One important part of electronics is converting the alternate current (AC) - the form of current delivered to our homes and offices - into a useful direct current (DC). Although I am just a newbie in this field, rather a hobbyist, making a full wave rectifier that takes an AC signal and rectifies it to a DC signal is just one minute project.
Background
Normally the AC current can be used only by those electric appliances which don't care about the electrons direction. A heater, a bulb or a resistor wouldn't care about it, they would heat as long as the electrons would flow, regardless or their direction. On the other hand, the direction of electrons flow is important for an electrical circuit which may use electrical components sensitive to polarity (see electrolytic capacitors, diodes, transistors, etc).
But why has the mains electricity current to alternate in the first place? Why can't it be direct in the first place and thus making everyone life easier? The reason is the way the electrical current is produced. The long story short: the current generator's rotor is rotating within the magnetic stator and this movement is what causes the electrons to have different potential energy at different points in time.
Look at the image below:
Source: www.electronics-tutorials.ws
From the graph on the right we see that we can describe the electron's potential energy at any point t in time using this function:
V(t)=Vm*sin(ωt+θ).
So this alternating current is not something that the scientist came up with, it's rather a consequence of the technology used to generate the electrical current which as I've said earlier is based on the rotation of an rotor within a magnetic core (the stator).
How to convert AC to DC
So how do we do it? How do we convert this AC to DC? Well, all we have to do is to rectify the alternating current, which half of time is above 0V and the other half of time is below 0V, into a current that has only one direction, that is either positive or negative. With other words we have to trick the electrons to take a path that will always have the same direction. We need a wave rectifier.
How do we trick the electrons to follow only one direction although they move in both directions? We could for instance chose to discard all the electrons that moves from 0V to negative or from 0V to positive. With other words we could choose to rectify only half of the wave. However, to use the current provided by the main at its full potential (ie. the full wave) it would be better to not discard anything, instead we should rectify everything, ie. to rectify somehow the full wave. Such a device/rectifier is called full wave rectifier.
How a full wave rectifier works
A full wave rectifier device consists of four diodes arranged in a such way that allow us to capture all the electrons that moves in both directions and to guide them in one direction only. Look at the image below:
Source: www.electronics-tutorials.ws
With this arrangement the current can only follow one of the following paths:
- the upper positive end of the AC wave -> D1 -> load -> D2 -> the lower negative end of the wave
- the lower negative end of the AC wave -> D3 -> load -> D4 -> the upper positive end of the wave
If you will follow the electron's path as described above you will see that while traversing the load the electrons always pass the load in one direction only (ie. from +V to 0V), regardless of the wave's side they originated from.
The resulted output will look like this:
Source: www.electronics-tutorials.ws
A little demo circuit that proves the concept:
It should work! Let's build our own circuit and test it with the help of an oscilloscope:
This is an ad-hoc circuit that takes a 10VAC as input and rectifies it to a 8.4VDC. I used a 470Ω in series with a 5mm red LED as a load then I measured the output across the positive and the negative half of the wave across D2 and respectively D3 diodes (the plot isn't shown here).
Below is a screenshot taken from the oscilloscope while probing the output across the D3 (cyan) and D4 diodes (yellow).
The probe taken across D3 (see CH2 on cyan) represents the rectification of the positive half wave. The probe take across D4 (see CH1 on yellow) represents the rectification of the negative half wave. The mean voltage on CH1 (yellow) is shown as 4.8VDC while the same parameter for CH2 (cyan) is 4VDC (measurement now shown here). However, when I measured with a DMM across the rectified output it shown 8.9VDC.
So, we've started with 10VAC and ended with 8.9VDC that is shown as 4/4.8VDC peak-to-peak. That's confusing, right? Well, we measured separately each rectification so on one of them we've got a mean 4.8VDC while on the other we've got 4VDC. By adding both we get 8.8VDC which is quite close on the 8.9VDC shown by my DMM. But why we've got 8.9VDC when we've started with 10VAC. Where did 1.1VAC gone? Well, each half wave consists of two diodes that drop the voltage by ~0.5V each, thus 1V together. So (10-1)VAC=9VDC.
But wait! On the CH1 measure table it's shown that peak-to-peak is 15.6VDC while you say that we've started with 10VAC. How's that possible? Good question! When we say 110/230VAC it doesn't mean that the instant voltage is 110/230VAC but instead it's its average/mean value. But not the average you might think of because between -160VAC and +160VAC the mean/average would be 0. Instead it is a Root Mean Square (RMS) value which is calculated by taking one measurement at a time, Square it, take the Mean of all these squares then extracting the Root of that value. So for instance if x1, x2 and x3 are three measurements taken at different points in time then the RMS=SQRT((x12+x22+x32)/3). Back to our problem, the 10VAC is the RMS while the 15.6VAC is the peak-to-peak value. For a waveform in pure sine form the relation between the RMS and peak-to-peak is as following:
RMS=Peak-to-peak / SQRT(2)
So 15.6VAC peak-to-peak should be equal with SQRT(2)*10 VAC ~ 14.14VAC. Since our wave was not a pure sine waveform there might be some differences, though. Hopefully it makes sense.
Since the output current has only one direction it can be called direct current. The only problem is that it's not regulated to a fixed value, it oscillates. Well, this problem can be easily fixed by using an capacitor in parallel with the load. The capacitor has the function of filtering these ripples, the larger its capacity the better constant output:
Source: www.electronics-tutorials.ws
When choosing the capacitor we should make sure that its rated voltage is always larger than our rectified output voltage otherwise...boom and smoke! The capacitor value on the other hand is directly proportional with circuit load current and inversely proportional with the ripple voltage and the ripple frequency (which is twice the input frequency):
C=I/fripple*Vripple
So for instance if our output is 8.9V, the ripple frequency is 100Hz and the maximum expected load is 2A then a 2/(100*8.9) farads capacitor should eliminate completely the output ripple.
Normally a 100μF capacitor should remove the most part of the ripples. The animation below shows how the smoothing capacitor filters out the wave ripples:
... and this is a demo circuit that proves the concept:
A simpler/better solution
It's useful to know the basics, to know how things work. However, sometimes it's simpler/better to pick up an off-the-shelf product which does exactly the same thing. For instance for only 0.9USD (or C$ 1.24) one can buy 5 pieces of full bridge rectifier (ie. 18 cents/piece) that would rectify a load 3A @ 700V. How cool is that?
These devices are very easy to use. Exactly as in our circuit there are 2 points for the AC input and 2 points for the DC output. Connect the DC output to the first last pin (the + and the - one) then connect the AC input to the middle pins. Please note that this bridge only rectifies the AC into a DC but does not remove the ripples. For that use a smoothing capacitor as described earlier.
What's Next
Once your power supply current is rectified from a AC waveform to DC form you would normally want to step-down/up the current to a specific voltage that is useful for your circuit.
For large currents there are specialized circuits (some encapsulated within a small silicone chip) that can do step-down the voltage to a fixed value. One example is the well-known LM78xx/MC78xx voltage regulator chip or its negative sibling LM79xx/MC79xx, with xx={5-24} volts. A similar chip that would allow to dynamically variate the output voltage is the ubiquitous LM317 chip. Keep in mind that these step-down voltage converters require usually a larger input voltage than their maximum output voltage. So if n volts is your input voltage source then the converted output voltage will always be smaller than n.
For smaller currents (let's say max. 250mA) there is a cheaper method which we could call it "the poor man step-down voltage converter". It consists of only one Zener diode in series with a resistor that aids in limiting the current to the level supported by the diode. In a nutshell, a Zener diode used in its reversed biased condition will start at a pre-determined voltage level (the breakdown-voltage) to conduct current while keeping the voltage across it to a pre-designed constant value. Thus it can be used to step-down the voltage to a pre-designed level given by the chosen Zener diode. There are BZX55 Zener diodes in the range 2.4-47V (with a max. power dissipation of 0.5W) and BZX85 Zener diodes in the range of 3.3-62V with a max power dissipation of 1.3W. You could also find the 1N47 Zener diode series within the range of 3.3-100V with a max. power dissipation of 1W.
Now, if you think that this article was interesting don't forget to rate it. It shows me that you care and thus I will continue write about these things.
Eugen Mihailescu
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