Circuit design help needed - analog
Typical fuel tank sending units have a resistance of 0 to 10 ohms in the full position and 90 to 240 ohms in the empty position. I have an old Weston 50uA meter I would like to use as a fuel gauge. I need help with designing an analog circuit that will drive the 50uA Weston meter from the tank sending unit (and not blow up the gas tank). V+ will be from the car's 13.5 V battery (running). Thanks for the help.
reachrkataI have this idea. Sorry that I have to draw a circuit and have no other way to do it that this way-
This is an opamp amplifier in the non inverting configuration. Just do following additions to this circuit.
1) In place of Vin, connect the battery with a 240 Ohm resistor in series and with the sending unit connected in parallel to ground.
2) At Vout, connect the Weston meter to the battery via a resistor with value such that it allows only 50uA current when opamp output is 0V.
3) Choose R1 = R2.
When you tank is full, Sending unit has a resistance of 0. Therefore opamp output is 0 and the Weston meter will show a reading.
When tank is empty, Sending unit will show 240Ohm. Therefore opamp output is = battery voltage and Weston unit reads 0uA.
Since the design uses battery for everything and also expects Battery level voltage at opamp output, be sure to use a rail-to-rail opamp.
There is one thing to note here.
Typical fuel tank sending units have a resistance of 0 to 10 ohms in the full position and 90 to 240 ohms in the empty positionThis design doesn't take care of the high resistance inaccuracy of the Sending unit. I assumed 0 Ohm and 240 Ohm.
Hope this helps.
johnerbesThanks for the reply Karthik.
I apologize for being unclear on the sending unit. What I meant to say was that the sending units vary with manufacturer and a particular unit could have resistance of 10 to 90, or 0 to 240, or anything in between. I haven't obtained a specific sending unit as yet. I do understand that the resistor in series with V+ should equal the maximum resistance of the sending unit.
If the sending unit happens to be 10 or so ohms in the full position (and not the 0 ohms you assumed) that will not be a problem as I can still select a current-limiting output resistor to read 50uA in the full position even if there is a residual difference between Vout and V+ battery.
I do have a couple of questions and comments.
Could I stick a voltage divider or zener circuit ahead of the 240 ohm series resistor to cut down the voltage at Vin? A desirable circuit design goal for safety reasons would be to minimize the current flowing through the sending unit resistor. If my sending unit and the series resistor are 90 ohms, there would be from 66 to 133 mA flowing through the sending unit. I would like to lower that as much as is practical.
What is an appropriate range of values when selecting R1, R2 (and R3) for gain=2 considering the extremely low load on Vout?
I'm not familiar with rail-to-rail. What are the numbers of common op amps that have that feature?
Could it hurt to put in a non-polarized cap across the sending unit resistor or from both ends of the sending unit to ground? I want to make the circuit as safe as possible.
What I meant to say was that the sending units vary with manufacturer and a particular unit could have resistance of 10 to 90, or 0 to 240, or anything in betweenAh Ok ! Now I understand the meaning of the resistance variation.
Could I stick a voltage divider or zener circuit ahead of the 240 ohm series resistor to cut down the voltage at Vin?Of course you can. It is also a good idea considering the fact that Battery is not a stable voltage. As you said it would also reduce the current through the sending unit.
What is an appropriate range of values when selecting R1, R2 (and R3)R3 would have to be same as the resistance of the sending unit. And as per the formula R3=R1*R2/(R1+R2). Since you want a gain of 2, R2=2R1. You can calculate accordingly.
considering the extremely low load on VoutThat should normally be no issue. In any case you can put a 10k to GND at Vout to act as a load and the circuit would still work.
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