Dark sensor - need 500mA output

Question

Here is my dark sensor circuit (below). However, my output current is very low. I get a maximum of 150mA to drive my strip LED. But I require atleast 450mA of output current.

I have tried many combinations. Tried emitter follower / tried replacing it with a N channel mosfet with not much improvements.

What can I do to achieve the desired 450mA?

I dont want to use a 3V relay in order to do this. Neither a solid state relay or a reed relay. Nor an optocoupler (low current). Any other alternate ways to do this?

Answer 1

Here's a "corrected" version of your original circuit, using a 30A-rated MOSFET with Vgs(th)=2.0V and a max Rds(on) of 31m(ohm) @ 4.5Vgs when passing 18A of current (according to the charts, average Rds(on) for ~4Vgs should be under 30m(ohm) when passing only 450mA)

Notice that R4 is a varistor...obviously this wouldn't be necessary in a finished design, but when experimenting with LED drivers, I find it best to use a variable series-resistor until I get everything working right. Then, it's a simple matter to measure the resistance of the "working" setting on the varistor & replace with a near-value fixed resistor in "production models."

^{simulate this circuit – Schematic created using CircuitLab}

-- -- -- EDIT -- -- --

Here's an idea for a circuit I personally would use preferentially to the one above; though its a somewhat different topography than you're using:

^{simulate this circuit}

In this circuit, when the resistance of LDR1 falls below the resistance of R2 (choose an R2 value for whatever "trigger point" resistance of LDR1 you want to switch at), the LM393 comparator's output will quickly transition from high-to-low; then when the resistance of LDR1 increases above that of R2, the comparator output will quickly transition from low-to-high.

The key advantages here are:

1. The comparator transitions quickly, eliminating the "gray area" where the LEDs would be partially energized, but ineffective & simply wasting energy (as well as the MOSFET being in a linear region & heating up with higher dissipated energy itself)

2. The comparator adds some hysteresis into the circuit, reducing the transitioning (flashing, or gray-area) that can often occur when light levels are very near the trigger point of a similar circuit that doesn't have enough hysteresis.

Answer 2

The BJT you're currently using should not be used for currents higher than 100mA.

The MOSFET you used has a threshold (Vgs-th) of 2 - 4V which does guarantee a drain current > 250µA. You need at least 4.5V to get a low enough Rdson -> high enough current for your application.

Use a MOSFET which has a lower Vgs-th given your conditions, then add a resistor in series with your LED strip to reach the desired 450mA. I do not suggest using the MOSFET in a specific operating point without feedback because its Rdson changes a lot with temperature and is not linear. You could however use a sensing resistor which you could use to control the LED current, but there is no need in my opinion to make this more complex if a series resistor meets your needs.