How to detect negative voltages into a diode bridge? - Page 1
Forum section: electronics / projects

author: sebmadgwick
author_rank: Regular Contributor
timestamp: November 30, 2016, 10:38:22 pm
content: I am building a circuit to interface to my door intercom system. The intercom system has just 2 wires. These wires normally supply a DC voltage between 9 V and 17 V. When a visitor rings the door bell, the wires instead supply an AC voltage of -20 V to 15 V (peak) at 50 Hz. Audio is AC-coupled on the DC voltages and the door latch is released by shorting the two wires.I have a 5 V digital circuit powered from the 2 wires via a diode bridge. I would like the digital circuit to detect when the door bell rings by detecting when the voltage into a diode bridge goes negative.My current solution is to use an optocoupler but this is not ideal because the peak AC voltage is unreliable and may be too low to switch on the optocoupler output. The optocoupler could be powered by a 7803 regulator but this feels ugly.How do you suggest my 5 V digital circuit detects a negative voltage the other side of the diode bridge?

author: danadak
author_rank: Super Contributor
timestamp: November 30, 2016, 11:57:50 pm
content: If you have a single supply circuit, and an opamp or comparator in it,you can use a R divider to an input then to Vdd to look at negative signals. UseOpAmp or the Comparator then to develop you logic signal. Here is acalculator to calc the divider. And a general discussion of negative signalinterfacing.https://www.dropbox.com/s/k1yn4m7heaixvmg/ADinput%20negative%20inputs.zip?dl=0Regards, Dana.

author: sebmadgwick
author_rank: Regular Contributor
timestamp: December 01, 2016, 01:25:47 am
content: Thanks for the suggestion but it does not address the problem. The negative voltage is not negative relative to the ground of the 5 V digital circuit. It is negative in the sense that the AC supply to the full-wave rectifier assumes either a 'positive' or 'negative' polarity.

author: sebmadgwick
author_rank: Regular Contributor
timestamp: December 01, 2016, 01:50:27 am
content: Perhaps the solution is just a transistor between one of the AC inputs and the DC ground (schematic attached).The base resistor needs to be calculated based on the peak AC voltage. The open-collector provides the binary output. Combing this with a high value pull-up resistor reduces the gain required so that the circuit is compatible with a wide range of AC voltages.

author: danadak
author_rank: Super Contributor
timestamp: December 01, 2016, 02:29:29 am
content: Your idea quite clever, see attached sim.Of course an LED >Diode (Reverse LED protection) > R on the AC side, photo transistor on the 5V side, would giveone approach. Essentially an opto coupler.Not sure in either case UL impact, never having done a UL design.Regards, Dana.

author: sebmadgwick
author_rank: Regular Contributor
timestamp: December 01, 2016, 03:05:00 am
content: Thanks for the simulation!

author: sebmadgwick
author_rank: Regular Contributor
timestamp: December 01, 2016, 04:34:40 pm
content: I tested the circuit using an N2222 with 1k base resistor and 1k pull-up. It works well. See the attached screenshotChannel 1 = AC input (differential probes)Channel 2 = AC zero-reference input (relative to DC ground)Channel 3 = Base voltage (relative to DC ground)Channel 4 = Collector voltage (relative to DC ground)