In the previous part I discussed most of the major components in the power supply. In this part I wrap up the discussion by talking about the smoothing capacitor and other components that will be needed.
The bridge rectifier produces a DC voltage but with a lot of ripple since as the AC voltage drops to zero so does the DC voltage. To produce a smoother DC voltage a smoothing capacitor is added to the circuit which props up the output voltage as the AC voltage drops. This has the effect of raising the DC voltage to a little below the Vmax voltage of the secondary winding. There are three important parameters when choosing a smoothing capacitor: working voltage, ripple current and capacitance
The working voltage must be greater than the maximum output voltage of the rectifier with some head room for back EMF – a working voltage of at least 100V should be chosen for this circuit. The ripple current should be greater than the maximum current drawn by the circuit – in this case 20A. Choosing the capacitance requires a little calculation:
C = I * f / V
Where C is the capacitance, I is the maximum current, f is the peak to peak time and V is the maximum permissible ripple voltage. The value of I in the example is 20A. To figure out f simply double the original supply frequency and take the reciprocal, in the UK the supply frequency is 50Hz so after rectification it is 100Hz which gives 0.01 seconds peak to peak. A typical value for V would be 10% of the peak supply voltage which in our case will be 7.07V. Plugging those numbers in gives a capacitance of 0.028288F or as it’s more normally written 28,288μF
The required capacitance can be supplied by either a single capacitor or a bank of capacitors wired in parallel. Looking up a suitable capacitor for this supply (33,000μF, 100V, 57A ripple) shows that it would cost about £80. An alternative would be three of these (10,000μF, 100V, 10.5A ripple) capacitors which would be about £60. It’s worth pointing out that some real bargains can often be had on eBay although the quality of the capacitor will be significantly lower for example a quick search turned up a pack of five 10,000μF, 100V capacitors for £23 including shipping.
Note 1: Capacitors come with a range of different connectors. Snap in are a bit of a pain to work with unless you are going to make a circuit board to mount them in. Screw terminals are much simpler for a build like this as it’s trivial to crimp a number of short connection leads.
Note 2: The grey band running up the side of the capacitor indicates the negative pin (the band often has something that looks like a minus sign in it). Take care not to connect an electrolytic capacitor up back to front – they don’t like it.
It is important that the power supply is left in a safe state once it is powered off and so bleed resistors should be fitted to the capacitors to drain them of power shortly after power is removed. Under normal circumstances, with stepper motors attached to the power supply the capacitors will discharge quickly through the windings of the steppers but the situation where there is no load attached to the power supply should also be considered for safety reasons.
With this set up a resistor is wired across the terminals of each capacitor. As soon as power is switched off the capacitor bleeds it’s power out through the resistor. To calculate the size of bleed resistor required it is first necessary to calculate the energy stored in each capacitor. The following equation is used:
W = (CV^2) / 2 = (0.01 * 70 * 70) / 2 = 24.5J
Where W is the energy held in the capacitor in Joules, C is the capacitance in Farads and V is the charge voltage of the capacitor. If a safe discharge time of 30 seconds is assumed then knowing that a Watt is a Joule per Second it is easy to calculate that a bleed resistor must dissipate 24.5 / 30 = 0.82W. The choice of bleed time is somewhat arbitrary, a very long bleed time will waste little power through the resistors but at the expense of safety. I plan on enclosing the power supply in a case which will be screwed shut, my selection of 30 seconds for the bleed time is based on the assumption that I won’t be able to get into the case in less than 30 seconds.
Using the following equation it is then possible to calculate the resistor required:
R = V^2 / P = (70 * 70) / 0.82 = 5975Ω
A 1W, 6200Ω resistor (example) would be a good choice for this project. The resistance is a little higher than the calculation calls for so the discharge will take a little more than 30 seconds.
The always connected set up is the safest approach but it has a slight downside, the resistors are bleeding power all the time the power supply is switched on. This means about 3.5W of heat is constantly being dumped into the power supply housing. While not a lot of power it does cause unnecessary heating that can be avoided by controlling the bleed resistor with a normally closed (NC) relay. The resistor required with this set up is calculated as follows:
W = 4 * ((CV^2) / 2) = 4 * ((0.01 * 70 * 70) / 2) = 98J 90J / 30s = 3.3W R = (70 * 70) / 3.3 = 1484Ω
A 5W, 1500Ω resistor (example) is an almost perfect match. Buying small quantities of items such as this can be difficult, eBay is often a good source but it may be necessary to use more than one resistor in parallel if an exact match can’t be found.
It is by no means necessary to fit indicator lamps having a light indicate the presence of a potentially deadly voltage is a good idea. On the primary side I’ll be fitting a neon indicator (example) and on the secondary an LED with suitable resistor.