I am building a 5KV DC low current power supply using a full wave rectifier setup (not a bridge rectifier) and a shell type 5kV step up transformer. But cannot operate it more than 2.5kV as the transformer produce humming sound. I am not sure whether the value of C (~17microFarad)and R(~5MegaOhm) I am currently using is correct. So how does one determine the value of R and C, and also the component required for smoothing the output voltage to obtain pure DC.
To get an idea of the currents you could use the free simulation software: https://www.duncanamps.com/psud2/
Perhaps your currents exceed the limits of the transformer.
I am not an electrical engineer, but I could imagine that a center tapped transformer is not always perfectly balanced. (That the center tap is not perfectly centered.)
It would be useful to give the electrical diagram and to specify two important elements: the frequency of the sinusoidal signal (I suppose it is 50 Hz) and specify p = DV / V, where V is the direct voltage and DV the peak-to-peak variation of this voltage. Indeed, the output voltage will not be rigorously continuous. Consequently, it is necessary to indicate your degree of requirement in a rigorous way (p = 5%? 2%? 1%?). If p decreases, C and R must have larger values.
Bridge rectifier would be natural choice for High Voltage, Low current situation. Note that bridge like 'full rectifier' also gives 100Hz riple. This way, you can have both half-windings in series and double the output voltage.
Regarding filtering : usually one needs several RC stages to achieve good filtering.
The RC time constant should be several times higher than period of ripple frequency is (50x2 =100Hz --> 10mSec). Start with a capacitor after the rectifier.
One can simulate the circuit using some free available online simulators e.g. Multisim.
W.R. Kampinga Thank you.
Gheorghe Zaharia . Yes, the frequency is 50Hz. Please find attached image for reference. I don't quite understand what you mean by direct voltage and therefore the peak to peak variation of the voltage. Does this mean the required DC voltage? The degee of requirement would be a near perfect DC voltage.
Nikolay Pavlov Please find the attached circuit diagram.
Samuel Lawitlang I would just disconnect the central tap and use bridge rectifier using some HV diodes.
Yes, I mean DC voltage. If you need to obtain just a DC voltage, R is not required. Just use the largest value of C in order to obtain a quasi-DC voltage. However, in a real situation, you need to use this DC value Vout to supply an electronic device. Therefore, RL is a simplified model of this electronic device. Its value con be computed: RL = Vout/Isupply, where Isupply must be measured (known). There is a problem: the maximum value of the current via diodes when they are ON. This maximum value must be known when you select the type of diodes to use. In fact, there are 2 important parameters to be known when you select the diodes: Maximum Recurrent Peak Reverse Voltage and Maximum Average Forward Rectified Current. When C increases, the diodes are ON less time but the maximum current via the diodes increases. This must be taken into account. Moreover Vout = VDC + vAC(t). If we note Vpp the peak-to-peak variation of vAC(t), the parameter p = Vpp/VDC characterizes the “quality” of Vout. When C increases, p decreases, so Vout get closer to VDC.
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