#Comparator relay circuit series#
#Comparator relay circuit free#

I've tried putting 6 to positive and 7 disconnected, 7 to positive and 6 disconnected, I've even tried connecting the positive of a AA battery to it (when doing so I connected the neg on the 9V and the neg on the AA, is that okay?). Pin 1: is connected to a resistor, and that resistor to the negative of a 9V battery It must not load the opamp more than it can handle. R1 depends on the relay coil current required and the gain of Q1. A simple low side NPN switch would to fine here: This is assuming the relay coil is meant to be driven from 9 V as in your original circuit. 4.3.3.3 The Comparator Circuit The comparator circuit is designed to. It's no surprise the opamp can't drive the relay directly. Here is the datasheet: **broken link removed** TABLE 4.3 State Table for the Relay Circuit Relay Coil Signals Relays S 1 S2 S3 RL. With access to the datasheet, I still can't figure out how I'm supposed to get it to work. Also as the resistive values are all equal, the voltage drops across each resistor will also be equal at one-third the supply voltage, 1/3Vcc. I went to Radio Shack and bought an LM339N comparator. Using our idea above of a voltage divider network, if we now use three equal value resistors so that R1 R2 R3 R we can create a very simple window comparator circuit as shown. My first thought was to research some type of thresholding device, so that 0V meant blow the horn, and 5V meant shut it off. Relay accuracy is discussed and the methods which.

#Comparator relay circuit free#

but the voltages given out by the alarm clock are insufficient to trip the relays I have, and I haven't found one online that could provide a dropout voltage of ~1.5V. The relay is rendered free from transient overreach by the use of a dual comparator circuit configuration.

The device to make the sound draws so I need to run it off a relay. I have soldered some points to the alarm clock so that when the horn should go 'beep' it outputs 1.5V, and when it should be off or intermittent, it is outputting 5V. (with the appropriate relay interface circuit from the comparator's digital low power output).I have an alarm clock which I am trying have beep a horn instead of the built in speaker.

This hysteresis would mess up the linear aspect of a follower. Another words, once the comparator has flipped the inputs must reverse back PAST the spot where the flip occurred. I have recently designed and simulated a system like this for the purpose of under-voltage lockout of a Lipo battery system, but the outputs are also suitable for triggering a power-path change via relays etc. Because a comparator usually needs hysteresis to work reliably.

#Comparator relay circuit series#

The set point for the comparator is 1.1V, the monitored input can be voltage-divided down so that when it reaches the required level, the comparator trips.īe aware that you will need to protect the comparator inputs from over-voltage, using Zener clamps and series resistors - if you have 12V being monitored, and very little or zero voltage dividers going on, the input to the op-amp will be damaged if you just directly connect it - the ESD protection diodes internally will burn out and the device will be compromised. The op-amp will need it's own supply to work properly all the way down to 1.5V, so you could try a small low power auto buck-boost converter to provide 3.3V all the way down to 1.2V or so input and up to say 16V input (hard to find, probably expensive, good luck). Use a 1.1V Reference Voltage Zener Diode as one input for the comparator, and the slowly reducing input voltage from the battery system to the other input. If you want to go the comparator route, You need a fixed voltage reference which works down to the 1.5V, and a moving voltage reference which trips the comparator at 1.5V.