# Input and output voltage of 7805

In the diagram below the input voltage to the ic 7805 is the output voltage of the capacitors (1000 micro Farad and 0.22 micro Farad).

The peak voltage arcoss the input of ic 7805 is the same as the peak rectified voltage or the voltage across the capacitors. This peak voltage at the ic input becomes 5v at the output of the ic 7805.

But I don't know how to calculate the peak to peak Ripple voltage of the input voltage to the ic 7805 and peak to peak Ripple voltage at the output of the ic 7805 voltage regulator.

I don't know is it even important to know this?

• Alex - Hi, Please note the site rule which requires that when a post includes content (e.g. text, image, photo etc.) copied or adapted from elsewhere, that content must be correctly referenced. As a minimum, for online material the source webpage or PDF etc. should be linked (see that rule regarding references for books / articles etc.). In order to help you, I found what I believe to be the source webpage link & added it. For the future, please remember it's your responsibility to do that :) Thanks. Commented Sep 14, 2023 at 21:40
• 12VAC out of the transformer will give nearly 17VDC when rectified and smoothed. That's a big drop across the 7805 and it will waste a lot of power and generate a lot of heat (2.4W of heat for every 1W your circuit consumes). Commented Sep 14, 2023 at 21:49
• For input ripple, see this previous question: electronics.stackexchange.com/questions/636422/… For output ripple, you need to dive into the spec of the exact 7805 regulator. Commented Sep 14, 2023 at 21:49

First, realize that this circuit is how people created 5 volts 50 years ago. Now, people use switching regulators. If you needed a lot of current, the transformer was huge. I worked on an early computer where the transformer weighed 5 pounds.

For this circuit, for a first order approximation, where I is the load current, f is the line frequency, C is the input capacitance.

$$\Vripple = \frac{I}{2fC}\$$

You generally want a ripple of a volt or two.

For the 7805 regulator to maintain regulation, the input must be greater than the output plus the dropout voltage. For a 7805, the dropout is about 2 volts. So your peak from the rectifier minus the ripple must be always be more than 7 volts. Don't cut it too close, you want the minimum to be 8 or 9 volts.

The peak at the input will be:

$$\Vpeak = 1.414V_{transformer} - V_{DiodeDrop}\$$

As Finbarr pointed out, if your voltage is too high, you will waste a lot of power in heat. A 12 transformer output is too much. You want about 8 volts RMS.

A long time ago, 6.3 volt transformer outputs were plentiful because they were used for vacuum tube heaters. These generally would work, but maybe not under worst case conditions, not recommended IMO.

A hobbyist generally doesn't worry about the output regulation. These parts were designed to power circuits that require 5 volts +/- 0.25 volts. A typical 7805 can easily do this (provided that the input exceeds the dropout). If you really need to know, the 7805 line and load regulation specifications will tell you how much the output will vary.

• Vripple = I / 2fc is the ripple to ripple voltage at the input of the ic 7805? What if I replace the I with Vpeak/Xc? And the equation now becomes Vripple = Vpeak/ 2FCXc. Is this equation valid?
– Alex
Commented Sep 15, 2023 at 3:23
• @Alex Yes the Vripple that I defined is at the input to the 7805. No, Xc is used for AC analysis. The maximum current drawn by your circuit is what is important. See the link in my comment. Commented Sep 15, 2023 at 6:45
• is Vripple means peak to peak ripple voltage at the input of the ic 7805 voltage regulator?
– Alex
Commented Sep 15, 2023 at 7:16
• @Alex Ripple is defined as the Vmax minus the Vmin. You could say that this is peak to peak, but ripple is a more precise term. Commented Sep 15, 2023 at 13:17
• @Alex I is actually the current into the 7805, but since the quiescent of the 7805 is usually negligible, I is basically the load current at the far right of the schematic. Commented Sep 15, 2023 at 13:31