Where to install RC circuit

Discussion in 'The Projects Forum' started by rwp289, Oct 6, 2010.

1. rwp289 Thread Starter New Member

Sep 30, 2010
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Even with some previous help, I still have a question about damping the needle movement of a gauge using a parallel RC circuit.

My goal is to slow down the movement of the gauge as the car stops, starts and goes around corners.

Attached is a schematic of the gauge circuit. The gauge is basically a calibrated voltmeter.

At Full the resistance is 10 ohms. At Empty it is 73 ohms. The battery has an average voltage of 12.65. Amperage is between 0.157 and 0.60 across the range of motion.

If I want T = 5 seconds

Q1. Where is the best location to install the capacitor? Across the variable resistor?

Q2. With a relatively low current flow, am I better off to go with a larger cap (1000mF) and smaller resistor or a smaller cap with a larger resistor to get T = 5?

Q3. Do I even need a resistor at all - other than the existing varistor?

Q4. An entirely different and better method?

Again, my goal is to slow down the movement of the gauge so that it, in effect, averages the reading of the last 5 seconds or so?

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2. wayneh Expert

Sep 9, 2010
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If you want to simply add a big cap, it would connect from the high side of the resistor and ground, so that it can help absorb or supply current (maintain voltage) as the resistance bounces around, thereby slowing the response to the bounces.

For RC=5 with R=10, C needs to be a whopping 0.5 Farad. That's 500,000 µF or 500 mF, ie. BIG. But you may not need a full time-constant, a full multiple of RC, to get the damping you need. I think you probably need to cover over short (<1 second?) excursions of the resistance. The same cap would give you RC=73*0.5 = 36 seconds as you approach empty. That's really highly damped but maybe OK.

3. rwp289 Thread Starter New Member

Sep 30, 2010
11
0

Wow. I guess I did the math wrong. I don't want anything that big. And I did write mF when I meant µF. So, if I went with 5,000 µF I would have .005*73 and I would get 0.36 seconds. Likely not enough to accomplish my goal and I don't ant to go bigger the 5,000 µF.

Any other suggestions on how to design a circuit that would slow down the change in resistance the gauge "sees"?

4. wayneh Expert

Sep 9, 2010
14,860
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Well, the only approach I can think of is to separate the sender and the gauge and situate a circuit in between them. That circuit would 1) Drive a current thru the sender and measure its voltage drop, 2) Smooth that signal, I think with an integrator, and 3) Produce a voltage at the gauge to make it "think" it's still seeing the sender. Steps 1) and 2) are pretty standard op-amp circuits. Step 3) is also, however the current may be higher than an op-amp can handle. I think you'll want a transistor and a shunt resistor for the last stage. The op-amp will control the transistor to give the same current across the shunt and the gauge that the gauge expects to see flowing across the sender.

I know this may sound murky if you haven't built what I'm describing, but it's fairly standard stuff. (Has to be, for me to know about it!)

PS: It would still be useful to know if the meter is responding to voltage or to the total current. If it's just voltage, than producing a voltage at the gauge would simplify step 3).

5. rwp289 Thread Starter New Member

Sep 30, 2010
11
0

Thanks for the reply. I called the gauge manufacturer (Autometer) and they confrimed it is a calibrated volt meter. How does that change step 3?

The current is low - less than two tenths of an amp across the variable resistor.

Your right about "murky" but that is part of the fun - the challenge of learning something new. Any and all help is appreciated.

6. wayneh Expert

Sep 9, 2010
14,860
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I'm no expert, but I think it means a low current source (simple op-amp circuit) could just deliver the right voltage, at low current, and the meter would display properly.
Everything's relative, and relative to a simple op-amp that's huge. For instance I've been using a "high power" op amp that's rated to a whopping 60mA.

But I think current level is moot if all you need is voltage. You could put, say, 10mA through the resistor and that would develop 100mV at full (10Ω) and 730mV at empty (73Ω). Then set an op-amp gain to convert those voltages to the corresponding voltages the gauge normally sees. Smooth it, and you're there.

7. rwp289 Thread Starter New Member

Sep 30, 2010
11
0

Thank you. I am learning alot and enjoying the process.