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Questions relating to designing PCBs
By djjohnson70
I am new to working on PCBs and so far the logic is very enjoyable. However, after a few trivial board designs I'm facing a challenge that must be very simple (I'm making a mountain out of a molehill). I'm trying to perform some logic in between 2 existing components. The first component is a circuit board which supplies 24VDC to the second component, a 24VDC PNP photoelectric sensor.

I want to invert the signal and perform a specified delay on the sensor. The delay is fairly straightforward using a delay line from DigiKey. The inverted signal is fairly straightforward as well. My problem is that I am using a 78 series voltage regulator to bring the 24VDC down to 5VDC for the delay circuit and inversion. After the delay, I have a 32mA 5VDC inverted signal. I need to provide 24VDC 100mA back to the circuit board. I am not sure how to get the voltage back up to 24VDC, and at the higher current. I originally considered driving a solid state relay, but the MTBF of a SSR is too small, and the component will fail too frequently. I am now considering a DC-DC converter (link below), which is powered by a transistor whose base is connected to the delay, and emitter is connected to another 78 series voltage regulator 5VDC. ... 5720586917

My questions are:

1) The circuit boards in I normally see are almost always operated using 12VDC or 24VDC power supplies, but I know that most circuit board components operate at 3-5VDC. I MUST be missing something obvious. How do I get back to 24VDC output from the board? I.E. - How do current circuit boards input and output 24VDC but perform logic at 5VDC? What component or circuit is common practice?
2) If there is no circuit that I'm missing, will my design work?

Even a quick response of the component I SHOULD be using, or a board layout that performs the task would be greatly appreciated. A small amount of guidance and I will gladly spend the time researching appropriate usage.

Dan J
By n1ist
You should be using the 5v signal to switch the 24v rail, not generate it. I would use a PMOS FET as the switching element. Using a 7805 with 24v input for any amount of current will get too hot very quickly. There are some switching regulator modules that look and work like the 7805 without the heat,

Do you need to delay powering up the sensor or to delay the output of the sensor? Can you give some more info on the source of the 24v, the delay you want, and a data sheet for the sensor - that would give people something to go on.

Thanks, Mike!

Here is a link to the sensor in use. ... 61630.html

The amount of current is only what is required for the circuit board to detect the sensor is active. Since the sensor is rated for 100mA, I expect that the board requires some amount less than 100mA for activation.

The existing circuit board is powered by a 24VDC switching power supply, and supplies 24VDC to the sensor. The sensor is active for approximately 35ms which is too short. Using a LTC6994-2 delay chip, I want to delay the trailing edge of the signal, so that the active time is approximately 60ms. The current circuit board needs a longer signal in order to reliably perform its function.

Eventually, I want to replace the photoelectric sensor with a Baumer XF100 series camera which will output a +24VDC pulse that will mimic the photoelectric sensor (but inverted). That signal will also need to be trailing edge delayed to function properly.

So... There is an existing circuit board that performs logic unreliably due to the duration of the signal from a photoelectric sensor. The circuit board is powered by a 24VDC power supply, and powers the 24VDC PNP photoelectric sensor. The sensor's active time is insufficient, so I want to add a board in between the sensor and the circuit board. I have access to the source 24VDC power supply if necessary. The board I add will increase the active time for the sensor using a delay chip by delaying the trailing edge of the signal. Eventually, I want to replace this with a Baumer XF100 camera which will provide a signal which is inverted from the photoelectric sensor (the need for inverting the signal), and will also require the delay chip for proper function.

Does this provide all the information?

Thanks again,
By n1ist
You are mixing up power and control. The sensor has a power input (always on) and a switched output (connected to 24V when on, floating when off). I would tie the sensor power to your 24v rail. Take the switched output, add a pulldown resistor, and have it switch a transistor to generate a 0 or 5V signal to trigger the delay. Have the delay's output drive a transistor to generate the 0 or 24V logic output. As for powering the delay, the 7805 will be fine as the current draw of the delay chip is so low,