w.r.t. gnd on all pins around opto and triac to find why switch does not perform to specs.
 

maryem

Member
the input of opto is ok , i have 1.2V which it has the marge 1.15V and 1.5V
but the output of opto is 18V and the output of triac is 18V
 

Tony Stewart

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Most Helpful Member
MOC3022 has specs for input trigger 5mA typ, 10mA max.

- If you design for nominal, some parts won't work.
- Therefore your current must exceed 10mA but be much less than 50mA abs. max.
- specs for Ift =1.5V max @10mA @25'C
- specs for PIC Out V_OL are 0.4V max @ 6mA @5V ...=> ESR=0.4mA/6mA=67Ω max (50 typ)
- assuming 5.00V For Ift=10mA R27,26 = (5-1.5)/10mA=350Ω
- subtract 67Ω due to driver... 350-67=283Ω
- change 470Ω to 270Ω
This is worst case at room temp with 5% margin, if you need more margin, consider R=240

Then check differential voltage output and input to Triac gate current against specs the same way for the quadrant used .

Note Quadrant 1 is better for triggering Triac, which means low side switching ( load between
 
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maryem

Member
Can i save the resistor 470Ohm and I change the resistance between yhe triac and optocoupler to 1000 Ohm like this ?:
upload_2016-9-13_13-46-24.png
 

Misterbenn

Active Member
If by save you mean not include the 470Ohm (R1) resistor, then no. That resistor is required to limit the current through the opto LED.
 

Misterbenn

Active Member
Also your motor needs a return path, at the moment you are only connecting to X4-1/2 to X3-1. I presume the motor has a neutral connection?
 

maryem

Member
i mean to change the 360 ohm resistot to 1K and i kept the 470 ohm resistor in her place .yes the neutral connection of the motor is arriving from the sector
 

Tony Stewart

Well-Known Member
Most Helpful Member
not for worst case, why? measure all voltages and currents from R drop

The triac needs twice the gate current in your configuration (quadrant 4 (IV))
Better is put 2kW load on other side of triac., A2, like in all datasheets examples.
 

alec_t

Well-Known Member
Most Helpful Member
i mean to change the 360 ohm resistot to 1K and i kept the 470 ohm resistor in her place
That will probably restrict both the LED current and the triac gate current too much.
 
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