# Industrial sounder control circuit

Discussion in 'Embedded Systems and Microcontrollers' started by LewisMF, Sep 14, 2015.

1. ### LewisMF Thread Starter Member

Nov 15, 2014
63
2
Hi everyone,

I am working on a project for college, where I have to activate an industrial sounder that works at 24V DC and consumes around 60mA. The line has to be monitored for open circuit so for that I though of using an end of line resistor in parallel, with and ADC channel from the microcontroller to monitor the resistor.

The problem is that in stand-by (sounder off) the voltage across the terminals has to be inverted, so that the sounder doesnt make a noise, but enough so that the ADC can monitor the end of line resistor. And when activated the voltage will be +24V DC.

I have been looking at some comercial models that do this and in stand-by mode, the voltage across the sounder output is around -3V DC and when active it is +24V DC.

What kind of circuit shall I use to be able to carry out this project? I am finding it very complicated to invert the voltage. Therefore I only have a +28V DC PSU. No negative voltages at all.

On the commercial products I have seen they use BJT and MOSFETS.

Any Ideas?

Best regards to you all

2. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,975
617
The invert part does not make sense.

When the sound source is On, is it continuous On or does it pulse On/Off?

3. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,975
617
I finally figured out what was bothering me about your resistor idea.

When the sound is On. There will be 24 volts across the sound source. If you put resistor in parallel with sound source, there will be 24 volts across the resistor. When you connect uC ADC across the resistor, you will be feeding 24 volts into the ADC... in my limited experience uC are either 5 volt or 3.3 volt systems which means that when you feed 24 volts to uC ADC, you will either destroy the ADC of the uC or destroy the whole uC.

What you want is to use two resistors. You use these two resistors to build a Voltage Divider. This way you can divide the 24 volts into (example) 4 and 20 volts. Put your uC ADC to read resistor that will have 4 volts across it. However, since you only have 4 volts now and the device you monitor is simply On or Off, you don't need ADC. Simply configure one of the Digital I/O pins of the uC to be input. When there is 4 volts on this pin, that is digital 1 and this tells you that sound source is On. When there is 0 volts on this pin, that is digital 0 and this tells you that the sound source is Off. There are two strategies to use this. One strategy is polling where uC constantly polls the pin to see if it is On (digital 1) or Off (digital 0). Another strategy is to use interrupt.

4. ### dannyf Well-Known Member

Sep 13, 2015
2,196
421
Put a switch on the sounder, either low side or high side.

put a led in serial with the sounder, and a light sensor on the led, to detect open circuit.

if you cannot find a 60ma led, put a power resistor in its place and parallel a r+led with it.

basically a DIY high current opto coupler.

5. ### LewisMF Thread Starter Member

Nov 15, 2014
63
2
Hi Guys,

I was aware that the uC's ADC can only take 3.3 or 5V max. (In my case it's 5V)

When the sounder is ON, it is on continuously, no pulsing at all.

The negative voltage is so that the sounder will not sound at all when the control device is in stand-by but so that the ADC can monitor the line, to check there is no short or open circit. The reason for the negative voltage is because even a minor positive voltage (i.e. 3V) across the terminals may make it sound a little, and this should not happen.

The only option I have to monitor the line is an end of line resistor, I cannot use any LEDs or opto-couplers.

Best regards,
LewisMF

6. ### strantor AAC Fanatic!

Oct 3, 2010
4,725
2,612
If you were use a low-side switch, you could monitor for open line as such...

consider what you would read with a DMM, if you were to connect the (+) lead of the sounder to a 24VDC source, and read the voltage between the (-) lead of the sounder and 24VDC ground. I would expect to read 24V, since as there is no current through the sounder and its (+) lead is connected to +24VDC, the potential of the entire internal circuit of the sounder is elevated to +24VDC. So in this way your DMM is telling you that the circuit is not open. Now, you just need to make your control device replicate what the DMM does.

The DMM uses a 100Mohm internal resistor, so you could use the same. but since your ADC only accepts up to 5V, you may need another resistor to form a voltage divider. It will depend on the input resistance of your ADC. You will need to size your resistors appropriately.

Then, when you want to turn on the sounder, you just short out the high value resistor to ground.