230 Volts - 50 Hz to 115 Volts - 60 Hz Converter
In reality very few devices today are frequency-dependent and many have switching power supplies which can be connected to a wide range of voltages. If only voltage is the issue then a transformer will solve the problem but there are still some devices which require a certain line frequency and frequency is much more complicated to convert. A friend has a shaver which resonates mechanically with the 60 Hz line frequency and will not work well at 50 Hz. (I still have and use daily a BRAUN shaver from about 1967 which resonates at 50 Hz and will not work well at 60 Hz even though the voltage can be switched between 110 and 220 V. I have another, newer shaver which does not depend on frequency for when I travel abroad.) Even motors which will work at 50 Hz by rotating slower need to have the voltage adjusted downwards due to the lower impedance at 50 Hz so sometimes it makes sense to provide the right frequency even if not strictly necessary.
In order to keep the peak and RMS values of the output equal to a 115 Vac RMS sine wave we need to have the output be:
1/4 cycle = 0 V (both transistors blocked),
1/4 cycle = +160 V (TR2 conducts),
1/4 cycle = 0 V (both transistors blocked),
1/4 cycle = -160 V (TR1 conducts),
It can be mathematically shown that this wave shape has the same RMS and peak values as a 115 V sine ac wave. Peak value is important for devices which charge capacitors to peak value and RMS is important for other devices. For this reason this waveform is the best rectangular approximation to a sine wave and is commonly called "modified sine wave". I, personally, dislike this marketing term because it is quite inaccurate. Rectangular wave or "modified square wave" would be more accurate. Some devices might require true sine wave. For a discussion of this see here. Many voltmeters measure average volts and correct by a factor of 0.707/0.636 to indicate RMS which assumes a sine waveform and will not be valid for other waveforms. Such an instrument when measuring a rectangular "modified sine wave" like this one will under-indicate by a factor of 0.5/0.636 and so the readings would have to be multiplied by 0.636/0.5 = 1.272 to find the true measurement.
In the following photo we can see the actual rectangular output (green) as seen on a 'scope and a mathematical sine wave (red) superimposed. Actual voltage from the mains have the top very much clipped due to all the rectifier-capacitor loads.
Special attention needs to be given to understanding the bootstrap capacitor which provides polarization for switching the high side MOSFET TR2. This capacitor is charged to 15 V through the diode, from the low voltage power supply which supplies the control section when TR1 is conducting and brings the capacitor to ground level. Then, when TR1 stops conducting the capacitor floats up with the source of TR2 providing the necessary voltage for the switching of the gate through the optocoupler (or whatever circuit is used in other cases). I actually used 47 uF which is more than enough.
Let us look now at the control section. It is formed by a classic 555 oscillator oscillating at 960 Hz, a four stage CD4029 divider (divides by 16) and three NOR gates of a CD4001. The waveforms shown are self-explanatory. It can be seen how T1 and T2 are alternatively positive for 1/4 cycle. T2 is applied to the optocoupler which shifts the level to that of TR2. There is a fourth, unused gate in the 4001 and it is good practice to connect the inputs to either ground or Vcc rather than leave them floating. I also placed a bypass capacitor between the power supply pins.
Pin 1 of the 4029 counter preloads a value set at pins 4, 12, 13, 3 into Q1-Q4 when connected to Vcc and the counter counts normally when connected to ground. This means we can stop the output by pulling it high. With the resistor and capacitor as shown the inverter output will take about a second to start working after the power is applied. I used this at first but I later removed the capacitor to speed up my testing and I never replaced it. You can use it or not or use a switch depending on your needs. It could also be used to implement an overload protection with a circuit that drives it high when it detects an overcurrent at the output.
Finally we have the 15 Volt power supply for the control section. I have not measured the consumption but I imagine it might be something like 10 or 15 mA. We need about 15 V for the switching of the MOSFETs and the control circuit ICs will also work well at this voltage. For such small consumption, rather than complicated regulator circuits, I always go for a very simple design as illustrated here. I just put a zener diode in parallel with the load and make sure the transformer has enough output resistance so the zener will not be overloaded. If I am designing a commercial product I can specify a transformer with the desired output resistance and this also helps reduce the cost as the smallest possible transformer is being used. But if I am using a recycled transformer taken from my junk box, as in this case, then I just put a resistor in series with the primary and I try several values until the resistor by itself is reducing the current to just a bit over what the circuit needs and the small excess is absorbed by the zener diode. You can't get much simpler than that. Note that the zener only conducts briefly during the ac input peaks. Also note that the resistor needs to be sized individually for each transformer and each circuit by testing. Transformers with the same nominal output values vary wildly in their actual open-circuit voltage and in their output impedance so you have to test for yourself. You can start with a large resistor value and decrease it gradually until you get the output voltage you need.
I assembled the circuit on perfboard as I designed it and tinkered with it until it worked quite well. With a resistive load it worked perfectly but when I connected an inductive load there was a problem because each time a transistor would cut off the opposite one would turn on instantly for a brief moment. Rather than trying to modify the circuit I solved this problem by putting a capacitor in parallel with the load and a small resistor in series with both. It is possible that an RC snubber in parallel with each transistor would have resolved the issue.
Page last revised 2008-03-18