I want to use a capacitor for an alarm clock but am not completey sure on what size to use?
Matt F asked:
This is a little project I am undertaking for the fun of it. My alarm clock doesn’t have battery backup so I wanted to install a capacitor in the event of short brownouts of a minimum of 3 seconds. I am not sure of the secondary voltage of the alarm clock but it is no more than10 volts as their are other capacitors rated for 10 volts. I measured the amperage during normal operation at .2 amps. So the total resistance should be around 50 ohms. I am thinking I will need a fairly large capacitor like in the milifarad range. Thanks!
This is a little project I am undertaking for the fun of it. My alarm clock doesn’t have battery backup so I wanted to install a capacitor in the event of short brownouts of a minimum of 3 seconds. I am not sure of the secondary voltage of the alarm clock but it is no more than10 volts as their are other capacitors rated for 10 volts. I measured the amperage during normal operation at .02 amps. So the total resistance should be around 50 ohms. I am thinking I will need a fairly large capacitor like in the milifarad range. Thanks!
This is a little project I am undertaking for the fun of it. My alarm clock doesn’t have battery backup so I wanted to install a capacitor in the event of short brownouts of a minimum of 3 seconds. I am not sure of the secondary voltage of the alarm clock but it is no more than10 volts as their are other capacitors rated for 10 volts. I measured the amperage during normal operation at .02 amps. So the total resistance should be around 500 ohms. I am thinking I will need a fairly large capacitor like in the milifarad range. Thanks!
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!READ THE DETAIL THAT IS DIRECTLY BELOW THIS ONE BECAUSE IT IS CORRECT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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This is a little project I am undertaking for the fun of it. My alarm clock doesn’t have battery backup so I wanted to install a capacitor in the event of short brownouts of a minimum of 3 seconds. I am not sure of the secondary voltage of the alarm clock but it is no more than10 volts as their are other capacitors rated for 10 volts. I measured the amperage during normal operation at .2 amps. So the total resistance should be around 50 ohms. I am thinking I will need a fairly large capacitor like in the milifarad range. Thanks!
This is a little project I am undertaking for the fun of it. My alarm clock doesn’t have battery backup so I wanted to install a capacitor in the event of short brownouts of a minimum of 3 seconds. I am not sure of the secondary voltage of the alarm clock but it is no more than10 volts as their are other capacitors rated for 10 volts. I measured the amperage during normal operation at .02 amps. So the total resistance should be around 50 ohms. I am thinking I will need a fairly large capacitor like in the milifarad range. Thanks!
This is a little project I am undertaking for the fun of it. My alarm clock doesn’t have battery backup so I wanted to install a capacitor in the event of short brownouts of a minimum of 3 seconds. I am not sure of the secondary voltage of the alarm clock but it is no more than10 volts as their are other capacitors rated for 10 volts. I measured the amperage during normal operation at .02 amps. So the total resistance should be around 500 ohms. I am thinking I will need a fairly large capacitor like in the milifarad range. Thanks!
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!READ THE DETAIL THAT IS DIRECTLY BELOW THIS ONE BECAUSE IT IS CORRECT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
January 15th, 2009 at 9:00 am
use yor RC time constant formula to determine how long the cap’s charge need to last to keep the clock powered and if a series resistor is needed to control the discharge
January 16th, 2009 at 1:11 pm
Better check your math first. If e= 120v and r = 50, you then have a 2.5 amp alarm clock….don’t think so. Once you know the true current consumption, you just need to calculate the capacitive time constant.
January 19th, 2009 at 8:02 pm
well, charge in cap = voltage x value of cap, so value of cap = charge / voltage. You want the charge to last 3 sec, and charge = current x time, so cap value = (0.2Ax3sec)/10v=0.06F, which is a large cap. Remember that you can use smaller caps in parallel if this is too large.