A few days ago I had the need to measure the load of an appliance, so I dug out my "Kill-A-Watt" power meter. The purpose of this device is to measure not only the load of the appliance in watts, but it will also measure the line voltage, frequency, and provide a running total of consumed power over time in kilowatt-hours (kWh). Usefully, this device will also measure things like power factor and volt-amps (Vars) - both things that can be useful in determining how much actual load something may be putting on your generator. (For more info on power factor and volt-amps, read the Wikipedia article linked here.)
I was both surprised and annoyed when I plugged in the unit and it informed me that the line voltage was about 146 volts so I grabbed a voltmeter and found it to be a more reasonable 123 volts - a typical voltage in the U.S. for the circuits powering lower-power household devices: Something was wrong! As I looked at the Kill-A-Watt's display I noticed something else: It seemed to be flickering slightly - something that I'd not noticed it doing before, but it was also a clue as to what might be wrong.
This was clearly due to a power supply problem within the unit - but since power supplies in these sorts of devices are often very simple, I figured that it would be pretty easy to fix and upon opening it up, I immediately recognized it as a typical "capacitive dropper".
A "capacitive dropper" power supply:
One of the simplest and cheapest ways to get a low current supply from mains voltage is to put a capacitor in series with it - and this is both smaller, lighter and less expensive than using a power transformer. If you aren't familiar with the use of capacitor droppers, it may seem strange - but it can be quite effective and safe if done properly. While a simple series resistance may seem more intuitive, it has the problem that it can generate quite a bit of heat. Let's take a simple example.
Suppose that we need 10 milliamps at 5 volts for a hypothetical oven clock, ignoring the "converting to DC" part for the moment. Doing the math we see that we need to drop from a nominal 230 volt mains (230 - 5 =) 225 volts - such as what might be found in the power supply for an electric oven. With 20 volts to drop, Ohm's law tells us that we need (225 volts / 0.01 amps = ) 22500 ohms - so let's pick the closest standard value of 22000 ohms (a.k.a. 22k).
While we could use an 22k resistor for this, knowing the voltage drop (225 volts) and current (0.01 amps we can see that we would be dissipating (225 volts * 0.01 amps = ) 2.25 watts. While this doesn't sound like much power, containing this much heat in a very small box would cause it to get a bit warm - and it would heat up everything else as well, probably shortening the life of these other components.
Alternatively, we can take into account the fact that our mains power will be a sine wave - 60 Hz in the U.S., 50 Hz in most other places - and use this to provide reactive current limiting - and by using the reactance of the capacitor, we can get the same voltage drop, but without any heat! This lack of heat has to do with the fact that unlike a pure resistance, a pure reactance - like that of a capacitor (or an inductor) is theoretically loss-less.
Comment:
Remembering that capacitors will block DC, it makes sense that as the frequency increases, more current can flow through a given capacitance and this is calculated using a simple formula:
Z = 1 / (2 * Pi * Frequency * capacitance)
Where:
For our purposes we can simply consider the "reactance" to be equivalent to resistance - the clue being that like resistance, it's value is in Ohms.
We can see from this formula that with frequency and capacitance in the denominator (bottom) of the fraction that if we increase either one, the resistance goes down in proportion. Because from the math above we already know that we need 22k of equivalent resistance, we can rearrange the formula, swapping the locations of capacitance and "Z" to solve for capacitance, as in:
capacitance = 1 / (2 * Pi * Frequency * Z)
So, for 60 Hz:
1 / ( 2 * 3.14 * 60 Hz * 22000 ohm) = 0.00000012 Farads = 0.12 uF (microFarads)
And for 50 Hz:
1 / ( 2 * 3.14 * 50 Hz * 22000 ohm) = 0.00000014 Farads = 0.14 uF
Recalling that the lower the frequency, the higher the effective resistance and the lower the current so if we want to make this work for both 50 and 60 Hz systems, we'll pick the closest standard capacitor value for the lower frequency, 50 Hz, or 0.15 uF (a.k.a. 150nF) to make sure that we can get the minimum current that we need.
At this point it's worth mentioning that neither the resistor or capacitor will actually reduce the voltage - it only limits the current: If you want to reduce the voltage to something useful some sort of regulator circuit is required - typically one that clamps it at or below the desired value.
A practical circuit for doing this is depicted in Figure 2, below:
Discussion:
While this may seem a bit complicated at first, it's easy to break down.
On the left we can see "C1" - the "dropping" capacitor that we calculated. Across this capacitor is R2, a 470k ohm resistor and the purpose of this high-value resistor is to bleed off any charge that might be across the capacitor if it happened to be disconnected at the instant the the sine wave of the AC mains was at its peak, preventing it from shocking the user with that stored charge. The resistor R1 is used to provide a bit of series current limiting: If the power switch were closed at the instant that the sine wave of the mains was at its peak, there would be a sudden surge through capacitor C1 and R1 limits this amount to prevent damage to other components.
The rest of this circuit converts the AC to DC and limits the voltage to a reasonable value: D1 is a full-wave DC rectifier and C2 filters the ripple to make it more pure. Resistor R3 provides a bit of current limiting so that IC1, a programmable Zener diode that in combination with R4 and R5 can do the job of clamping the current to no more than 5 volts.
What was wrong with the Kill-A-Watt?
By now you have probably figured out where I was going with this discussion.
The display reading a high voltage indicated that it was possible that the computer within the Kill-A-Watt was getting too-low a voltage, causing its internal voltage reference to also be low which, in turn, caused its reading of the voltage to be too high. The flickering display was also a clue indicating that there was AC mains ripple on its power supply. Carefully checking the voltage at the input of the onboard 5 volt regulator of the unit I saw that it was just 5.2 volts instead of the needed 7+ volts, with about 300 millivolts of ripple - far too low a voltage for the regulator to work properly and the ripple explaining why the display was flickering.
Both of these pointed at the high likelihood of its "dropping capacitor" being bad, and upon opening it up I spotted the capacitor on the rear board. Removing this capacitor, marked "0.47 uF" and checking it, I noted that it read 0.31uF - about 2/3rds of its original value.
On discovering this I knew that I would need another 0.47uF capacitor of the same voltage rating (e.g. 250 volts AC) and that it should be an "X1" or "X2" safety capacitor. This last point is important as these types of capacitors are designed to fail open rather than short out - something that would certainly result in spectacular destruction of the device! Its worth noting that for capacitive dropper power supplies, one should only use the appropriate safety-rated capacitors, such as an X1 or X2. (Note, the original X1 capacitor was rated for somewhat higher peak voltage than the X2 with which it was replaced, but both are rated for this particular type of service at this voltage.)
This "safety" property is also a clue as to why its value reduced from 0.47 to 0.31uF. One of the reasons why these capacitors can fail "open" is that their internal conductors, when subject to a fault - say a power surge, a significant spike, or just age - will fail open. Often this doesn't happen with the entire capacitor, but small portions of the thin, conductive metal film within the capacitor will degrade, the result being that the value of the capacitor will gradually drop. In this case, as the capacitor failed the Kill-A-Watt's circuitry was no longer able to get enough current, causing the voltage to be pulled below the threshold at which it would operate properly, resulting in erratic operation.
Fortunately I was able to rummage around in my box of film capacitors and find a 0.47uF film unit that was similarly rated and was a safety capacitor (with an "X2" rating). Even though it was the same height as the old one (a "taller" capacitor would not have fit in the case) it was about 25% longer - but the designers of the circuit board had left a series of holes to accommodate several different sizes of capacitors.
Success!
With the new capacitor installed I plugged the unit in and it worked normally!
Many devices uses these "capacitor dropper" supplies such as appliance timers, the electronic controls of stoves and ranges and nightlights. If you have one of these devices that has stopped working - or slowly "faded out" over time - there is a good chance that this is because the main dropper capacitor in its power supply has reduced in value - but if you do attempt a repair, extreme care must be taken if you test it as the entire circuit will be "lit up" by and not isolated from the power mains and posing a possible shock hazard. If you do replace its capacitor it must not only be of the marked value of the original capacitor, but it must be an X1 or X2 type safety capacitor of equal or greater voltage rating!
[End]
This page stolen from ka7oei.blogspot.com
 |
| Figure 1: 147 volts on the power mains? I don't think so! Click on the image for a larger version. |
I was both surprised and annoyed when I plugged in the unit and it informed me that the line voltage was about 146 volts so I grabbed a voltmeter and found it to be a more reasonable 123 volts - a typical voltage in the U.S. for the circuits powering lower-power household devices: Something was wrong! As I looked at the Kill-A-Watt's display I noticed something else: It seemed to be flickering slightly - something that I'd not noticed it doing before, but it was also a clue as to what might be wrong.
This was clearly due to a power supply problem within the unit - but since power supplies in these sorts of devices are often very simple, I figured that it would be pretty easy to fix and upon opening it up, I immediately recognized it as a typical "capacitive dropper".
A "capacitive dropper" power supply:
One of the simplest and cheapest ways to get a low current supply from mains voltage is to put a capacitor in series with it - and this is both smaller, lighter and less expensive than using a power transformer. If you aren't familiar with the use of capacitor droppers, it may seem strange - but it can be quite effective and safe if done properly. While a simple series resistance may seem more intuitive, it has the problem that it can generate quite a bit of heat. Let's take a simple example.
Capacitor dropper supplies have, by their very nature, a very poor "power factor" - that is, the waveform of the voltage is not in phase with the current through them. What this means is although our example may be pulling about (235 volts * 0.01 amps =) 2.35 watts, because the voltage and current aren't following the sine wave at the same time, if you were to put a Kill-A-Watt on such a circuit, it would not read 2.35 watts - in fact, it may not read anything at all! Why? Remember that watts is "volts * amps" - but if the current and voltage are out of phase far enough, the voltage may be zero while the current is at maximum - and later in the sine wave's cycle, the voltage may be at maximum but the current may be zero - and in either case, the math tells us that that would be zero watts. The Kill-A-Watt can also read "volt-amps" - which will correctly indicate the "volts * amps" - even when they don't happen at the same time. If the "power factor" is 1.0, watts and volt-amps will be the same, but if it is something other than 1.0, the volt-amps will be higher than the watts. Why do we care? A generator or inverter can deliver only so many amps - and the "watts" rating that they have always assumes that the power factor is perfect. If your generator load has a terrible power factor - say 0.5, that means that the "watts" reading is about half of the "volt-amps" reading - but since the amps is the same in either case it may appear that the generator cannot supply the power. In other words, if your generator is trying to power, say, computers that have a terrible power factor, you may find that it will trip out at a lot lower wattage than you might expect! |
Suppose that we need 10 milliamps at 5 volts for a hypothetical oven clock, ignoring the "converting to DC" part for the moment. Doing the math we see that we need to drop from a nominal 230 volt mains (230 - 5 =) 225 volts - such as what might be found in the power supply for an electric oven. With 20 volts to drop, Ohm's law tells us that we need (225 volts / 0.01 amps = ) 22500 ohms - so let's pick the closest standard value of 22000 ohms (a.k.a. 22k).
While we could use an 22k resistor for this, knowing the voltage drop (225 volts) and current (0.01 amps we can see that we would be dissipating (225 volts * 0.01 amps = ) 2.25 watts. While this doesn't sound like much power, containing this much heat in a very small box would cause it to get a bit warm - and it would heat up everything else as well, probably shortening the life of these other components.
Alternatively, we can take into account the fact that our mains power will be a sine wave - 60 Hz in the U.S., 50 Hz in most other places - and use this to provide reactive current limiting - and by using the reactance of the capacitor, we can get the same voltage drop, but without any heat! This lack of heat has to do with the fact that unlike a pure resistance, a pure reactance - like that of a capacitor (or an inductor) is theoretically loss-less.
Comment:
An "inductive dropper" power supply is also possible - and in many instances, it would be preferable - but it is far easier and cheaper to make small, low-loss capacitors than low-loss inductors, so it is done in only very special circumstances.
Remembering that capacitors will block DC, it makes sense that as the frequency increases, more current can flow through a given capacitance and this is calculated using a simple formula:
Z = 1 / (2 * Pi * Frequency * capacitance)
Where:
Z = reactance in ohms
Frequency is in Hertz
capacitance is in Farads
For our purposes we can simply consider the "reactance" to be equivalent to resistance - the clue being that like resistance, it's value is in Ohms.
We can see from this formula that with frequency and capacitance in the denominator (bottom) of the fraction that if we increase either one, the resistance goes down in proportion. Because from the math above we already know that we need 22k of equivalent resistance, we can rearrange the formula, swapping the locations of capacitance and "Z" to solve for capacitance, as in:
capacitance = 1 / (2 * Pi * Frequency * Z)
So, for 60 Hz:
1 / ( 2 * 3.14 * 60 Hz * 22000 ohm) = 0.00000012 Farads = 0.12 uF (microFarads)
And for 50 Hz:
1 / ( 2 * 3.14 * 50 Hz * 22000 ohm) = 0.00000014 Farads = 0.14 uF
Recalling that the lower the frequency, the higher the effective resistance and the lower the current so if we want to make this work for both 50 and 60 Hz systems, we'll pick the closest standard capacitor value for the lower frequency, 50 Hz, or 0.15 uF (a.k.a. 150nF) to make sure that we can get the minimum current that we need.
At this point it's worth mentioning that neither the resistor or capacitor will actually reduce the voltage - it only limits the current: If you want to reduce the voltage to something useful some sort of regulator circuit is required - typically one that clamps it at or below the desired value.
A practical circuit for doing this is depicted in Figure 2, below:
 |
| Figure 2: A typical "capacitive dropper" power supply. This image is from the Wikipedia article "Capacitive Power Supply" - link This supply was designed for use with 220-240 volt mains so a lower- value capacitor could be used for C1 and still obtain 10 milliamps. Click on the image for a larger version. |
Discussion:
While this may seem a bit complicated at first, it's easy to break down.
On the left we can see "C1" - the "dropping" capacitor that we calculated. Across this capacitor is R2, a 470k ohm resistor and the purpose of this high-value resistor is to bleed off any charge that might be across the capacitor if it happened to be disconnected at the instant the the sine wave of the AC mains was at its peak, preventing it from shocking the user with that stored charge. The resistor R1 is used to provide a bit of series current limiting: If the power switch were closed at the instant that the sine wave of the mains was at its peak, there would be a sudden surge through capacitor C1 and R1 limits this amount to prevent damage to other components.
These capacitive dropper circuits, while simple, have a drawback: The cannot and must notever be used on circuits that can come in contact with anything that might be referenced to ground such as a body. Because it is directly connected to the mains it is possible that if a person or animal touches any part of the circuit - or even touches something powered by it - it could result in a dangerous or fatal electrical shock! The other caveat is that these circuits rely on the fact that a mains supply is a pretty good sine wave. If someone plugs this into an AC power supply that is not a nice, clean sine wave - such as an inexpensive 12 volt power inverter - it may be fed with a square-type wave - sometimes called a "modified sine wave" - which, unlike a true sine wave, has a lot of energy at higher frequencies. What this means is that the reactance at these higher frequencies will be lower and too much current will flow through the capactive dropper and the device that it is powering - including a Kill-A-Watt - will probably be damaged/destroyed! |
The rest of this circuit converts the AC to DC and limits the voltage to a reasonable value: D1 is a full-wave DC rectifier and C2 filters the ripple to make it more pure. Resistor R3 provides a bit of current limiting so that IC1, a programmable Zener diode that in combination with R4 and R5 can do the job of clamping the current to no more than 5 volts.
What was wrong with the Kill-A-Watt?
By now you have probably figured out where I was going with this discussion.
The display reading a high voltage indicated that it was possible that the computer within the Kill-A-Watt was getting too-low a voltage, causing its internal voltage reference to also be low which, in turn, caused its reading of the voltage to be too high. The flickering display was also a clue indicating that there was AC mains ripple on its power supply. Carefully checking the voltage at the input of the onboard 5 volt regulator of the unit I saw that it was just 5.2 volts instead of the needed 7+ volts, with about 300 millivolts of ripple - far too low a voltage for the regulator to work properly and the ripple explaining why the display was flickering.
 |
| Figure 3: The "bad" capacitor. Although marked 0.47uF, it measured 0.31uF. This is an "X1" type "safety capacitor, specifically designed for this use. Generally speaking, this could be replaced with either an X1 or X2 type of appropriate voltage rating. Click on the image for a larger version. |
On discovering this I knew that I would need another 0.47uF capacitor of the same voltage rating (e.g. 250 volts AC) and that it should be an "X1" or "X2" safety capacitor. This last point is important as these types of capacitors are designed to fail open rather than short out - something that would certainly result in spectacular destruction of the device! Its worth noting that for capacitive dropper power supplies, one should only use the appropriate safety-rated capacitors, such as an X1 or X2. (Note, the original X1 capacitor was rated for somewhat higher peak voltage than the X2 with which it was replaced, but both are rated for this particular type of service at this voltage.)
This "safety" property is also a clue as to why its value reduced from 0.47 to 0.31uF. One of the reasons why these capacitors can fail "open" is that their internal conductors, when subject to a fault - say a power surge, a significant spike, or just age - will fail open. Often this doesn't happen with the entire capacitor, but small portions of the thin, conductive metal film within the capacitor will degrade, the result being that the value of the capacitor will gradually drop. In this case, as the capacitor failed the Kill-A-Watt's circuitry was no longer able to get enough current, causing the voltage to be pulled below the threshold at which it would operate properly, resulting in erratic operation.
 |
| Figure 4: The new capacitor on the board. While the new capacitor was the same height as the old - important for being able to fit in the case - it was longer. The circuit board has a series of holes on one side allowing different-sized capacitors to be used. Click on the image for a larger version. |
Fortunately I was able to rummage around in my box of film capacitors and find a 0.47uF film unit that was similarly rated and was a safety capacitor (with an "X2" rating). Even though it was the same height as the old one (a "taller" capacitor would not have fit in the case) it was about 25% longer - but the designers of the circuit board had left a series of holes to accommodate several different sizes of capacitors.
Success!
With the new capacitor installed I plugged the unit in and it worked normally!
Many devices uses these "capacitor dropper" supplies such as appliance timers, the electronic controls of stoves and ranges and nightlights. If you have one of these devices that has stopped working - or slowly "faded out" over time - there is a good chance that this is because the main dropper capacitor in its power supply has reduced in value - but if you do attempt a repair, extreme care must be taken if you test it as the entire circuit will be "lit up" by and not isolated from the power mains and posing a possible shock hazard. If you do replace its capacitor it must not only be of the marked value of the original capacitor, but it must be an X1 or X2 type safety capacitor of equal or greater voltage rating!
[End]
This page stolen from ka7oei.blogspot.com