Tuesday, 5 June 2012

Power On Indicator

Some types of electronic equipment do  not provide any indication that they are  actually on when they are switched on.  This situation can occur when the back-light of a display is switched off. In addition, the otherwise mandatory mains  power  indicator  is  not  required  with  equipment  that  consumes  less  than  10 watts. As a result, you can easily forget  to switch off such equipment. If you want  to know whether equipment is still drawing power from the mains, or if you want  to have an indication that the equipment  is switched on without having to modify the equipment, this circuit provides a solution.
One way to detect AC power current and  generate a reasonably constant voltage  independent of the load is to connect a  string of diodes wired in reverse parallel in series with one of the AC supply  leads. Here we selected diodes rated  at 6 A that can handle a non-repetitive  peak current of 200 A. The peak current  rating is important in connection with  switch-on  currents.  An  advantage  of  the selected diodes is that their voltage  drop increases at high currents (to 1.2 V  at 6 A). This means that you can roughly  estimate the power consumption from  the brightness of the LED (at very low  power levels). The voltage across the diodes serves as  the supply voltage for the LED driver. To  increase the sensitivity of the circuit, a  cascade circuit (voltage doubler) consisting of C1, D7, D8 and C2 is used to double  the voltage from D1–D6. Another benefit  of this arrangement is that both halve- waves of the AC current are used. We use  Schottky diodes in the cascade circuit to  minimise the voltage losses.
The LED driver is designed to operate the LED  in blinking mode. This increases the amount  of current that can flow though the LED when  it is on, so the brightness is adequate even  with small loads. We chose a duty cycle of pproximately 5 seconds off and 0.5 second  on. If we assume a current of 2 mA for good  brightness with a low-current LED and we can  tolerate a 1-V drop in the supply voltage, the  smoothing capacitor (C2) must have a value of  1000 µF. We use an astable multivibrator built around two transistors to implement a  high-efficiency LED flasher. It is dimensioned to minimise the drive current of  the transistors. The average current consumption is approximately 0.5 mA with a  supply voltage of 3 V (2.7 mA when the  LED is on; 0.2 mA when it is off). C4 and  R4 determine the on time of the LED (0.5  to 0.6 s, depending on the supply volt-age). The LED off time is determined by  C3 and R3 and is slightly less than 5 seconds. The theoretical value is R × C × ln2,  but the actual value differs slightly due to  the low supply voltage and the selected  component values.
Diodes D1-D6 do not have to be special  high-voltage diodes; the reverse volt-age is only a couple of volts here due  the reverse-parallel arrangement. This  voltage drop is negligible compared to  the value of the mains voltage. The only  thing you have to pay attention to is the  maximum load. Diodes with a higher  current rating must be used above 1 kW.  In addition, the diodes may require cool-ing at such high power levels.  Measurements on D1–D6 indicate that  the voltage drop across each diode is  approximately 0.4 V at a current of 1 mA.  Our aim was to have the circuit give a  reasonable indication at current levels  of 1 mA and higher, and we succeeded  nicely. However, it is essential to use a  good low-current LED.