Detecting an open circuit by sensing current flow using a shunt and comparator

Question

I need to detect if a load resistor is taken out of circuit. The load resistor is a photoresistor and can vary from 50k\$\Omega\$ up to 2M\$\Omega\$ and is driven by +5V to ground.

The approach I thought of was to put a small sense resistor between +5V and the load, and use a comparator to detect the voltage drop across the sense resistor. When current is flowing, there will be a drop across the sense resistor, one comparator input will be higher than the other and so the comparator output will go to (say) +5V.

When the load is disconnected, there is no drop across the resistor, and so the comparator inputs are (very close to) equal. However, I can't be sure if the output will switch to ground, since the output state of the comparator is undefined when the inputs are equal.

Is there a simple way to add a tiny offset so that the comparator switches at (say) \$ V_+ > V_- + offset \$? That would allow me to guarantee the comparator output when the inputs are equal. I don't want hysteresis here but just an offset to the point where the transition happens.

Alternatively, is there some other way to solve this problem? Basically, I want to detect the difference between even a small current flowing and no current flowing.

Edit: Especially to those who suggested the wheatstone bridge, what if the photoresistor varies from 50k\$\Omega\$ to 50M\$\Omega\$? At the higher end of that range, the two inputs to the comparator become very close to being equal again... If I use a gigantic (i.e ~50M) resistor (R3 in @markt's diagram) in the static arm of the bridge to compensate for the larger load range, then V- is going to be very, very close to rail; when the load is disconnected, the two comparator inputs are again very close to being equal to each other.

Answer 1

A good cookbook for op-amp designs (Application Note 31 - Op Amp Circuit Collection) can be found here from the former National Semiconductors (now part of Texas Instruments). On the lower part of page 26, you can find a current monitor circuit. It uses a small value sense resistor in series with the load, like your idea, to measure the current flowing through the voltage being dropped across the sense resistor.

Answer 2

^{simulate this circuit – Schematic created using CircuitLab}

Try this.

Answer 3

You probably need to use what is called a bridge circuit. This uses two limbs of resistors.

Here's a picture I just found that is similar to what you probably need: -

The comparator inputs connect to the mid points of the 2 limbs. It's called a wheatstone bridge if you want to look it up. Here is a link where I took the picture from.

This needn't run from +/-15V supplies providing you choose the right comparator or op-amp BUT watch out for leakage currents into the op-amp/comparator inputs if you are using large values of resistance.

Answer 4

With a 50 MΩ (presumably) dark resistance it will be very hard to differentiate between sensor present and an open circuit.

One option that might be worth considering is to mount an LED beside the sensor and pulse the LED periodically while simultaneously checking for a drop in resistance. If there is light on the sensor already the resistance will be low so it won't be necessary to blink the LED.

This solution requires at least one extra wire to drive the LED and due to the switching and checking it probably suits a micro-based project rather than a continuous analog-only circuit.

^{simulate this circuit – Schematic created using CircuitLab}

You should probably check for short-circuits too while you're at it.