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.

- Unregulated Power Supply Design – Andy Collinson
- Power Supply Tips –
- Capacitors – Energy Stored – The Engineering ToolBox

The toroidal transformer is the component that actually deals with transforming the mains 230V down to a useful voltage for the CNC machine. Toroidal transformers are easy to source from eBay or from dedicated electrical component retailers (for example here, here and here). Taking the optimum voltage of your machine and working backwards gives the required secondary winding voltage for the transformer. The calculations are quite simple and for the example machine are shown below.

70V / 1.4 = 50V

The initial 70V is the desired supply voltage for the steppers which is divided by √2 (or 1.414) because the machine is fed DC but the transformer produces AC. The conversion from AC to DC (called rectification) gives us a DC voltage which is a little lower than the maximum voltage of the secondary winding. This gives a target secondary winding voltage of 50V. While it is certainly possible to buy a toroidal transformer with a single secondary winding dual secondary windings are more common and this then gives the choice of series or parallel wiring. If the two secondary windings are wired in series the voltages are additive so divide the result of the calculation by two to give a desired secondary voltage of 25V. Alternatively a dual 50V could be used and wired in parallel.

Justification for using the value 1.414 Transformer Secondary: 50Vrms Simple rectification gives: Vdc = 2Vmax/π = (2Vrms√2)/π ≈ 0.9Vrms or 0.637Vmax (note that 0.637 * √2 ≈ 0.9) After just rectification Vdc would be: 50V * 0.9 = 45V Smoothing, however, lifts Vdc towards Vmax (which is Vrms√2) by discharging a capacitor during the low points of the cycle. Vdc can never quite quite reach Vmax which is captured by using a value of 1.414 which is a little less than √2.

We now know the current required and the voltage we want to achieve so it is possible to calculate the amount of power it is necessary to supply:

70V * 12A = 840W

This is the peak power required by just the steppers but the drives will require some power as well so a reasonable choice for the transformer would be 1000VA. Since this power supply will only be providing power to the drives and steppers it is probably slightly easier to wire the two secondary windings in series so I’ll pick a dual 24V part. The specification below is for a transformer from Airlink Transformers (link) and would be a good choice for this project:

Part number: | CM1000225 | ||

VA: | 1000 | ||

Input vac: | 230 | Input current amps: | 4.30 |

Output vac: | 24+24 | Output current amps: | 20.80 |

Regulation: | 4% | ||

Dimensions: | OD: 165 Ht: 75mm | Weight: | 7kg |

Type: | Safety Isolating transformer Conforms to EN61558-2-23 CE & RoHS | ||

Input Termination: | 200mm Flexible leads 16awg | ||

Output Termination: | 200mm Flexible leads 14awg | ||

Fixings: | Hardware 2x135mm Neoprene pads + steel dish washer with bolt M8 90mm | ||

Price: | £60.00 |

On paper this transformer will give provide 66V DC @ 20A after rectification and smoothing. Airlink also supply a dual 25V supply with would provide 69V DC at 230V AC which looks like a better match, the problem however is that if the supply voltage is at it’s 253V limit the the system would supply 76V DC which gives little room for back EMF.

Toroidal transformers are susceptible to large (and I really mean huge) inrush currents when first powered up which can easily be enough to trip a breaker or blow a fuse.The magnitude of the inrush is dependent on where in the AC cycle the transformer is switched on so the transformer can appear to have a fault only sometimes when it’s powered on. Since the duration of inrush current is short, a few AC cycles at most, a slow blow (type T) fuse is normally a good choice.

For the primary side of the transformer a fuse 50 to 100% greater than the expected current is a fair choice. In this build, with a 1000VA transformer, that means a fuse between (1000 / 230) * 1.5 = 6.5A and (1000 / 230) * 2 = 8.7A – the nearest standard fuse is a 7A type T fuse (example). The secondary side doesn’t have to be fused but if doing belt and braces, a 25A type T (example) would be a fair choice. The secondary fuse should be between the transformer and the rectifier, alternatively you could fuse each driver separately.

If you want to use an MCB rather than a fuse then you’ll almost certainly need a 6A Type D for this power supply. The inrush current is going to be around 4.3A * 8 = 35A for approximately 0.1 seconds (see here) which means a 6A Type C MCB will probably trip at least some of the time. I initially tried a 6A Type C (as they are easy to get hold of) but I found it tripped about 25% of the time on power on. A Type C will instantly trip at between 5 and 10 times it’s rated current (e.g 30A to 60A) where as a Type D will instantly trip at between 10 and 20 times it’s rated current. I’m guessing I must have got a fairly sensitive breaker, the probably was worse with the smoothing capacitors connected as they also provided their own draw.

A bridge rectifier is a device for converting AC power (as produced by the transformer) into DC power (as used by the steppers). Essentially it is just a matter of picking a bridge rectifier that can handle the expected current and voltage for the power supply being built. For the power supplies most DIY builders will be putting together this is a very simple task as even a massively over specified bridge rectifier will only cost a few pounds.

The key specifications to check when picking a bridge rectifier are peak forward current and peak reverse voltage. The forward current must simply be greater than the current provided by the power supply but peak reverse voltage is a little more complex. For the reverse voltage there will typically be two values peak reverse (repetitive) voltage and peak RMS reverse voltage and both should be higher than Vmax (50V * √2 = 70.7V). You should also allow for additional voltage head room on the reverse voltage to ensure the rectifier isn’t damaged by back EMF from the steppers. A good choice for this project is shown below (part number 629-6320 / GBPC3504-E4/51 from RS):

Bridge Type | Single Phase |

Dimensions | 28.8 x 28.8 x 7.62mm |

Maximum Operating Temperature | +150 °C |

Minimum Operating Temperature | -55 °C |

Mounting Type | Screw |

Package Type | GBPC |

Peak Average Forward Current | 35A |

Peak Forward Voltage | 1.1V |

Peak Non-Repetitive Forward Surge Current | 400A |

Peak Reverse Current | 0.005mA |

Peak Reverse Repetitive Voltage | 400V |

Peak RMS Reverse Voltage | 280V |

Pin Count | 4 |

As can be seen this bridge rectifier (costing just £3.60) is more than capable of handling the current and voltage in the example system. Before rushing out to buy one though there is one other consideration. The table gives a value for peak forward voltage, this is the voltage drop that occurs across each diode in the bridge when current is flowing. Unfortunately that voltage isn’t just magically lost into the æther it gets converted into heat which will need to be dissipated. The amount of heat that will be produced at the transformers peak current is:

1.1V * 20A * 2 = 44W

Where 1.1V is the voltage drop across the diode, 20A is current limit of the power supply and 2 is the number of diodes are in use at any one time in the bridge. Forty four watts doesn’t sound like much heat to dissipate but trust me that’s quite enough to have to deal with when it’s buried inside a case next to other components that are generating heat – a heat sink may be required.

To calculate the performance of the heat sink that will be required you first need to find out the thermal resistance of the rectifier, this will be available on the data sheet for the component and for the part selected above it is 1.4K/W. Now decide on a maximum working temperature for the rectifier and determine the ambient temperature. The rectifier will, according to the data sheet, work up to 150oC but it’s probably wise to run it some what below this to extend it’s life so set the working temperature at 130oC. Room temperature is typically around 20oC but it is safe to assume the environment the rectifier finds itself in will be warmer so assume 30oC. Plugging these figures into the equation for heat resistance gives us the total maximum thermal resistivity the system (rectifier plus heat sink) should have:

(130 - 30) / 44 = 2.27 K/W

The rectifier has a thermal resistance of 1.4K/W so the heat sink must have a maximum resistance of 2.27 – 1.4 = 0.87 K/W. That sort of thermal resistance is possible to achieve with a standard computer CPU heat sink and fan which can be bought for a few pounds or salvaged from an old computer for free.

In reality though the rectifier won’t reach anywhere near this sort of heat output since the maximum draw on the power supply will only be 12A and that will only occur for brief periods. Plugging the figures in a second time gives us a heat dissipation of 26.4W which means a total thermal resistivity of 3.78K/W and a heat sink with a thermal resistivity less than 2.38K/W which is very easy to achieve with passive cooling. In fact the power draw will, on average, be so low just bolting the bridge to the aluminium plate that makes up the power supply mounting will provide enough cooling.

In the next part of this article series I will be discussing the smoothing capacitor(s) and other components required for the power supply.

- Transformer Basics – RAF Tabtronics
- What is forward and reverse voltage when working with diodes? – Electronics Engineering, Stack Exchange
- What is the RMS reverse voltage for diodes? – Electronics Engineering, Stack Exchange
- Thermal Resistance – Wikipedia
- Transformer Inrush – Open Electrical