This is a small device that the manufacturer of our underfloor electric heating provided. It measures the resistance of the wires to determine whether the cable is intact and not shorted.
The inside shows a slot for a standard 9V battery, a speaker and a small PCB.
This is the PCB, front and back
And this is the schematic, reproduced by me in KiCad
The first thing I notice is R1 which I first identified as a 20k resistor (now corrected in schematic). This is probably much too high to allow LED1 to light up. Did I read the colours the wrong way? They are: red-black-black-red-brown. This would mean it can either be a 20 kO ├▒1% resistor or a 120 O ├▒2%. The latter seems much more reasonable to put in series with a LED so I will assume that is the actual value.
After connecting my lab supply, adjusting it to 9V and toggling the switch both LEDs turn on. The buzzer remains quiet (read below, the reason being I resoldered it improperly when it came loose earlier).
I first identified this resistor as 6.8k but this also feels as too high, also considering the LED lights up pretty intensely. The colouring is bleu-grey-red-gold though, with the gold assuring the colour-read-direction this should indeed be a 6.8k resistor.
The current through R2 and LED2, ignoring Q1, should be $I=V/R=9/6800=0.0013A=1.3mA$. While low, this current might be enough for such a small 3mm red-coloured LED.
This resistor measures 20k in circuit. With the transistor in off-state I believe this resistor can be measured in circuit and the value is in accordance with the colouring.
Next I wonder when the buzzer BZ1 is supposed to make a sound. Supposedly the buzzer should sound when a DC voltage is applied over it because the circuit is not set up to provide AC.
I notice the component is is labelled '3-24V DC' and after applying 9V over the buzzer contacts a loud beep sounds.
Because the emitter of Q1 is connected to ground this transistor is in the 'Common Emitter' configuration.
The datasheet of the 2N3904 shows a V~BE~ at saturation between 0.65 and 0.95V. This means that when the voltage between pins 1 and 2 of Q1 reach this value the transistor will be in saturation.
When the switch is ON but G/L1/L2 are unconnected. Both LEDs are on in this state.
Why does the buzzer/beeper not sound? According to the schematic there should be 9V across it. ~~I notice there is probably an error in my schematic (and annotated image of the back of the PCB) because in the image the buzzer connections are not on a vertical line while they are on the front.~~ The reason is that I soldered the buzzer wrong after the black cable came loose earlier. Resoldering it properly and it does indeed sound.
Or in other words: How to set the transistor Q1 to the cut-off state. To test that, let's lower V~BE~ to say 0.5V.
Test 1: I will place a resistor between L1 and L2 of 1k. ~~Not working, the sound still goes off~~ (I connected G and L1 instead of L1 and L2). Both sound and LED2 remain off.
Test 2: A very low value resistor (say 10 ohms). Both sound and LED2 remain off.
Test 3: Connect a potentiometer between L1 AND L2 and determine at what values the state changes, and this appears to be from around 1250 ohms. Any higher and the beep quickly gains in volume (as the transistor moves through the active state to saturation).
Test 4: Is there also a lowest-boundary? When L1 and L2 are shorted both sound and LED2 remain off.
This image is from the Aliexpress product page
After another test the alarm does indeed sound then G is shorted to L1 and L2. Why is that?
To L2 is simple because then the transistor is bypassed entirely.
To L1 is because then the voltage at G is applied to the base of Q1, quickly moving it to saturation.
What remains unclear to me is why the Open Circuit shows 'N/A' in the table, because when the leads remain unconnected (open) the alarm does sound.