220VAC to TTL converter

Question

I'm building a circuit that sends TTL level signal to uP whether AC signal exists or not. The schematic is as follows:

Actually, the circuit almost works as I expect; however, there occurs some coupling effect on the emitter side of phototransistor; which cause a delay at the output signal. I also added SPICE simulation of Vout and Vb of BC847B.

Green is Vout, and blue is VbQ1. Vout of the circuit is directly connected to I/O pin of my uP. Two questions arise here:

1) How can I prevent this coupling effect? (I've already tried to couple Vcc to emitter pin of optocoupler.)

2) What is the reason of periodic 90mV increase observed on the output signal? I don't think that it will affect the low-level sensing of uP (Vol = 0.3Vcc); however, it is better to supply a clean zero volt to the uP, I suppose.

Thanks in advance.

Answer 1

You're probably fine in regards to the uP's input, but if you're really concerned, you could use a schmitt trigger buffer in between. They're designed for noisy inputs. Or if your particular uP has a built-in comparator as part of its analog stuff, then you could use that also.

As for the ripple, it's because the circuit is only getting pulsed once per AC cycle. Between pulses, it's no different from having lost it entirely. So you have a tradeoff between analog ripple and response time, neither of which can be zero. Such is the problem with sensing AC.

Another approach might be to intentionally pulse the uP at the AC frequency, then have two interrupts:

1. External interrupt clears a timer and sets the AC_present bit.
2. Timer overflow interrupt clears the AC_present bit.

If you care about line frequency, you could measure it pretty easily by recording the timer before you clear it and then doing some math with that number.

Answer 2

Let's take the questions in reverse order:

2) What is the reason of periodic 90mV increase observed on the output signal?

The simplest explanation goes like this: because you are only driving the optocoupler in one direction, the output of the optocoupler/transistor chain is (essentially) a 50 Hz square wave. This is then filtered by R3/C2, but the time constant is only 4.7 msec. Since the on and off periods of 50 Hz is 10 msec each, there is plenty of time for droop to occur on the output during the time when the transistor is off. If you get rid of C2 and look at the output you'll see what I mean.

1) How can I prevent this coupling effect? (I've already tried to couple Vcc to emitter pin of optocoupler.)

The simplest way is to replace R1/R4/D1 with

^{simulate this circuit – Schematic created using CircuitLab}

This will provide a much longer drive time with only small dropouts as the AC goes through zero.

The other thing you can do is to recognize that you are grossly underdriving your optocoupler. Your peak LED current is about 0.7 mA (220 x 1.414 / 440 k), while the nominal limit for sustained operation is 20 mA. Admittedly, with more current you get more power dissipated in the resistors, but as it stands you're only producing .06 watts per resistor. At the very least, get rid of one of the resistors.

Answer 3

Although what you have presented above uses fewer components and even has the added benefit of mains isolation, another detection circuit that I know of is presented below. Perhaps with added mains isolation between the second stage amp and the peak detector then this circuit could be of use.

^{simulate this circuit – Schematic created using CircuitLab}

This circuit is generally used as a small signal detector by using an ideal diode op-amp configuration, amplification stage, and a peak detector fed into a buffer. You could possibly make the second stage an attenuator instead.

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Or something as simple as a comparator, assuming you have appropriately conditioned the mains signal (AC-DC conversion, for example), and a PNP transistor to convert the positive and negative rail voltages that you would get at the output of an op-amp comparator:

^{simulate this circuit}

In this case, when the conditioned signal is greater than the reference the Op-Amp will rail high and turn the PNP off. In the off state, the TTL_OUT node will be pulled-high through Rc. Likewise, when the signal is less than the reference the Op-Amp will rail low and turn the PNP on. You must design Rb and Rc such that in this situation the transistor will saturate and thereby drive the TTL_OUT node to saturation (~200mV).