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I've build this circuit on the breadboard and with 5V AC I'm getting 2.5V at the output:

schematic

simulate this circuit – Schematic created using CircuitLab

The problem is in the simulator it outputs 1.678V, which calculates to:

(1.678 x 2) / 0.636 = 5.27V

Link to the simulator.

Why the results are different?

ElectronSurf
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2 Answers2

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The problem is in the simulator it outputs 1.678V, which calculates to: (1.678 x 2) / 0.636 = 5.27V

The theoretical average value of a ½ wave rectified signal is the peak voltage (5 volts) divided by \$\pi\$. So you should get 1.59 volts at the mid point of the ripple on your falsetad simulation. And it looks like you do.

The peak of 1.678 is the wrong place to look because it is meaningless in the context of rectification and smoothing <-- it's the mid-point average of the ripple voltage where you look. If the simulation didn't auto-scale (very annoying) it would be easier for you to see.

So, if your real circuit output is 2.5 volts you might be making the same mistake and looking at the peak ripple voltage when the capacitor value you have used is significantly less than 10 uF.

Andy aka
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  • Found the problem; It needed a 10K as a load at the Vout to output proper value. – ElectronSurf Jul 09 '23 at 09:40
  • You might be swallowing a spider to catch a fly. – Andy aka Jul 09 '23 at 09:41
  • Adding/removing a capacitor don't have an effect on the output voltage, neighter increasing or decreasing the value of it. the only thing that happens to be the problem is that output is not loaded. so where's the fly. – ElectronSurf Jul 09 '23 at 09:45
  • BTW how can I not look at the peak value when the capacitor only stores the peak value? – ElectronSurf Jul 09 '23 at 09:54
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    Where you have your capacitor (across the 10 k resistor in the feedback loop, it will average the rectified signal. If you want the peak value then the capacitor need to be from the output to 0 volts @ElectronSurf (this might be the fly) – Andy aka Jul 09 '23 at 10:38
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Essentially, the cap, resistor, input voltage peak and frequency are determining the peak voltage here. Since you confirm that resistor value is correct on bench (I hope you confirmed this by using a multimeter). Then, either frequency is less than 50Hz on bench or cap used is less than 10uF on bench.

Use an oscilloscope and confirm that the input waveform is as expected and use a multimeter to check the cap value.

sai
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