# How to construct this simple circuit properly?

I am a total newbie in electronics and circuit design. I am working on a project that involves a AC motor and four 1W LEDs. I found a motor that works for me and it comes with a AC/AC power adaptor (Adaptor's Output: AC 6V 500mA). I need to add 4 1W LEDs to the circuit that share the same power source.

After doing some research on the web I know I need to use a bridge rectifier to convert the AC to DC so the LEDs will work, but I don't know which rectifier to use. Also, I don't know which capacitor to use (to smooth the output) and if any resistor(s) need to be in the circuit. I included a schematic of the circuit, but I am not sure if the schematic is correct or not. Any help will be much appreciated. If you use four leds in parallel then the total current you'll need is 4x the current of each led (and I doubt that your supply will be able to provide that anyway).

Since you have a 6v AC source that will be rectified to a level of about 6v * 1.414 = 8.48v I would suggest you connect the leds in two groups of two leds in series each, assuming that the forward voltage drop of each led is about 3-4v (which is normal for 1-3W leds).
By doing that you reduce the power consumption of the power supply (you'll need half the current) and reduce the heat dissipation on the resistor connected in series with the leds (will have to drop a much lower voltage). simulate this circuit – Schematic created using CircuitLab

I have omitted the motor because I didn't find a symbol for it in the editor

What capacitor should I use?

The ripple voltage when using a full wave rectifier can be calculated from the following equations:

$$V_{ripple} = \frac {I_{load}}{(2 \times f \times C)}$$ where $I_{load}$ is the output current, f is the 50 or 60 Hz mains frequency, C is the capacitor value in Farad. The result is in volts peak to peak.

As a rule of thumb a 8000uF capacitor will have a ripple of 1v p-p with 1A load, if you half the load current you half the ripple, if you half the capacitor size you double the ripple.

• This is just a comment not a criticism because I don't know myself, but I wonder if there's a good reason the cap is needed at all? After the bridge the 120Hz of flicker would presumably be way too fast to be visible, although I'm not sure if it would have any negative effect on the life of the LEDs. – PeterJ Jan 24 '14 at 12:20
• @PeterJ I find flickering kind of annoying but ignoring that, wouldn't the led provide a higher luminosity when there is a rezervor capacitor keeping it constantly on rather than turning on/off with every cycle? – alexan_e Jan 24 '14 at 13:28
• The average voltage would be the same, it'd stop the drop-outs from the voltage hitting zero but also lower it on the way back up as it recharged. But at the same time it makes not exceeding the maximum current much easier to calculate. Anyway just made me wonder, might be a good new question at some stage not suggesting it's something worth addressing for this question. – PeterJ Jan 24 '14 at 13:41

Bridge rectifier is just 4 diodes connected like in the schematic. For simple circuits, any will work as long as the maximum current and reverse voltage doesn't exceed the rated value. 1N4001 should do it for this case. There are also rectifier in one complete package, also take note of the rated current and reverse voltage when buying one.

Capacitor is just for smoothing the voltage output. Ripple voltage is calculated from Vripple = I/(2fC). Due to low input voltage and high current draw, ripple will be very high. For this particular case, driving the LED directly with the AC should give you about the same result (given that the LED can withstand the reverse voltage, which it should), or you would need a very big capacitor. Unless you use some regulator like 78xx ICs.

For the LEDs, you should put a current limiting resistor in series with each too. Subtract input voltage from the forward voltage of your LED to find the voltage drop on the resistor. Then you can calculate the value of the resistor from your required current. Do take note of the maximum power of the resistor you use to prevent frying it.

But for the given schematic it may not work as expected since

Input 6V*sqrt(2)*500mA = 4.2W

But you are trying to connect 4*1W LEDs = 4W plus current drawn from the motor which should well exceed the input power. So your LED might not shine as bright if you want this to work.

• This could be more helpful if you explain the criteria for selecting components instead of providing a part number and saying "xxxx should do it." – JYelton Jan 24 '14 at 6:12