Diode bridge, how to choose the right capacitor value ???

Question

Electronics

I know that there should be a smoothing capacitor in parallel with the diode bridge
- how to choose the right value of the smoothing capacitor in microfarads "mF" in parallel with the diode bridge ???
- intuitively suspecting that the capacitor should be of large capacity ..... nothing was at hand as three 250mF (400V) capacitors connected in parallel ..... at the output after the diode bridge from 12V it turned out 16V (which is very undesirable in my case , we need stable 12 and 24VDC)
QUESTION; How does the capacitance of the capacitor affect the output voltage after the bridge? Is there a formula for calculating?
and another question on the topic; - why is it that when 6 hydraulic valves are connected in parallel to one transformer, the power in the network rises from 12V up to 17VDC, and if you connect 24V to the transformer winding, the power rises up to 34VDC ??? where to look for rake? can separate the valve coils with additional diodes for each winding ???
Thank you !

Answer 1

In fact, you should read some book on electrical engineering.
In short, the AC voltage on average in the AC network is zero, because it constantly changes sign, and changes from one amplitude value to another. Therefore, in an alternating current network, it is customary to measure the effective value of the voltage, which is , because such a value is consistent in terms of energy characteristics with a constant voltage of the same magnitude.
However, after rectifying the current, the output is a pulsating voltage with an amplitude . If you charge the capacitor with such a voltage without connecting the load, then the capacitor will be charged to this very amplitude value. In your case , what you are seeing.
The calculation of the capacitance of the filter capacitor depends on the magnitude of the load and the permissible level of ripple. Your rectified current can be roughly represented as the sum of DC and AC. The variable component can pass through a capacitor, which represents some resistance for it. In order to effectively suppress the variable component, it is necessary that the resistance of the capacitor for alternating current be much less than the load resistance that consumers create.
Here you can calculate the capacitance, you need to keep in mind that after the diode bridge, the frequency of the variable current component will be 100Hz
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Without a detailed description of the circuit, it will be difficult to say something further. If you are not familiar with circuitry, it is better to buy a ready-made power supply.