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Background

Below is the circuit I designed to open a solenoid valve after a delay from C2 (represented by L1).

S1 is a float switch which closes when a resevoir goes below a given level. The reason S1 is not directly in line with the source is the school of hard knocks lesson that the reed switch (inside the float switch) fuses from the current through the solenoid.

My problem is that the reference voltage being provided by R3 will constantly drain current, even when the valve isn't activated by the bias at Q1. I thought about putting both the inverting and non-inverting inputs on S1, so that neither the inverting nor the non-inverting inputs will have a voltage, but what would the effect be here?

Questions

I vaguely recall that if the two inputs are at equal voltage the output will be zero? Is that correct?

If not, are there other suggestions for stopping wasteful current in the circuit? enter image description here

PS

This the first circuit with a specific function in mind I've done from scratch, so ... other design flaws/suggestions you see, feel free to mention :)

The LM386 is just the op amp I have handy. I didn't feel the need to get a comparator because it's a delay of 7+ seconds, so the slow switching isn't really a concern.

Also R3 is actually 240K.

Edit: Per comments, modified image Would this be the correct way to implement the PMOSFET?

enter image description here

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  • 1
    \$\begingroup\$ Sounds like you want a PMOSFET. \$\endgroup\$ – Ignacio Vazquez-Abrams May 25 '14 at 2:33
  • \$\begingroup\$ Oh, you mean like from the non-inverting input to a pmosfet connected to R3? How would that be different from just putting both in series with S1? Or do you mean in place of the darlington/comparator? \$\endgroup\$ – Daniel B. May 25 '14 at 2:52
  • \$\begingroup\$ A PMSOFET on the power rail, controlled by the switch. \$\endgroup\$ – Ignacio Vazquez-Abrams May 25 '14 at 2:53
  • \$\begingroup\$ And by PMOSFET you mean P channel right? \$\endgroup\$ – Daniel B. May 25 '14 at 2:57
  • \$\begingroup\$ P-channel enhancement MOSFET. \$\endgroup\$ – Ignacio Vazquez-Abrams May 25 '14 at 3:02
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Ok, since I did not understand your need for a delayed turn-off, I've fixed that. Rather than modify my previous answer, I've provided this one instead.

schematic

simulate this circuit – Schematic created using CircuitLab

When the switch is closed, D1 provides a path for voltage to turn on Q1, while R2 charges C1. When the switch opens, the voltage on C1 through D2 keeps Q1 on until C1 has discharged to about 5 volts, at which time Q1 turns off. Using a MOSFET with a gate threshold (Vgth) of 4 volts provides a delay of 7 seconds for the values shown. The easiest way to change the delay is to change R3.

Turnoff is quite gradual, taking ~.5 seconds, so some heatsinking is a good idea. On the other hand, the gradual turnoff suggest that the flyback diode is unnecessary.

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  • \$\begingroup\$ I think you misunderstood ... I need a delayed ON, and a (near) instant off. I'd used a larger resistor for R1 than R2 in my schematic because I want the discharge to be faster than the charge, so that if the switch is opened and closed for a brief (< 1s) period, there will still be some delay before the power transistor is biased again. \$\endgroup\$ – Daniel B. May 26 '14 at 0:24
  • \$\begingroup\$ Arrgh. I'm not doing well today, am I? Oh well, the answer is simple. Remove D1, and while you're at it, remove R1 as well. Add a 1k resistor in series with the gate of Q1. Now get a SPDT reed relay with a 12-volt coil, such as a Magnecraft W172DIP-7 mouser.com/ds/2/357/105A_172DIP-4841.pdf Use the Common/NC contacts to connect Q1's gate to battery negative. That's all it takes. While the switch is open, the gate of Q1 is grounded, turning off the transistor. Meanwhile, C1 is being discharged. \$\endgroup\$ – WhatRoughBeast May 26 '14 at 1:29
  • \$\begingroup\$ For the low low cost of 5 times a comparator like I had in my circuit (that worked), or the op amps I have lying around. The P channel mosfet on the input suggested by Ignacio seems like a better solution ... What would be the benefit of your solution over mine (plus the PMOSFET on V+)? \$\endgroup\$ – Daniel B. May 26 '14 at 5:16
  • \$\begingroup\$ Oh, and to be more clear ... the comparator ensures that the solenoid only actuates near the peak of the capacitor's charge. This way if the switch turns off and on rapidly, there will still be some delay as the cap recovers to max. At least, that's how it works in multisim >.> \$\endgroup\$ – Daniel B. May 26 '14 at 17:35
  • \$\begingroup\$ I take it back. It wouldn't be cheaper, I forgot I'd disassemled an old UPS a month or so ago and salvaged like 5 12VDC relays. \$\endgroup\$ – Daniel B. May 27 '14 at 1:02
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You're doing this the hard way.

First off, R3 is not your problem. 12 volts / 240k is 50 uA. If you look at the data sheet for an LM386 http://www.ti.com/lit/ds/symlink/lm386.pdf p.4 "Quiescent Current", you'll see that for a 12 volt supply the LM386 will draw about 5 mA, or 100 times more than R3. This is with no load on the output.

So the question you need to ask is "How much current can I afford to supply in the off condition?". If the answer is 1 uA, then the following circuit will do you fine. Just be careful about picking Q1 - the attribute you're looking for is Idss with Vg = 0. This varies with model, but I've seen it at 1 uA.

If 1 uA is too much, then I suggest using your float switch to activate a small relay, and use the contacts of this relay to drive a more powerful relay which will actually drive the pump.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ The purpose of the comparator is to extend the "off" period until the capacitor has nearly fully charged, rather than slowly increasing the voltage. It isn't necessary for biasing the darlington, it's because I need a long delay (10s or so), so that the infil won't outpace the drain from the tank and cause the circuit to turn of and on repeatedly as it compensates for the tank draining. \$\endgroup\$ – Daniel B. May 25 '14 at 5:41
  • \$\begingroup\$ Your circuit is actually pretty similar to my original design, before I added the extra bits though I had a darlington rather than a MOSFET. \$\endgroup\$ – Daniel B. May 25 '14 at 5:42

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