DIY buzzer alarm query

DTismyname · May 1, 2021

Hi I’m new to the forum and have no electrical background other than changing plugs and light fittings at home. My house was recently targeted by burglars attempting to get in the house by snapping the lock whilst I was asleep. So I have had an idea to fit a solar powered PIR light pointing towards the door so that when someone approaches close to the door the light will come on and the buzzer will sound in the house giving me a chance to catch them in the act before they snap the lock (i have anti-snap locks). My only problem is after hours of trial and error there is only 1 point i can connect to, that controls the buzzer with the PIR but it isn’t as loud as it is when connected straight to the battery so I’m thinking that something is restricting the power at some point on the circuit.
Any help would be appreciated even if you think I should bin the idea

Thanks and sorry for the novel of a post

Spoon · May 1, 2021

Welcome to the forum mate.
Maybe the Solar Panel doesn't have enough power to power the PIR and the buzzer.

Lucien Nunes · May 1, 2021

What voltage is the buzzer rated at? The available voltage seems to be 3.7V from the battery (actually cell, as there's only one). If the buzzer is rated for a higher voltage it will be quieter at 3.7V.

The LEDs are switched by transistor Q1 and current-limited by R10 and R11 in parallel making 0.25Ω. If the buzzer is suitable you would need to connect its positive lead to L+ and its negative to the junction of Q1 and R11. This is assuming the IC does not use PWM drive to the LEDs (impossible to tell from here) in which case additional steps will be needed to make the output suitable for the buzzer.

One option is to power the buzzer from an external source of a suitable voltage and use a transistor to switch it in response to the falling voltage on Q1's collector. If you wanted to try this we could perhaps guide you through it. I assume as much of your interest is in the DIYing aspect as achieving the result.

DTismyname · May 1, 2021

Spoon said:
Welcome to the forum mate.
Maybe the Solar Panel doesn't have enough power to power the PIR and the buzzer.

Thanks for your reply pal. The solar panel doesn’t actually power the unit, it just charges the 18650 Li-ion battery but you could be right

DTismyname · May 1, 2021

Lucien Nunes said:
What voltage is the buzzer rated at? The available voltage seems to be 3.7V from the battery (actually cell, as there's only one). If the buzzer is rated for a higher voltage it will be quieter at 3.7V.

The LEDs are switched by transistor Q1 and current-limited by R10 and R11 in parallel making 0.25Ω. If the buzzer is suitable you would need to connect its positive lead to L+ and its negative to the junction of Q1 and R11. This is assuming the IC does not use PWM drive to the LEDs (impossible to tell from here) in which case additional steps will be needed to make the output suitable for the buzzer.

One option is to power the buzzer from an external source of a suitable voltage and use a transistor to switch it in response to the falling voltage on Q1's collector. If you wanted to try this we could perhaps guide you through it. I assume as much of your interest is in the DIYing aspect as achieving the result.

I really appreciate the detailed reply, thank you.
the back of the buzzer says it is 3-24v DC. When I connect it straight to the battery it is significantly louder than when connected to L+ and the junction of Q1 and R11. Now that I have had the idea and worked out what I want to do in my head I am determined to see it through… All I need is a kick in the right direction so any guidance would really be appreciated

Lucien Nunes · May 1, 2021

There are two possible reasons for the difference in sound level you describe. One is simply the voltage drop in Q1, the switching transistor. If this is a normal bipolar transistor driven into saturation, this will be around 0.2-0.3V and unlikely to cause a major change in volume. The other is that the LED brightness might be regulated by the control IC driving Q1 with rapid pulses and the average voltage is lower than the peak voltage.

Do you have a multimeter?

DTismyname · May 1, 2021

Lucien Nunes said:
There are two possible reasons for the difference in sound level you describe. One is simply the voltage drop in Q1, the switching transistor. If this is a normal bipolar transistor driven into saturation, this will be around 0.2-0.3V and unlikely to cause a major change in volume. The other is that the LED brightness might be regulated by the control IC driving Q1 with rapid pulses and the average voltage is lower than the peak voltage.

Do you have a multimeter?

Sorry I don’t have a multimeter.