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Need help with relay circuit

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Here is my problem. I have a circuit that I need to build using a relay - the switch will activate and ground out another circuit. The problem is the control signal I have to activate the relay coil with goes from 0 - 1.5V max. I can't seem to locate a relay that will pickup at such low voltages - any suggestions (will I need to build a DC-DC converter to get a 5V relay to work? Also, the control line is not dedicated to this usage - so I can't draw a ton of power out of it....)

Do you have a 5V supply rail available? Use a transistor or mosfet to operate the relay, feed your control signal into the base/gate.
do you have to have a relay? Why not just use a opto isolator.. The LED inte opto-isolator will light with the 1.5 volts and you still have both circuits isolated

Check this one out **broken link removed**

An optoisolator won't work at 1.5v. The LED needs current limiting, but the max forward drop on the LED (1.5v) leaves no headroom for a current limiting resistor.
Rworth, tell us more about the circuit you want to short out. What is the voltage? How much current must the relay (or some other type of switch) contacts handle?
Where does the control signal come from? How much current can you draw from it?
More details

The circuit that is connected to ground is low current, etc. What this is for is an in-car phone setup, where for the integration to the radio circuit I must connect a 'mute' input to ground for the phone to mute the radio and activate the incar setup. The control signal is coming from one of the wires from the phone kit to the phone, and it's the only signal that varies when the phone is in use of ringing (when ringing or in use it goes from 0 to 1.5V at the signal wire).

So, I need to isolate/switch the mute signal to ground based upon the signal from the phone spiking to 1.5/1.6v. Sounded like a perfect application for a relay, but the voltage available is too low.
I'll eat my words, with apologies to ivancho. You might get the optoisolator to work if you only need a few milliamps of current through your switch, and you are only going to build one of them. I looked at the NTE3041 datasheet, and it says typical fwd voltage for thr LED is 1.15v at 10ma. It will be less at lower current. The minimum current transfer ratio is 100%, so, let's say if you only need to switch a milliamp to mute the radio, you will only need a milliamp through the LED. If the LED is typical, it might only have 1v of fwd drop, so you could use 470 ohms in series with the optoisolator input for current limiting from your phone signal. If you need more current, use a smaller resistor, so long as the phone can tolerate the additional current.
The reason I said "you are only going to build one of them" is because the max fwd drop on the opto input is 1.5v, which would, as I said in the previous post, leave you no headroom for a current limiting resistor. You couldn't deal with this on a production product.
I picked this optoisolator at random as an example. You would need to look at specs to see if others would work.
Ron H Apologies accpeted :lol: no problem.

I believe that if you have a optoisolator it will work. The optoisolator is not more than a LED and a phototransistor (or alike). IF you are to drive any LED at their forward driving voltage you are not going to need a limiting resistor because you know what current the LED is going to consume at that voltage. As the opto isolator I told you (MCT6)the led has a forward voltage of 1.5V so you know that the current it will consume at this voltage is 20mA according to the datasheet.

Since you need to ground your "mute" signal, that tells me that the signal input is pull high and to make it mute you pull this low. What it does not tell me is what current will it be pulling to low (or sinking). If the current available from the optoisolator is not enough use the opto to drive a transistor to increase you power capabilities. But since it is a signal I guess that the current isn't an issue and it won't be more than 100mA.

Good Luck

Ivancho, I wouldn't attempt to drive the MCT6 with a 1.5v battery. The LED would probably be destroyed. Driving it with a 1.5v logic signal would probably be OK, because the logic signal will probably have enough internal resistance (although the logic level may be degraded unacceptably). You seem to have interpreted the datasheet as a guarantee that, if you put 1.5v across the LED, the current will be 20ma. This is far from reality. The datasheet only says that if you force 20ma through the LED, the voltage will not exceed 1.5v.
Take a look at the Fairchild MCT6 datasheet, which has more information. I have pasted the typical voltage vs current curve below. Note that, at 25C and 1.5v, the current typically is over 100ma! This exceeds the maximum allowable current for the LED.
Keep in mind that the LED is a light emitting diode, and as such, a small change in forward voltage will result in a large change in forward current. Another way of looking at it is, LEDs should be thought of as current-controlled devices. They should never be driven from a voltage source.
The reason I said that this part might work can be seen from the curve below, keeping in mind that this is a curve of a typical device. At 20ma and 25C, the voltage is only about 1.2v, so there is voltage headroom for a current limiting resistor. However, you might buy a device which has 1.5v drop at 20ma, so the same resistor would give you much less current, and therefore much less light output.


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So - If I understand this correctly, I need to attach the diode interface to the signal wire/ground and then the circuit which I will be grounding out (with only about 1ma current through that circuit) to the 'switched' interface.

My main concern is how much current the diode will draw from the phone/signal wire - I have no specs on how much this will supply but I expect that it will be very low..

A transistor BC 547 can do the job. simply connect your relay in series between the Vcc ( atleast 6 V ) and collector of BC547. Apply the 1.5 V control signal directly to the base of the transistor ( or through a current limiting resistor if necessary). Practically to use a transistor as a switch it is a good practice to maintain Ib = Ic/10 to keep the transistor in hard saturation. connect the emitter terminal to GND. 6V is necessary as 5V is required for the relay and 0.2V is the drop across the transister during saturation. This transistor can easily withstand 150mA of current flowing through it during saturation.
Ron H: You are right..... oops sorry about that. A 15 ohm resitor will solve that. 1.2V at 20mA will be enough to light the LED and bias the phototransistor.

You have to find out what amps can you get from the signal. You don't want to overload the line :?

But while the phone is ringing, I suppose the o/p will not remain steady, ie the radio may be muted and put on in rapid succession, whereby causing harm to the radio cktry. What can be done about that? ( A phone ring is not a contineous signal whereby causing the variation in output from ) to 1.5V). Will there be a necessity of a latching ckt?
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