Voltage rectification. What am I doing wrong?

Question

Electronics

Good day. In the course of learning circuitry, I came to the issue of voltage rectification. In theory, everything seems to be clear. With a diode bridge, we turn negative voltage values ​​\u200b\u200binto positive ones and we straighten the whole thing with a capacitor. In Multisim, I sketched a circuit, and was a little confused that the voltage curve more or less starts to approach a straight line with a capacitor larger than 300mF. I started digging deeper and came across this article . The capacitor is used three orders of magnitude smaller. Please help me figure out what and where I do not understand correctly. Here is the same circuit from Multisim with a capacitor of "only" 5mF:

UPD: I'll clarify that in the article I mentioned, a 300uF capacitor is used, and in the simulation in Multisim, my curve is smoothed with a capacitance of more than 300000uF. In the diagram from the screenshot, there is a 5000uF capacitor, which is 6 times more than in the article, and smoothing is not even close.

Answer 1

What exactly is annoying? That's right: the larger the capacitance of the capacitor, the better it smoothes out the pulsating voltage, turning it into "almost constant".
In the article in. the resistance of the oscilloscope is large enough, and the capacitance of 300uF is enough for this "almost" to be neglected. If it were smaller, a larger capacity could be needed to get the same "straight line".
UPD: https://www.youtube.com/watch?v=ARmNQrwgG0A

Answer 2

You just don't know yet that it's not the absolute values ​​that matter, but the ratios. For example, if you specify ideal elements (an ideal diode with an infinitely large reverse resistance, an ideal oscilloscope with an infinitely large input resistance), then it turns out that ideal smoothing (i.e., the complete absence of ripples) is obtained for any capacitance of the capacitor - simply because it will charge to peak values ​​and will not discharge at all.
If the elements are not ideal, then the capacitor is not only charged, but also discharged. Then a 10 microfarad capacitor loaded with a 10 kilo-ohm resistor will have exactly the same ripple as a 100 micro-farad capacitor loaded with a 1 kilo-ohm resistor, and the same as a 1000 microfarad capacitor with a 0.1 kilo-ohm resistor. This is called the "time constant" - the product of the capacitance of the capacitor and the resistance of the resistor, the same for all of these listed options. If we want to reduce the ripple, we must increase this "constant", no matter how - by increasing the capacitance, resistance, or both.

Answer 3

Quote from your article

We take our third conder. Its capacitance is 330 microfarads. I can’t even measure it with an LC meter, since I have a limit of 200 microfarads on it.

Answer 4

nehrung and pi314 said everything correctly. But from my practice of explaining electronics to people on the fingers, I know that very many people are confused at the very beginning by the words about the input resistance of the measuring device. Therefore, I will add something for clarity.
First, when you are modeling or building a circuit with a power supply, it is always useful to include in this circuit the consumer of this power, that is, the load. Resistor, LED, something else. This is safer (in real life), otherwise the source can, in fact, be short-circuited.
Secondly, if you still do not have any load in the circuit, the measuring device acts as the load. Accordingly, in a situation with a smoothing capacitor, the more current the measuring device can pass through itself (the lower its input resistance), the faster the charge accumulated in the capacitor will be used up, the more capacitance is needed for smoothing.