In the previous part of this series I looked at the calculations needed to determine the current and voltage the stepper power supply should be built to provide. In this part I start to discuss the components that the power supply is built from. From the previous part we know that we need to provide 70V at 12A to power the stepper motors.
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:
|Input current amps:
|Output current amps:
|OD: 165 Ht: 75mm
|Safety Isolating transformer Conforms to EN61558-2-23 CE & RoHS
|200mm Flexible leads 16awg
|200mm Flexible leads 14awg
|Hardware 2x135mm Neoprene pads + steel dish washer with bolt M8 90mm
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):
|28.8 x 28.8 x 7.62mm
|Maximum Operating Temperature
|Minimum Operating Temperature
|Peak Average Forward Current
|Peak Forward Voltage
|Peak Non-Repetitive Forward Surge Current
|Peak Reverse Current
|Peak Reverse Repetitive Voltage
|Peak RMS Reverse Voltage
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.
References and Further Reading
- 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