Can someone please explain to me how we choose the voltage and capacitance of a capacitor when building a DC and AC circuit. In a full wave rectifier application for example a smoothing capacitor after the diode bridge, how would we choose the voltage and capacitance is there some rule of thumb or some formula? Let's say we have a 110 V AC to 12 V AC step down transformer, then a diode bridge, then a smoothing capacitor. What Voltage would the capacitor need to be the same as the secondary winding 12 V? 20 % higher than 12 V so 14.4 V? OR double 24 V? And what about capacitance in this example I really have no idea how I would choose capacitance 50 nano farads, 100 nano farads, 3000 micro farads? Is there some formula? For AC capacitor applications I can't think of an example please provide one if possible. Is the capacitance that a capacitor needs to be rated at somehow equal to the current it needs to be able to pass through? Like the higher the current that must pass through the capacitor the higher the capacitance? This isn't homework or anything I'm just a hobbyist trying to understand why I see electronics with different capacitors rated at different voltages and capacitance.
But see after the full wave rectifier diode bridge we have DC since it doesn't alternate am I right? So how would we then choose how much capacitance we need? Also I forgot let's say the transformer can output max 1 Amp so 12 V and 1 amp.
So, if my understanding is correct, the capacitance value of a capacitor is essentially how much current it draws into itself, stores, and then when it reaches it maximum current/charge it automatically discharges? In other words the capacitor will not discharge until reaches it's maximum capacitance value?
- Markus ImhofLv 71 month agoFavorite Answer
It's as easy or complicated as you want it to be, and reading a few capacitor data sheets can already give you some ideas - or at least point at directions for further research.
Start with the voltage, that's the easy part. The capacitor's rated voltage should be higher than the maximum voltage you can get across the capacitor. In your example, if the circuit can at any point be unloaded, that would be 1.4 times the unloaded AC voltage of the transformer, which in turn would depend on the specific transformer. Fortunately, the capacitors that would be suitable in this place come in discrete steps - common voltages are 16V and 35V, and the 16V would be pretty marginal, so I'd choose a 35V type.
Capacitance is a bit more difficult. First of all, for a smoothing capacitor, more is better. Nope. Not always. Yes, you want to use that capacitor to reduce the remaining ripple after the rectifier, so the larger the capacity, the less ripple you get. In first order theory. Now take the next order into account: the capacitor will only be recharged if the (periodically oscillating) output voltage from the rectifier exceeds the instantaneous voltage across the capacitor, which depends on the load current. The larger the capacitor, the less it gets discharged by a specific load current, and the shorter the time for the recharge becomes. So, with increasing capacitance, you are also shortening the recharge pulses, while increasing the recharge current. Guess what short, high current pulses will do? Create ripple :-). So, there's a reasonable upper limit to what the capacitance should be. Usually, for 50Hz, that turns out to be in the range of several hundred to a few thousand uF - mostly determined through experience, although you can also calculate or model it.
And then, there's another charateristic of oyur capacitor that can (will, for some applications) become important - the ESR, equivalent series resistance. Yes, and ideal capacitor won't show this - but an ideal capacitor would be a vacuum plate capacitor, which becomes a bit unwieldy for applications like smoothing a 50 Hz ripple. Real capacitors have losses - not too bad in the case of e.g. polyester film capacitors, potentially really bad in the case of electrolytics. Bad enough, if you pass too high a current through the capacitor (here's your AC application) to boil off the electrolyte and, since the capacitor was sealed until this point, make a mess of your lab bench. Also, the ESR will describe the voltage drop across the capacitor when charging/discharging, and for a lot of applications (including AC smoothing in case of audio equipment) you want that ESR as low as feasible - otherwise, any high current demand from your circuit will induce a drop in the supply voltage, which you usally do not want.
Going a bit further into undesired side effects, the dielectric in your capacitor may also be non-linear. Again, not for a vacuum plate capacitor, but any other type of technical capacitor will show that to some degree, worst of all electrolytics. That means, the capacitance of your capacitor will depend, to some extend, on the voltage applied to it. Not that critical if you're dealing with a smoothing electrolytic that has a -50/+100% specification on its capacitance in any case (and would still work with twice that), but a bit more important if that capacitor is a part of an audio circuit and you want to pass your music through it. This is where you then not only will have to decide on capacitance and voltage (and possibly ESR), but also on the finer characteristics of the dielectric. And, after reading what the capacitor you would ideally use in that place, costs - and how big it is - also on whether you perhaps should go back a couple of steps and use a different approach to your design :-)
edit: if you had DC, you wouldn't need a capacitor. But you don't have DC - what you have is an oscillating voltage between 0 and (roughly, if the transformer is loaded to specified current) 16 Volts at twice the mains frequency.
edit regarding your second update: nope. Whether a capacitor charges or discharges is entirely up to the voltage you apply across it. If the external voltage is above the voltage of the capacitor (as described by Voltage = Charge/Capacity), the capacitor is charged. If the external voltage is below the voltage of the capacitor, it is discharged. The current flowing in each case is determined by the resistance of the external circuitry and, for real life capacitors, by the internal resistance of the capacitor and, for real life circuits, the inductance of the whole kaboodle. The last term doesn't usually become important unless you get to radio frequency - but once you get there, even a simple piece of wire (like e.g. the wire on the capacitor) can become an important part of your calculations.
- PhilomelLv 71 month ago
The capacitor on a power supply is meant to reduce the noise (Ripple voltage) on the supplied voltage output. If the ripple it too high increase the capacitance.
- billrussell42Lv 71 month ago
" In a full wave rectifier application how would we choose the voltage and capacitance is there some rule of thumb or some formula? Let's say we have a 110 V AC to 12 V AC step down transformer, then a diode bridge, then a smoothing capacitor"
The voltage should be higher than the DC output, which in this case is 1.4x12 = 16-17 volts. And add to allow for high line voltage, say 130 volts, which gives you 20 volts DC. So you need a 20 v cap at a min, I'd use a 25 v unit to improve reliability. Voltage number is just a rating number, the voltage the cap can withstand without dying. You could use a 500 volt cap in this application, but it would be big and expensive.
The C value can be found from the load current and the ripple voltage allowed. If we allow 1 volt PP of ripple and 1 amp current, or 16 Ω load, with 60 Hz frequency, that means the cap loses 1 volt of charge, which is 1/16 of the charge, in 8 ms (1/2 cycle) at 1 amp.
voltage on a cap, discharging
v = v₀e^(–t/τ)
τ = RC = 16C
15 = 16e^(–8m/16C)
e^(–8m/16C) = 0.94
–1/2000C = -0.065
2C = 1/0.000065
C = 8 mF or 8000 µF
This is only an approximation, as the discharge time is less than 8 ms, more like 6.
this seems complicated, but you can use simple rules. Just by scaling the above you can get to 800 µF if I = 100 mA, for example.
AC applications, I assume you mean coupling or bypass caps. Each one is different, you have to analyze the circuit over the frequencies of interest to determine the value.
If you mean motor caps, that depends on the motor, and you should use the value the manufacturer recommends.
Note that I never mentioned AC current thru the cap. Usually that is irrelevant.
voltage on a cap, discharging
v = v₀e^(–t/τ)
v₀ is the initial voltage on the cap
v is the voltage after time t
R is resistance in ohms,
C is capacitance in farads
t is time in seconds
RC = τ = time constant
re your update: So, if my understanding is correct, the capacitance value of a capacitor is essentially how much current it draws into itself, stores, and then when it reaches it maximum current/charge it automatically discharges? In other words the capacitor will not discharge until reaches it's maximum capacitance value?
that is wrong on so many levels, it is difficult to answer. Study some basic physics and electronics please. Caps do not automatically discharge. The capacitance value is fixed and does not change. The capacitance value is a measure of how much charge the cap can store. That is all.
A capacitor discharges when a resistor or some other conductive component is connected across it, so the charge has a way to discharge. Electrolytic caps have internal resistance due to their construction, and discharge on their own, but it takes seconds to minutes for that to occur. A non electrolytic cap can hold its charge for a long time, even weeks.
- cosmoLv 71 month ago
The "voltage" of a capacitor is just the highest voltage that capacitor can handle without being damaged. So any nominal voltage spec that is higher than that of your circuit is OK. BTW, that voltage spec is just when the capacitor is relatively new --- capacitors age differently depending on how they are made and cheap ones may fail years earlier than a really good one. This is especially true of electrolytics.
The actual capacitance must be large enough to absorb the varying current from the rectifier circuit and smooth it out over an AC cycle. So
C > 1/(fR) where f is the frequency (60 Hz) and R is the characteristic impedance of the circuit. R ~ V/I where V is the voltage and I is the current.
Putting that together,
C > I/(fV) , and so the bigger the current the bigger the capacitor needed, as you thought. With higher circuit voltages and frequencies, less capacitance is needed. If I is in Amps, V is in Volts, and f is in Hz, then C is in farads. A farad is huge, most capacitors are micro or nano farads.
So you might think that a really big value of C is best, and that's true during the continuous operation of the circuit. But during turn-on and turn-off, a huge capacitance can put out huge currents that may damage other components, and that must be taken into account. If you are designing the circuit yourself, it's always a good idea to stick in a fuse, so if there are large transients at turn-on and turn-off, it's the fuse that breaks. Also, this big transient current is why you usually want a "slo-blow" fuse.
Another thing about capacitors is that they have some internal resistance which results in them having a characteristic time constant, faster than which they cannot operate. This becomes dominant at "radio" frequencies. So if you're trying to filter out some high frequency, a big "slow" capacitor may not work. One way around this is "big capacitor little capacitor", where a big capacitor is used in parallel with a small, fast capacitor. This provides some effective capacitance over a wider range of frequencies.