In this content we are going to see about the circuit that aids for high voltage generation needed for operation of microwave oven.The magnetron needs a high negative voltage to be applied to its cathode for correct operation. Before we commence with a description of the high voltage circuit and its operation, it is essential that we point out the Dangers that exist when working with such a circuit
                                                       The high voltage circuit when operating produces a High Voltage from a low impedance source, the High Voltage or Power Transformer. It is capable of maintaining a high voltage of about 2000V A.C., with a high current of between five and six amps. It should therefore be noted that touching any part of this circuit while it is operating would result in Electrocution.  Working on this circuit live, even with well-insulated equipment is not recommended. Also, after the circuit is switched off, a large electrical charge remains in the high voltage capacitor. There is therefore a danger of Electric Shock even when the oven is no longer connected to the mains supply.

A typical high voltage circuit used to generate the magnetron supply is shown below.
                                      The circuit consists of a leakage transformer, usually referred to as the High Voltage or Power transformer, which has three windings. The primary winding to which the mains supply is applied. The high voltage secondary which steps up the mains input to 2000V A.C. and a filament winding that produces a low voltage with high current supply for the filament of the magnetron. The other components in the high voltage circuit are the high voltage capacitor and high voltage diode. A compensative(bleed) resistor is added to the circuit, usually incorporated in the can of the high voltage capacitor, which discharges the capacitor when power is removed.

                                         It is essential that the sharp terminal connections on the magnetron are secure. If these connections become loose or oxidised the resistance of the joint will become high. Then because of the high current flow localised heating will occur. This heating will cause further oxidisation and will also loosen the joints further, due to expansion, resulting in higher resistance in the joints and further heating of the connections and possible arcing. The result can be damaged terminals or insulation material on the magnetron requiring its premature replacement and damage to the connecting leads. When re-fitting the sharp connectors to the magnetron after service, always ensure they are clean and tightly secured.

                                           As previously stated the power transformer filament winding produces a low voltage A.C. It should however be noted that the HIGH VOLTAGE circuit is connected to one filament connection. This means that all of the wiring and connections in this part of the circuit will have a high negative potential with respect to ground. The voltage across the filament 2.5 to 3V A.C., the negative potential from all parts of the filament circuit to ground is Connected to the high voltage secondary winding is the remaining circuitry, the high voltage capacitor, high voltage diode, bleed resistor and of course the magnetron. The capacitor and diode form what is often referred to as a 'Half Wave Doubler', it carries out D.C. restoration of the A.C. input from the transformer secondary. 

                                     The type of transformer used in the high voltage circuit of a Sharp microwave oven is known as a leakage transformer. It is designed to have more magnetic flux leakage than the standard type of transformer, that is to say that not all of the magnetic flux produced by the primary winding finds a magnetic path to affect the secondary windings. There is a magnetic shunt built into the assembly that allows the magnetic flow to take an alternative path. 
                                      As far as the operation is concerned, this has the effect of allowing the voltage to vary in order to maintain a fairly constant current through the magnetron anode circuit. The reactance of the transformer effectively changes with the load.
                                 It is important where a magnetron is being used that its anode current is maintained at a constant value during operation or the output power will vary dramatically. The leakage transformer operation will achieve this. 
                               If the high voltage A.C. across the secondary winding were to be measured it would be seen to vary considerably as the load in the oven moved around the cavity on the turntable. This is in response to the changing load put on the magnetron. It would also be noticed however, that because of the action of the transformer and half wave doubler circuit, that the high negative D.C. voltage across the magnetron would remain stable, as does the anode current.

The circuit below shows a simplified schematic diagram of the 'Half Wave Doubler’ circuit. To be more precise its action on the input waveform is more consistent with a D.C. restoration circuit. The doubler label comes from voltage measurements made on this circuit with no magnetron load. A voltmeter connected in position 'A' in the diagram below would show 2000V ac., and -4000V D.C. in position B. When analysing these results, it could be said that a doubling of the input voltage had occurred. However if the same measurements were made using an oscilloscope, it would be seen that the waveform at position 'B' is the same as at position 'A'. The circuit has shifted the waveform with respect to the zero line (ground) this is also shown in the diagram below.

                                          The reason the meter shows a doubling of voltage is that the same reference point is used, that is ground. One lead of the meter is connected to ground when taking both voltage measurements. In the case of waveform 'A', the meter is set on the A.C. voltage range. This will show the RMS value of one half-cycle, with respect to ground, this would be approximately 2000V. 
                                            In the case of waveform 'B' the meter would be set to measure D.C. voltage, because the waveform is unidirectional, e.g. negative only. The meter would read -4000V, which is the RMS of the peak to peak voltage on this waveform, which is also equal to the peak to peak value of waveform 'A'.  
                                             When the magnetron is connected to the circuit, it will affect the waveform at position 'B'. The change is due to the current being drawn by the magnetron. The magnetron will begin to conduct at about -2,000 volts, it then acts as a shunt regulator and clamps the waveform at that point. Now when we check with a meter the voltage would be in the order of -2000V D.C.
                                            We can see from the diagram on the previous page that the magnetron is pulsed on and off at mains frequency therefore the RF output will be present for only 50% of the time. The RF energy is pulsating at 50Hz and is only present during the negative half-cycle of the mains supply. This means we only get about 50% efficiency from the magnetron, the result is that the output power is only half of the power consumed by the oven. However, this pulsing helps to prevent damage occurring to the magnetron caused by overheating.


                                            The circuit has two different time constants. When the input signal produces a forward bias potential on the diode, the total resistance is the load resistance in parallel with the diode forward resistance, therefore the total resistance is low. When the diode is reverse biased, the total resistance is the load resistance, which is high.
                           Time = Resistance x Capacitance
                                            Using the above equation, it can be seen that when the diode is conducting the time constant is very short and when the diode is not conducting then the time constant becomes much longer. The capacitor charges quickly through the diode and discharges slowly through the load resistance. 
                                            To understand the operation of the circuit assume that the capacitor is initially uncharged. During the first quarter cycle the diode is forward biased. Current will now flow via the diode and the capacitor is charged. The polarity of the charge is shown in the diagram below. The level of the charge will be a little under the peak value of the input.

                                             As soon as the voltage on the input side starts to fall, the diode will become reverse biased, with the charge on the capacitor holding the diode off. The input circuit will now return the positive plate of the capacitor to ground, therefore the other plate of the capacitor will be taken negative. In fact the negative plate of the capacitor will always remain at a voltage equal to input less the peak value. When the input approaches its next positive peak any lost charge will be replaced. 
                                             With the magnetron connected and drawing current the circuit will maintain a voltage nearing the peak to peak value of the input waveform. The output does however change wave shape due to the action of the leakage transformer and the current consumption of the magnetron. 
                                              As already stated the charge on the capacitor is effectively added to the input waveform, to give an output that is shifted up or down depending on the polarity of charge. In the case of a microwave oven the shift is negative, as a negative potential is needed for the cathode. 
                                              The charge gives us the difference between input and output. If the capacitor starts to discharge, the diode action ensures that on the next positive peak the lost charge is replaced. In other words if the capacitor charge is reduced by an increased load, the diode current increases. This fact means that diode current will be equal to the current being drawn by the load on the doubler circuit. In the case of the microwave oven the load is the magnetron, and therefore diode current will be the same as the magnetron anode current. 
                                                There is one further point to be considered while dealing with this circuit, that is the value of the capacitor. The larger the value, the more current available for the magnetron to draw on. Therefore, the value of the capacitor will affect the power output of the oven. The larger value the greater the output.

                                    When replacing capacitors in the high voltage circuit of a microwave oven it is very important that the same value is used as the original one fitted during manufacture. It should be noted that different models with similar output powers, may have different values of capacitors, therefore different transformers will be used. Also models with different output powers may use the same magnetron, with the output being controlled by the capacitor value.
                                     From the safety aspect it should also be noted that a bleed resistor is built into the capacitor, which will discharge the capacitor when the power is removed. In some models the bleed resistor is built into the diode. To prevent a potentially hazardous situation developing, when replacing the capacitor or high voltage diode, always use the correct type by part number. 
                                     When checking these components, if it is found that the bleed resistor has gone high in value or open circuit, the component should be regarded as faulty and replaced.

                                         The Asymmetric Rectifier or Short Protector, as it is sometimes called, is included in the circuit to prevent damage to the high voltage transformer and high voltage capacitor, in case of a short circuit occurring on the magnetron or high voltage diode. The short protector will cause the line fuse to blow in the above fault conditions. 

                                                                As can be seen in the diagram at the top that the device will not conduct in any direction, therefore has no effect on the circuit under normal operating conditions. The peak inverse breakdown voltage of D1 is 6kV and D2 is 1.5kV. Under fault conditions when a short circuit occurs the peak inverse voltage of D2 will be exceeded, this will cause the diode to go short circuit. This in turn will cause a very high forward current through D1, causing it to go short circuit also. This will cause a dead short to appear across the high voltage transformer secondary, which will blow the line fuse, removing power to the high voltage transformer.

                                                                   We have seen that very high voltages are generated in the high voltage circuit, and therefore during operation the high voltage diode is subjected to high peak inverse voltages. This means that a specially constructed diode is needed to withstand such voltages. However it is not possible to manufacture a single diode with the required characteristics, this is achieved by connecting twenty diodes in series. Therefore the peak inverse breakdown voltages of each of the diodes are added, which provides the required overall peak inverse breakdown voltage. This needs to be borne in mind when carrying out resistance checks on high voltage diodes.
Next Post »