a circuit that will indicate when a contact is broken?

[Randybb]

Hi, I'm trying to figure out how design a simple circuit that will give a visual indication (either turning on a light, or changing the color of a light) when contact between a metal plate and a metal foot is broken (when the foot is lifted up off the plate). I'm getting confused trying to think about it... can anyone give me some hints?

 

[OLIVE2222]

You can use a classic led+resistor circuit. Just short the led with the metal plates. When the metal plates contact is broken the led will bright.

 

[Randybb]

That circuit would constantly drain the battery though, any way to prevent this?

 

[OLIVE2222]

Such a circuit doesn't pop up, what is your application? is a wall plug power supply not suitable?

 

[KrisBlueNZ]

Yes, although you can't completely eliminate the drain on the battery. You need some current flowing through the plate and foot. Here is a circuit that has a very low current consumption in the OFF state.



This circuit uses two transistors connected as a Darlington pair. This configuration has very high current gain - the current gain is the product of the current gains of the individual transistors. So it needs very little base current to turn the transistors ON and illuminate the LED.

You can make a Darlington pair with two separate transistors, as I have shown in the diagram, or you can buy a Darlington pair in a single package - for example, the MPSA14.

This base current comes from R1, a 1 megohm resistor. When the contacts are closed, the bottom end of R1 is held at 0V and there is no voltage available to bias the transistors into conduction. In this state the current drain is 1 µA multiplied by the battery voltage in volts; for a 9V battery, this will be 9 µA (microamps). This is a pretty low operating current and a standard PP3-type 9V battery should last pretty much its shelf life.

When the contacts are broken, R1 pulls the voltage at R2 up towards the positive supply and this causes Q1 and Q2 to conduct, illuminating the LED.

Q1 and Q2 do not turn on "hard", i.e. they don't saturate, so there will be some voltage lost across Q2. There is also about 2V dropped across a typical red LED. The value of R3 that I've chosen gives an LED current of about 10 mA from a 9V battery.

If you want to learn more about this circuit, and basic electronics in general, Google Darlington pair and Ohm's Law.

 

[BobK]

This circuit brings up something that has always bugged me about the Darlington configuration. As Kris said, it gives you the product of the current gain of the two transistors, so that is great. But, the second transistor cannot saturate because it's collector (shared with that of the first) would then be at too low a voltage to put enough current through the first transistor. Typically, Darlingtons will have a voltage drop of much higher than a single saturated transistor.

However, a slight modification, as shown in the circuit below, can give you the same high current gain, while allowing the second transistor to fully saturate. The blue trace is the voltage at the collector of Q4, the Darlington one. The green trace is the voltage at the collector of Q2, the modified version. Clearly, Q2 is in saturation and is dropping only 90mV while Q4 is not and is dropping 860mV. This could make a big difference in power dissipation if the currents were higher, requiring a heat sink where it would not be needed with the modification.

So why would one use the Darlington and not the circuit on the right, which one might call cascaded transistors?

 

[KrisBlueNZ]

But, the second transistor cannot saturate
[...]
This could make a big difference in power dissipation if the currents were higher, requiring a heat sink where it would not be needed with the modification.

Yes, all agreed.

So why would one use the Darlington and not the circuit on the right, which one might call cascaded transistors?

The Darlington has two characteristics I can see that are advantages in some situations:

• The Darlington connection forces all of the current through the load path, so no current is "wasted";
• The alternative circuit doesn't limit the base current into Q2. This current is limited only by the base current into Q1 and Q1's current gain.

There will also be some difference related to the fact that in the Darlington configuration, Q1's collector is swinging around along with Q2's collector, whereas in the alternative circuit, Q1's collector is at a fixed voltage so Q1's V_CE will also be almost constant and it won't suffer from Miller effect.

There could be other differences as well. Those are the ones I can figure out myself.

You can avoid the second problem to some extent by adding a resistor in series with Q1's collector or emitter. The collector is normally the better option.

I agree that your circuit is more appropriate in some cases. I've used it myself when it's more appropriate. But it isn't always better.

 

[OLIVE2222]

The only no drain solution I see is a mechanical arrangement with a reed contact and a magnet.