Looking for a low/no current alternative to a SPST relay

[zeb]

I have a circuit with a 13.8v DC output that powers a solenoid. I'm not sure what the current draw of the solenoid is, but it's probably not much. I would like to have a 3W light come on when the solenoid is energized. The problem is that the circuit is very old and is known to have very low tolerances for current. I'm afraid of the bulb drawing too much current and blowing components in the circuit. (In fact, there are 2 other outputs on this circuit with known current limits - the acceptable range for those is 160-200 mA. If you draw more than 220-230 mA, it will blow the circuit.)

So my first thought was to use a normal SPST electromagnetic relay to isolate the light from the output of the circuit, but then I got to thinking that the relay probably has it's own current draw. Someone recommended that I look into a CMOS Flip Flop to use instead of a relay. He said that it has practically no current draw. I started reading up on those but feel like I've gotten a little over my head.

Are there any simple solid state relays, or other circuits that will perform like a SPST relay, but with absolute minimum current draw?

 

[Phrak]

Do you have a circuit drawing?

 

[zeb]

does this help?

 

[uart]

You'll easily be able to get a small relay with a coil current well under 200mA, so that will work.

But you'd also be able to use the old circuit output to drive the gate of a mosfet and then the current draw (on the old circuit output) would be close to zero.

 

[zeb]

If the output that powers the solenoid is anything like the other two outputs, then a 200 mA draw from a relay would definitely be too much to add to the existing draw of the solenoid.

• output 1 (engage solenoid): current draw: unknown, threshold: unknown
• output 2 (engage motor - forward): current draw: 160-200 mA, threshold: 220-230 mA
• output 3 (engage motor - reverse): current draw: 160-200 mA, threshold: 220-230 mA

This old circuit is rare and expensive to replace, so I don't want to risk blowing it. I think a transistor based solution would be best, but I don't know anything about them. What type of mosfet would be best here?

 

[AlephZero]

If you only want to switch a low power bulb, you could use a reed relay instead of a conventional relay. 12V reed relays with 1kohm coils, that will only draw about 12mA are readily available. For example http://www.jaycar.com.au/productView.asp?ID=SY4038

Note, you wire the relay coil in parallel with the solenoid, not in series with it as in your diagram. Otherwise, the relay will energise and switch on the light, but the solenoid will not!

 

[zeb]

Note, you wire the relay coil in parallel with the solenoid, not in series with it as in your diagram. Otherwise, the relay will energise and switch on the light, but the solenoid will not!

I don't get it. The solenoid will not energize? That doesn't make sense. I would think that in either series or parallel both the solenoid and the relay would energize.

 

[Phrak]

I don't get it. The solenoid will not energize? That doesn't make sense. I would think that in either series or parallel both the solenoid and the relay would energize.

You may want to recheck your circuit. You have "old circuit" in series with the relay coil and a solenoid to ground. If it is not drawn in error, then you would have to provide some details about the elements in the old circuit block.

 

[Bob S]

Use an NPN Darlington, like the RadioShack TIP120-D.

http://www.radioshack.com/product/index.jsp?productId=2062617

Connect the Darlington collector directly to the 13.6 volts, the darlington emitter to the 12 volt, 3-watt bulb, and the Darlington base to the hot solenoid terminal. The base will draw less than 1 mA.

Bob S

 

[zeb]

The current circuit has no relay. The relay depicted in the red-dashed box is an example of what I want to add. As it is now, the "old circuit" has 14 pins... pin 1 is power input, pin 5 is power output to the solenoid, pins 12 and 14 are ground, and the other pins do other things. The solenoid is in an actuator with an input for power from the "old circuit", a pin for ground, and other pins that do other things. For simplicity, I have only included the relevant parts of the circuits in my diagram.

So, are you saying that if I were to add a relay (which I'm not), then it would have to be connected in parallel in order for both the relay and the solenoid to work off the power output of the "old circuit"? What I want to add is a transistor (or transistors) to perform the switch. It makes sense to me to have the transistor (or transistor circuit) be connected in parallel with the solenoid.

 

[Bob S]

Because the solenoid is inductive, it could have an L dI/dt voltage spike when its power is turned off, which could harm the base of the Darlington transistor mentioned above. So the base should be protected from overvoltage using a diode and series resistor as shown in the thumbnail. The output of the Darlington will be about 12.6 volts. If this is to low, then use a single NPN transistor.

Bob S

 

[zeb]

Thanks Bob for your input. I have a few questions about your suggestion.

1) Why did you choose a 100 ohm resistor for R2?

2) How does the diode help with the spike? Why did you connect it back to the power supply?

3) Why did you use two transistors?

4) How did you calculate 12.6v?

5) how much current does this setup draw?

6) Would a N channel mosfet perform the same function as the Darlington, but with the added benefit of no current?

 

[Fish4Fun]

Zeb,

Is the function of the light simply an "indicator" to let someone know if the solenoid is engaged? If so, you might consider something like this:

This circuit will:

1) Clamp the Solenoid Voltage Spike
2) Consume a maximum of 12mA in addition to the solenoid
3) Light the LED when the solenoid is on.

You will want to select a diode capable of handling the solenoid spike, and select an LED that operates @ 12mA.

Obviously your existing circuit will replace the battery & switch.

Fish

 

[zeb]

Thanks Fish for your suggestion. Unfortunately, I want to avoid changing the type of light. The location of this bulb requires a special mounting bracket that takes only this specific type of bulb. I believe this bulb comes in a few different wattages, but since it is alongside other bulbs changing it's wattage will result in an inconsistent display - which is important to me. Besides, this project is a modification to an existing system, and as such I want it to be as easy as possible to implement and to reverse.

 

[Fish4Fun]

zeb,

I can't get the image to post right. I will try again later.

Fish

 

[Bob S]

Thanks Bob for your input. I have a few questions about your suggestion.

First, the inductance of the solenoid creates a large voltage spike (like in conventional (pre-1970) ignition systems) when the switch opens. This spike, often over 100 volts, could damage solid state circuits. So some protection is required.

1) Why did you choose a 100 ohm resistor for R2?

The 100-ohm resistor is to limit the current surge through the diode during the voltage spike. Assuming a 1-mA current requirement for the solid state switch, the dc voltage drop is only 0.1 volts (negligible). The voltage drop for a 10-mA current is 1 volt, marginally acceptable. The resistor could be reduced to 20 or 50 ohms if necessary.

2) How does the diode help with the spike? Why did you connect it back to the power supply?

The purpose of the diode is not to eliminate the voltage spike across the solenoid, but to eliminate the voltage spike across the solid state switch input. With the diode, the voltage to the solid state switch is never above 13.8 plus 0.6 = 14.4 volts.

3) Why did you use two transistors?

The reason for two transistors (Darlington) rather than one is to reduce the required current to less than 1 mA. if only one transistor were used, the current could be as much as 10 mA. With one transistor, the voltage drop is only 0.6 volts (plus the voltage drop in R2), not 1.2, like in the Darlington. Is 10 mA too much current?

4) How did you calculate 12.6v?

The 12.6 volts is 13.8 volts minus two diode (base-emitter junction) drops.

5) how much current does this setup draw?

With two transistors (a Darlington), the current is less than 1 mA. With one transistor, it is less than about 10 mA.

6) Would a N channel mosfet perform the same function as the Darlington, but with the added benefit of no current?

No, the N-channel mosfet works on the principle of voltage diffferences, while the bjt works on the principle of currents. In this circuit, minimizing the total voltage drop (V_BE or V_GS) is important. With the n-channel mosfet, the voltage drop could easily be over 2 volts.

A special note here. Your configuration requires a non-inverting circuit with a current source (positive voltage) output, because the other end of the bulb is grounded. So a simple circuit could be only a common-collector transistor or Darlington, or a common-drain N-channel mosfet. A better circuit might be a common-emitter npn transistor (or common source N-channel mosfet), but in this case the bulb could not be grounded.

I hope this helps.

[added] See corrected thumbnail with second protection diode.

Bob S

 

[Fish4Fun]

I know this has been mentioned, but this circuit:

With a G6L-1F DC12 relay it would only add 15mA to your existing circuit (assuming it is 12V), and it would offer 100% isolation between your supplies. The obvious advantage is that it would be very simple to implement.

Bob's circuit certainly draws less current.

Fish

 

[zeb]

@ BOB: Why two diodes? Wouldn't the second one be the only one needed anyway? doesn't the current flow from the solenoid and need a path to ground? Why keep the other diode? An explanation of the flow of the electrons would be helpful.

@ Fish: 15mA is a little too much for my comfort. If I were to discover somehow that 15mA is safe to add to the old circuit, then I would certainly go with a low draw relay like the one you suggest.

 

[Bob S]

@ BOB: Why two diodes? Wouldn't the second one be the only one needed anyway? doesn't the current flow from the solenoid and need a path to ground? Why keep the other diode? An explanation of the flow of the electrons would be helpful.

The Darlington base is very sensitive to overvoltage spikes of either polarity, so using two diodes to protect it from spikes of both polarities is cheap insurance. A simple inductance + resistance coil model would need one diode, but a solenoid coil includes distributed capacitance which could make it resonant (with a low Q). Furthermore, the complete inductive voltage equation for the solenoid is V = L·dI/dt + I·dL/dt, because the magnetic circuit inductance L changes when the current is turned on or off.

Bob S

 

[zeb]

Bob, so I was out getting the parts to build this circuit and the guy at the store asked what wattage I needed for the 100 ohm resistor, and what current I needed for the diode. What do you recommend?

 

[Bob S]

Hi zeb-

You need
1 ea 1/2 watt, 100 ohm resistor
2 ea 1N4001 (or 4002 or 4003 or 4004) diodes
1 ea TIP120 Darlington transistor

Bob S