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I’m new to EE, I’m a machinery mechanic.

I want to design a hydraulic oil radiator fan circuit by using a built-in temperature sensor. The sensor have 3-wire 24 V GND and signal wire. The signal wire will generate a 0-9 V signal to ECU represent-30-130 degree C. I want to turn on the fan around 50°C which around 5 V and above. So I searched on Google found that an LM393 comparator should be able to do the job.

I have never designed any IC circuit before. But here after a week Googling and searching YouTube I made this schematic; I don’t know if it would work or not. Please give any suggestions.

Here is the idea: 24 V power draw from the fuse box, I assume 5 A is enough, than it passes through TVS and LC filter feed it to LM7809 LDO to power LM393 comparator.

And than a 10 kohm potentiometer feed to the -input of LM393, so I can adjust the turn on threshold of my desire temperature. And the sensor signal also passes through an EDS/TVS and RC filter to the +input of LM393.

So when the sensor gives out 5 V or above, LM393 will have a 9 V output to drive 4N60C gate to energize the relay coil.

enter image description here

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  • \$\begingroup\$ L1/C2 resonate, therefore a plug-in event will briefly cause C2 to overshoot approaching 2x supply voltage (presuming the supply itself has substantially larger output capacitance than C2, which it often would). Can probably just add another TVS to eat it. Also lesser issue, the P6KE24 is cutting it too close on a 24V nominal supply (ie Vbr(1mA)=23V-25V). Smth like P6KE30 might be a better fit (relying on your C2=10uF cap to assist due to the now higher Vcl=41V+) \$\endgroup\$ Commented Nov 18 at 20:33
  • \$\begingroup\$ also look into how to do hysteresis (ie slight positive feedback) with the comparator, to keep it from switching too often when it's near the trip point \$\endgroup\$ Commented Nov 18 at 20:39
  • \$\begingroup\$ LM7809 is not an LDO regulator, it's a regulator, but it does not feature low drop-out, which is what the acronym "LDO" is referring to. I do not understand why you require this regulator at all, when the LM393 would work perfectly well with 36V. And even if you did need a regulator, it certainly doesn't need to be LDO. The 4N60C is a MOSFET, not a darlington pair as you have pictured it. Which is it, BJT or FET? It is also rated for 650V operation, well beyond what you actually need, with a horrendous 2Ω on-resistance. There are much better options. \$\endgroup\$ Commented Nov 19 at 1:20
  • \$\begingroup\$ Also, that 5μF capacitor C4 is incorrectly placed. As it stands, it AC couples the sensor voltage, ensuring that the comparator's non-inverting input will be 0V, not the sensor voltage you expect. I think you meant to swap C4 and R6. The LM393 has an open-collector output, requiring some kind of "pull-up" which is absent from your design. D5 is the wrong way round, it will clamp the input to a maximum of +0.7V. \$\endgroup\$ Commented Nov 19 at 1:26
  • \$\begingroup\$ The usual sloppy mix-up of terms is to say "LDO" when we actually mean linear regulator. Yeah I'm guilty of that all the time too. Although many LDO aren't very low drop-out by modern standards. \$\endgroup\$ Commented Nov 19 at 9:11

3 Answers 3

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Here are some concerns I have with your design, and some ways to solve them. I'll leave it up to you to build and combine these modules, and test them together.

First, a clean supply of +15V to provide up to 100mA without needing a heatsink on U1, easily sufficient for a modest load like the comparator and its periphery:

schematic

simulate this circuit – Schematic created using CircuitLab

Your L1 hints at a dirty supply, but an inductor in that position would likely cause more problems than it solves. I achieve a similar filtering function with R1 and C1 here, which will mitigate momentary spikes or dips on the 24V supply, and the +15V at P should be clean and stable. Install C2 and C3 very close to regulator U1's terminals.

I've chosen 15V because it's well under the \$V_{GS(MAX)}\$ limit of ±20V of most common MOSFETs, so that the comparator output can drive an N-channel MOSFET directly, but is large enough to switch the transistor fully on. By the way, your transistor symbol is a darlington pair, which could also work, but it is an inappropriate symbol for a MOSFET, and very misleading.

Your arrangement of C4 and R6 is a high-pass filter, when you should be passing a DC potential; swap them, to form a low-pass filter instead. Also, 500Ω is a heavy load on the sensor output, you can increase R6 significantly, and lower C4 commensurately. D5 looks like it's supposed to prevent excessive voltage at the comparator input, but it's the wrong way around. This is what I think you meant, for illustration, but will need further modification still so don't use this as shown:

schematic

simulate this circuit

Instead of low-pass filtering the sensor signal, with a capacitor, a better solution to mitigate noise and oscillation is hysteresis, which you implement with a small amount of positive feedback:

schematic

simulate this circuit

C5 isn't optional, it's a supply decoupling capacitor to keep the comparator's supply voltage steady while currents change during switching of states. Install it close to the IC's power pins. Positive feedback is via R7 and D2. The values shown will produce about 0.5V of hysteresis (the "gap" between switching thresholds, which can be seen here, as \$V_{IN}\$ (blue) crosses up and down through those thresholds. Orange is the comparator output \$V_Z\$, which requires pull-up resistor R8 due to the open-collector outputs of the LM393:

enter image description here

The 0.5V threshold gap (between the green markers) is probably too much for your application, but you can reduce it by increasing R7. Doubling R7 will roughly halve the hysteresis gap, you should experiment with the simulation to find the exact gap you require.

R8, the pull-up, needs to be as small as possible, to source as much current as possible to charge the MOSFET's gate capacitance quickly, but not so small as to overload the comparator output. The value 2.7kΩ should produce about 6mA, which the comparator is able to sink, according to the LM393 datasheet.

It will be difficult to interface an indicator LED to this open-collector output comparator, without an additional transistor, because that output cannot source current to illuminate an LED - it can only sink current. Therefore I suggest placing the LED across the relay coil instead. Usually I would recommend a 1N4001 fly-wheel diode to protect the transistor when the relay switches off, but with a fragile LED there too, I think you'll need something faster, like a UF5401, which is D4 below, replacing your own D2.

The MOSFET you choose, the 4N60C is not recommended, since while it is robust, it has significant on-resistance, around 2Ω. If the relay coil requires, say, 2A of current, this transistor will dissipate a lot of power, requiring heat-sinking:

$$ P = I^2R = (2{\rm A})^2 \times 2{\rm \Omega} = 8{\rm W} $$

There are dozens of better choices, with on-resistance \$R_{DS(ON)}\$ under 100mΩ, like the IRFZ44, which claims \$R_{DS(ON)}\$ under 30mΩ:

schematic

simulate this circuit

Strictly speaking, D5 isn't necessary, but without knowing more about the 24V supply, and the relay, I feel it prudent to protect both MOSFET Q1's gate and the comparator, from any shenanigans going on at Q1's drain.

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It is a good start. First D5 needs to be flipped around. Then the Temp Sensor input...I guess...is meant to be lowpass filtered by C4 and R6. C4 and R6 needs to be swopped. Right now the form a high pass filter.

LM393 have what is called an open collector output. That means the output can only pull to GND or be "open"/high impedance. So you need to add a pull-up resistor from the output to LM7809, when LM393 need to turn ON Q1.

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LM393 will have a 9 V output

It won't because the output of LM393 is open-collector. It can only sink current, not source it, and you are lacking a pull-up resistor on the output.

So, at the very least adding a pull-up resistor on LM393's output is necessary for there to be any chance that the circuit will work.

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