Need help with AC switching and PWM

[pebe]

Here is the circuit. As your Arduino board will be grounded to the coffee machine, it needs an optocoupler to the triac. Your HCPL817 will be suitable. I have replaced the ST micro in the AppNote with a CMOS 555 timer, wired as a bistable, to keep the load on the optocoupler to a minimum. The triac is switched on when the gate is negatively pulsed. So when the LED in the optocoupler is powered, the triac is triggered on. The 555 is powered by a diode pump. It works like this.

It is easier to explain if the ‘Line - A’ is considered to be at a fixed voltage. The ‘Neutral - B’ is then a 220V sinewave relative to it. 220VRMS is about 300Vpk so consider that B goes 300V above and below A.

C1 and C2 are both discharged at the start. When B is +300V relative to A, C1 will charge via R1 and Z1 (Z1 acting as a normal diode) to 300V and its RH side will be negative.

As the polarity changes, the voltage on both sides of C1 starts to fall, turning off Z1 and turning on D1. As the voltage on B goes from +300V to –300V, C1 discharges to zero then charges with the opposite polarity through D1 and C2. The charge that was on C1 will now be split between C1 and C2. As C2 is 500 times larger than C1, the voltage across C2 will increase by 600/500 = 1.2V, with point C being negative to A.

As the process is repeated in successive cycles, the voltage across C2 will get progressively larger until it reaches 5.6V, at which time Z1 will bleed away the charging current. So C1 effectively ‘pumps up’ C2. The current that can be drawn will be a function of the sizes of C1, C2 and R1 but I don’t know the law it will follow. The original was intended for 50mA pulses of 50mS duration.

The 555 needs to be the CMOS version, because the output of the standard bipolar type will not swing as high as V+.

I hope that may be useful for you.

[Bhante]

Many thanks for your help.

Can I substitute the ACS120 in here? In this case maybe I don't need the 555 because the gate triggering current is less than 10mA? I'll still have to get a couple of parts before I can try it though (R1 and C1).

Does it matter if L and N are switched? I think in the UK you can usually rely on which terminal is live and which is neutral, but here in Germany not. The plugs are symmetrical, and an electrician once told me they don't bother to distinguish in the socket wiring.

Bhante

[pebe]

Yes, you can use the ACS120, and it doesn't matter about thepolarity of the input.

[Bhante]

[quote="pebe"]Yes, you can use the ACS120, and it doesn't matter about thepolarity of the input.[/quote]

With the ACS120 the gate current is less (<10mA), so this should be OK should it?

[pebe]

I have modified yoiur circuit slightly so that the optocoupler switches the gate to -5V relative to COM, via 140ohms (can be 120ohms) as recommended in the data sheet .

[Bhante]

Many thanks Pebe.

[quote="pebe"]I have modified yoiur circuit slightly so that the optocoupler switches the gate to -5V relative to COM, via 140ohms (can be 120ohms) as recommended in the data sheet .[/quote]
Ah I see, my circuit wouldn't have worked because of the direction of the transistor, and because the gate voltage would be almost +5V. No. Wrong in both cases I think.

[quote="datasheet"]This ACSâ„¢ switch is triggered with a negative gate current flowing out of the gate pin G.[/quote]
This means electrons flow out of the gate, or conventional current flows into the gate. "Negative current" is a somewhat unusual phrase isn't it? It is a little confusing actually because in my circuit the gate is nearly +5V relative to A, whereas in yours the gate is nearly 0V relative to A. What have I misunderstood?

Another practial question - would you advise putting a capacitor across the terminals of the Ulka pump or something similar to reduce transients? By the way the Ulka has a built-in diode, so that it only uses half the AC cycles.

Bhante

[pebe]

Yes, this direction of current flow is always ambiguous. But unless electron flow is specifically stated, I always assume that the flow of current is the 'conventional' flow from the +ve pole of the supply through the load to the -ve pole of the supply.

As I read the diagram on page 4 of the manual, an ST72 is shown so the inference is that its Vdd is connected to the neutral terminal and Vss would be at -5V. So the gate fed from the ST72 output would be either 0V or -5V.

Long ago when I started in electronics, it was the accepted thing to draw a circuit with voltage graded from +v at the top to -ve at the bottom, and inputs on the left and outputs on the right. A resistor drawn vertically would carry current and one drawn horizontally would not - it would merely be a signal feed.

If ST had used those rules in their drawings then any ambiguity could have been avoided. The line to Vdd could have been shown entering the ST72 box at the top and the Vss taken out at the bottom. Its output to the gate could have been shown leaving the box on the right hand side.

I don't know how the pump operates. It might operated by some form of vibrating plate where the diode is imperative, but if may be best to consult the manufacturer.

[Bhante]

Many thanks for all your help.

Bhante

[miketaylor]

Are you feeding your circuit with an AC sine wave generator or a DC square wave generator?