339-darlington-relay

I have a ckt with an LM339, output voltage is 6v. Rail voltage is 8V. I need to control a 12V relay which switched at 8V. I am using a darlington transistor and i tried to follow what SGTWOOKIE mentioned in this thread.

http://forum.allaboutcircuits.com/showthread.php?t=11450

It doesnt work for some reason. Is it necessary that I use a diode. Will the ckt work without one.
Thanks for the help.

A relay designed for operation with 12v will be very uncertain to operate at 8v.
If you used a Darlington TIP120, you also reduced the available voltage by ~0.8v or so due to the saturation voltage of the Darlington.

Why don't you post the circuit you have so far, and we will go from there?

Here is the circuit I was trying to build. The MESFET doesnt work for some reason and I do not know why. Right now I have an IRF510 and a TIP105 to try to switch the relay circuit. The rail voltage is 8V. and the output from the lm339 is around 6V. The relay is a 12V 10A. The rest of the circuit wrks fine other than the relay part.

Will your relay trip when directly connected to the rails (ie. not using the MOSFET switch)? It's dicey using a 12v relay at only 8v, as Sarge has pointed out. It's made worse by the choice of MOSFET - the 510s won't be fully "on" until the gate gets well over 10v. So you've got even less than 8v across your relay. Try measuring it, and you'll see. It should change when the MOSFET triggers, but if it's only 6v or so, that might not be enough to trip the relay. A logic level FET will be full on at 8v, and that might be enough to make the difference.

ps: Your diagram is right but it's a common mistake to mis-wire the MOSFET, getting the pins wrong. Your testing should show a voltage change. If not, recheck again that you've gotten the pins right.

Do you have any other opinions on how I could trip the relay with my the lm339 output? The relay trips when I use any voltage greater than 8V. I did try using the darlington pair so that the rail voltage will trigger the relay, but somehow I think the connection and resistor values I am using are not right.

If the relay trips when you use any voltage greater than 8v, and your power rail is only 8v, I don't see how you will be able to cause the relay to engage.

Add a 390 Ohm resistor from the GND terminal of the 7808 to GND. That should give you about 10v out of the 7808, and you will have enough voltage to cause the relay to engage.

nathomas said:

The relay trips when I use any voltage greater than 8V.

Click to expand...

Well there's your answer. Can you not supply the full battery voltage to the high side of the relay, instead of the 8v rail? The MOSFET - even though the gate is only seeing ~7v - will open a path to ground and trip the relay.

will the lm339 be able to take that that amount of voltage. What is the maximum voltage the LM339 can take. I have a 15V rail and if possible I can avoid the 8v step down all together.

could you show me how the mosfet connection would be. for some reason when I make the connection the mosfet just heats up and lots of current passes through. my relay is a four pin relay. i am using the 10k and the 100k resistor as shown in the circuit. and will 6V be enough to trigger the mosfet? i just need to know how the connection would be from the lm339 to the mosfet and to the relay.

The LM339 can handle more voltage (at least 30v) than the 4000 series CMOS components.

Don't go above 16v, or the 4000 series CMOS IC may be damaged.

If you want to regulate the output to around 12v, use a 750 Ohm resistor from the 7808 ground terminal to ground instead of the 390 Ohm resistor.

You have LEDs on the outputs of the 4001 - this will not work well at all, as the 4000 series CMOS has a limited current source/sink capability. It may prevent the MOSFET from turning on.

You have 10k resistors between the 4001 outputs and the IRF510 gates. Reduce that to around 1k Ohm. If you are using a Darlington instead, then use 10k.

I cannot tell what the value of pull-up resistors you are using on the outputs of the LM339. Use this formula: Rpullup ~= Vdd/0.004. If your Vdd will be 12v, then Rpullup = 12v/0.004 = 3k Ohms.

Make certain that you have unused INPUTS to the 339 and the 4001 connected to ground.

one thing i forgot to mention is that I am using a 74hc02 instead of the 4001. does that make any difference? thank you both for the help. will try it out and see what happens.

thanks alot. it works at 15V and the relay switches without any transistor or mosfet..

Ack! 74HC02's are not rated for that high of a voltage! 6V is the maximum!

You will burn it out rapidly.

nathomas said:

... when I make the connection the mosfet just heats up and lots of current passes through..

Click to expand...

Not sure what's happening here. I'm surprised enough current is coming through the relay to heat up the mosfet.

But I just wanted to pass along a general rule: If the MOSFET is heating up and your current is not excessive, the problem is the MOSFET is not fully on, because its gate voltage isn't high enough. Once you get the gate up, the resistance of that MOSFET should be so low that little or no heat is produced until you approach the specified max current for that MOSFET.

You've now eliminated using any transistor to switch your relay? You'd better check the current sinking ability of your device, in addition to the voltage issue.

wayneh said:

You've now eliminated using any transistor to switch your relay? You'd better check the current sinking ability of your device, in addition to the voltage issue.

Click to expand...

Do you mean the whole device..My dc supply shows that I am using around 100mA. Do you think it is not right to use direct supply from the 7402 to switch the relay? What are the disadvantages?

SgtWookie said:

Ack! 74HC02's are not rated for that high of a voltage! 6V is the maximum!

You will burn it out rapidly.

Click to expand...

Is there some other part that I could use that could handle the voltage..?

nathomas said:

Do you mean the whole device..My dc supply shows that I am using around 100mA. Do you think it is not right to use direct supply from the 7402 to switch the relay? What are the disadvantages?

Click to expand...

I think switching that current with the 7402 will toast it, especially since you're over its max voltage spec. You'd be better off putting a MOSFET back in to act as your switch (7402 output at the gate). With only 6v at the gate, it won't turn fully on unless you find a logic-level FET. But at 100mA, I think you'll probably be OK anyway. Watch for heat on the FET. If you can hold your finger on it, it'll likely be OK.

nathomas said:

Is there some other part that I could use that could handle the voltage..?

Click to expand...

The 4001 (CD4001B, HEF4001B, MC14001B, etc) will handle up to 16v.
The 74HC02 is only rated for up to 6v.

In the circuit layout I have put up, at the IRF510 output to the relay, how does the connection go to the relay. Does this connection go to one terminal of the relay and the other end of the relay to the positive of the rail?

Yes, how you described it in the 2nd sentence.

There should also be a diode across the relay coils' terminals to absorb the reverse-EMF when the current through the coil is turned off. Use a 1N400x (where x=1 to 7) diode, cathode (the end with the band) towards the positive rail, anode towards the MOSFET drain.