Very useful and informative Brian, thank you.That is a BAD design. Shorting the probes together will damage the transistor and possibly R2 as well.
The current through the LED is limited by two factors, the value of R1 and how much the transistor is conducting. Good design practice says take the worst case scenario of the transistor fully conducting (as though it has a short between collector and emitter) and set R1 to a value that protects the LED. I'm assuming from your calculation that the battery is 9V and R1 is 470 Ohms so if we assume the LED drops about 1.75V and the transistor is fully conducting, the LED current will be (Vbattery - Vled)/470 or about 15mA which is a perfectly acceptable amount.
How much the transistor conducts between collector and emitter will depend on the current allowed to flow into the base pin. The idea behind the circuit is the resistance between the probes ( = the amount of conduction through the water) provides the base current from the positive side of the battery. Some of the probe current will also go through R2 instead of the base so varying it's value will change how much is diverted away from the base and hence how 'sensitive' to conduction the probes appear to be. Note that if R2 is set to minimum resistance and the probes touch each other it shorts out the battery!
I would do the following:
1. check the orientation of the transistor pins, don't assume they are C B E as in your photograph, there are variations.
2. tell us the value of R2. As a guess it should be between 100K and 1M.
3. Add a 1K resistor in series with the probe connected to Q1 & R2. This will not significantly alter operation and it will protect Q1 and R2 if the probes touch each other.
Brian.
That explains things well Brian, thank you.The calculations are wrong I'm afraid.
The current through the collector can at most be (supply voltage - LED drop - Vce) / R1.
Assume the transistor is in saturation (conducting as much as it can) so Vce is zero, the current is (9 - 2)/470 = 0.0149A or about 15mA.
Vbe doesn't feature at all in the collector current calculation. For most signal LEDs like the one you used about 1.7V is a more typical drop.
The base voltage cannot go more than about 0.7V above the emitter voltage because there is a forward conducting diode junction between those pins. If more than 0.7V is applied, the base will sink excessive current and the transistor will be damaged, that's why I said shorting the probes would kill it because it applies 9V from the battery directly.
Also consider that the value of R2 is 10K when set to maximum resistance, as you adjust it the value drops and it can go down to zero. At zero with the probes touching it shorts out the battery but at low values the current through it can burn out R2 itself. Because it can have battery voltage across it, current will flow through it. At low resistance settings the current can be quite high and Ohms law states "power dissipation (W) = Current squared (I^2) times the resistance (R)" so it might run very hot. Bear in mind that the power rating of small potentiometers is typically 0.25W across the whole length of the carbon track. It become progressively less as the length of track carrying the current is made shorter.
Brian.
You’re right Dana – the absence of R2 doesn’t seem to make any difference at all!Circuit in serious need of help. First and foremost R2 is almost pointless because
it sets sensitivity only for a specific battery Voltage. Circuit is hugely T dependent
on performance. Also led will vary in brightness depending on T and ionic contect of
water and beta of transistor.....
Are you trying to detect a specific level or just the presence of water ? If presence liquid
state or gaseous (humidity) state ?
Regards, Dana.
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