Hi guys I am pretty new to making my own circuits and I wanted to ask how can I increase the output amps on my darkout sensor (pictured) it's basically a 6volt circuit with a BC547 transistor and a 100K resistor running a 20ma Relay (but for some reason won't run the relay?)

Can I simply use a different resistor to let more power through to help run my relay? What value resistor do you recommend?

Thanks in advanced.
Paul.

 

Datasheets for the transistor and relay?
Is the relay supposed to pull in when it's light or dark?
What is the load you want to switch?

ak

 

Data sheet for transistor:
https://www.google.com.au/url?sa=t&...RIoZ14hq3L2YikVUA&sig2=SjVMvHxp-TeeUg27b7Xx1w

I couldn't find a data sheet for the relay it's from jaycar CAT.NO:SY4092

The relay will pull at night to activate a PIR motion sensor, the load isn't much just a 10 SMD LED light, not sure on the exact current draw though.

 

I found the attached diagram that is the same circuit as mine but is 12 volt and uses a 680K resistor, I am thinking I am using the wrong resistor for 6 volts maybe?

 

The fixed resistor value depends on the LDR value at the brightness (or darkness in your case) when you want the relay to pull in. The transistor is acting as a comparator. Its Vbe (base-emitter diode forward voltage) is the trip point since the base is 1 Vbe above GND. The two resistors divide down the battery voltage. As it gets darker, the LDR resistance increases, increasing the voltage at the base. When that voltage is high enough for base current to flow, the relay is activated. For the same LDR in both circuits, a higher battery voltage means a larger fixed resistor to get the battery divided down to Vbe. But it isn't automatic that doubling the battery voltage requires doubling the resistor value.

A complication here is that the turn-on action of a transistor is not crisp. The transistor starts conduction when Vbe is only about 0.4 V, and conduction increases as Vbe increases. Above around 0.6 to 0.7 V, Vbe stops increasing (or continues to increase at a much lower rate). So the exact point where the transistor has enough base current to incite enough collector current to pull in the relay is not as cook-booky as most books indicate, an unfortunate consequence of real-world devices.

By "increase the output amps", I assume you mean changing to a larger relay with higher-rated contacts and a higher coil current requirement. Again, need to know the coil current for the new relay. Depending on that value, one method is to change the transistor to one that can sink more current. Another is to keep in there the relay that works, and then use it to switch on a second, larger relay. Which of these do you want to try first?

Another change would be to replace the transistor with a voltage comparator IC, but you'd still need a transistor driver for the relay.

ak