12V input on microcontroller pin

Question

I'm trying to count pulses/sec. on a microcontroller pin in the ~5 to 100Hz range. The µC can operate at 5V input, so I have to get the voltage level down safely.

A simple resistor comes to mind, yet that leaves any surges open directly to the µC pin - meh.

I've come across this answer, but the question remains if that circuit is capable of "fast" 100Hz changes.

Is there a proven, reliable way (by means of an IC maybe?) of contacting 5V or 3.3V pins to "dirty" 12V inputs? I have the 12V and 5V available to drive any "ready made" IC.

Answer 1

Use a circuit like this:

^{simulate this circuit – Schematic created using CircuitLab}

R1 and R2 determine the voltage range, and perform the initial division. These resistors must be capable of some power. Typical is MELF 0.4W. All other can be chip resistors/capacitor.

R3 prevents any surges to cause harm to the schmitt trigger. R4 an R5 are optional to prevent any floating signals.
However, the combination R3/R4 can also be used to adjust the threshold, if necessary.

C1 and C2 determine the maximum speed. Combination R3/C2 can filter slow. C1 filters transients.

A separate schmitt trigger is used since you can get them really small and cheap. And it prevents routing a weak signal over long traces. Whilst also being a sacrificial part on major surges.

I've designed this circuit based on what I have seen inside PLC's. Above circuit is for 24V. Adjust resistors to match 12V according to IEC61131-2.

The concept of the standard is to ensure the input has to sink a minimum amount of current before considering it a '1'. The three types specify how much, and are applied based on environmental noise. This prevents glitches from touching it or nearby relays. The drawback is that R1/2 have to be of decent power rating and low resistance.

Answer 2

I would try a resistor divider solution like shown below.

Select the resistor ratio so that the divided voltage is at the proper level for the MCU when the input is at its nominal voltage. The zener diode voltage is selected to clamp the MCU input when the input goes above the max input. The zener will also protect the MCU if the input happens to go negative.

This solution will work great for the relatively low frequency range that you have specified.

Answer 3

If the signal levels are GND and 12V (or >5V), the most simple and 100% safe way is this:

^{simulate this circuit – Schematic created using CircuitLab}

If it really serves your purpose depends on the actual impedance of the 12V signal (should be way below R1) and what you mean by "dirty".

Also, as @MichaelKaras correctly points out, the low level on the µC's input may be shifted up to the low level of the 12V signal plus Vf of the diode (up to about 0.7V). You should check if this is a problem in your case or not. If it is, you can still try and use a Schottky diode with a Vf of about 0.35V.

Answer 4

I would use a resistor divider and then protect the uC with a 5.1v Zener

If you put the zener between the pin and ground in parallel with, say, a 10k pull down resistor, then feed your voltage divided signal in then... zener is more than fast enough, and cheap / easy.

I often do this and divide the signal before the zener bit with a pot.

Other option is as linked, if your really worried an opto could be used, if its not a safety issue I would go with the above or have the pin normally high from 5V Vcc and pull it low with a fet (off top of my head 2N7000 should work) - but its less simple than the zener option.

Answer 5

I'd use an opto-isolator, 100Hz is easily within the range of any decent one. 4n25 springs to mind as a common part number, and I know that's capable of much better than 100Hz.

Answer 6

The method selected partially depends on what the input signal does, how it behaves and how that might effect the input circuit and the code that reads it?

e.g. Is it always 12V? Does it have spikes or noise? How much current can it drive? Can current be driven into it? Will taking current from it affect anything else? Is it safety critical?....

Because of this there can never be a universal answer to this question as the 'correct' solution depends on what the rest of the system does. The chosen solution that meets the requirements will have differing cost and complexity.

That said, as nobody else has yet suggested it, I'll go for a FET input.

A JFET or MOSFET can be used and either could be common source or common drain modes. For example, common drain:

^{simulate this circuit – Schematic created using CircuitLab}

The advantage of the common drain mode is that it allows the input to be connected to both an analogue (e.g. ADC) or a digital pin. If the signal is truly digital I would enable the schmitt trigger on the CPU input (if it has one), or add an external schmitt buffer to the CPU's input pin.

Advantages

• Very high input impedance
• Partially isolated input (can withstand +/- 30V, depending on FET selection)
• Analogue possible
• Minimal effect on external signal

Answer 7

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. Opto-isolated interface. Use internal pull-up on GPIO.

An opto-isolator solves several problems.

• Complete electrical isolation between the 12 V circuit and the 5 V logic.
• Handles dirty 12 V signal without risk.
• Simplicity.

Answer 8

^{simulate this circuit – Schematic created using CircuitLab}

R1, R2 and C1 forms a voltage divider with a 1kHz low pass filter. Any unwanted high frequency signal travelling on the 12V can be filter away. The calculation for the filter frequency is 1/(2 pi R2 C1). Note: The Base requires at least 0.7V to function properly, take care when adjusting the resistor.

BJT is being use because it's very common compare to mosfet. In the event the 12V is still active but the 5V for your uC is down, the BJT will not pass current into the pin and cause damage.

For uC programming, use a high to low trigger to count your pulse. As this circuit will reverse the pulse.

Answer 9

You can opt for a LM7805/LM7803 voltage regulator for 5V and 3.3V respectively.I am assuming the uC is isolated from a current demanding load, if any.