Circuit to convert 24VAC digital signal to 5V DC digital signal

Question

I am designing a circuit to convert 24VAC signal to 5V DC signal. The 24VAC signal is designed where 24VAC is on and 0VAC is off. The signal needs to be fed to an MCU which is 5V DC.

The circuit designed has to be very cheap so more expensive ICs and full wave rectifiers are not options.

There are two designs that I came up with

The first design outputs a 5V DC signal for high and about 0VDC for LOW.

The second design eliminates the capacitor but outputs a 5VDC for low and 5VDC square wave for high which will be accounted for in code.

I would like to know if there is an advantage for one over the other considering that the circuit needs to be as cheap and safe as possible since it will be used in an industrial environment and will be heavily affected by dust, debris, and humidity.

Answer 1

Figure 1. R4 is redundant as R1, R3, Q1 and R2 shunt it.

Similarly in the second scheme R1 is redundant.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 2. Solution with component count of 2.

If you use a bi-directional opto-isolator and series resistor you can cut the component count to two. The opto-isolator output transistor will pull low every half-cycle if you use the MCU internal pull-up resistor. Your software will need a timer to check that AC has been lost for 10 ms or so before determining that AC is off.

I have written more on opto-isolators here.

Answer 2

Industrial environments can be harsh from an electrical point of view. If the 24VAC isn't close to the input, the signal can pick up noise and transients, also possible differences in ground potentials, all of which which can cause bad readings or even damage the input. For these reasons, galvanic isolation is often used.

Having said that, for lowest cost, look into using a basic resistor divider with a couple of diodes to clamp to +5V and ground. I would add a small capacitor to reduce noise and help out with transients and ESD.

Functionally, there isn't much difference between the two circuits, except that the first would be better at sourcing current, the second better at sinking current. But since there's a 10k resistor in series with the output, this probably isn't a concern

Answer 3

Assuming a \$5\:\text{V}\$ DC supply rail and assuming that a broad approach is satisfactory, then something like this might be okay:

^{simulate this circuit – Schematic created using CircuitLab}

It's mostly just a AC high pass filter and resistive divider (\$C_1\$, \$R_1\$, and \$R_2\$), followed by \$D_1\$ to protect \$Q_1\$'s base-emitter junction from reverse voltages, followed by a 2-BJT shaper (\$Q_1\$, \$Q_2\$, \$R_3\$, and \$R_4\$), followed by a simple RC lowpass filter/integrator (\$R_5\$ and \$C_2\$), followed by a simple saturated switch that normally is OFF unless there is a sufficient number of low-going pulses at the collector of \$Q_1\$ to pull down the voltage on \$C_2\$. When \$Q_3\$ is active, it pulls the output high. Otherwise, \$R_8\$ pulls the output low.

I've no idea what your source impedance is, but the values of \$R_1\$ and \$C_1\$ can be adjusted per need there.

The design is more "slapped together" than it is carefully crafted, because I have no idea if this achieves any of your goals. If the broad strokes are okay, then fine... more could be done to craft it better. If not, then at least this provides enough for you to clearly say so.

Answer 4

So, I finally got around to drawing up a simple circuit using a bridge rectifier.

As mentioned in the comments, there are SMD bridge rectifiers for low current available for about the price of a single transistor. I can order a bridge rectifier that costs less than a 2N3904 from the same supplier.

The blue section represents whatever is generating the 24VAC.

The red section represents the microprocessor input, with clamp diode to 5V and 100 kOhm as the input impedance.

The AC-detector is really just the three resistors, the rectifier bridge, and the capacitor.

The three resistors reduce the 24VAC to something that will be a little over 5V when rectified.

They also limit the current that can reach your processor in case you get more than you expected on your 24VAC.

This is the simulation:

You can't see it, but the current through the clamp is less than 1mA (red line across the bottom.) So, probably safe enough to let the clamp diodes on the processor's input handle it. If not, a schottky diode is pretty cheap.

The voltage builds up a little slowly. That's about 75 milliseconds before it hits 5V. Ought to be fast enough, though. The decay when the 24VAC disappears ought to be similar.

That's a cheap, dead simple AC detector that offers at least some protection against over voltage.

Probably as cheap as if not cheaper than the transistor based circuits.