Hi my friends
I have problem with understanding the output of the following circuit

can you please explain the output for me

Regards

 

Please Guys , Just a little bit explanation will help a lot

 

You have a 3 V battery. The diode has a potential drop
of around 0.7 V. 3 - 0.7 = 2.3 Hence, the diode turns
on at 2.3 V (in your case -2.3 because of the direction in
which you hooked up the diode and the battery), which is
just what you see in your plot.

 

In power supplies, diodes are used to rectify an AC output. A diode lets current flow one direction but not the other. As an example, look at the picture I have on my oscilloscope. In the first picture, you'll notice just a standard sine wave. Next, I have a diode with the cathode to the negative ternimal. Lastly, I have the cathode to the positive ternimal which chopped off the positive AC. If we were to put a capacitor it would charge from the peak of the cycles and discharge on the 0V. This leaves a steady DC output, although there will still be some ripple.

If you switched the diode to the opposite polarity the waveform would flip upside down.

 

Thank you guys for answering me
I understood the 2.3 volt part
but how the voltage increase to 10 volt , That i dont understand

 

Thank you guys for answering me
I understood the 2.3 volt part
but how the voltage increase to 10 volt , That i dont understand

The -10 volts you see is due to the fact that during that period, the diode is reverse biased; therefore no current flow (except into the measuring device) and no voltage drop through the current limiting resistor. During the positive half cycle of your source, the diode is forward biased and is allowing current flow into the battery. The remainder of the positive 10 volt half cycle is dropped across the current limit resistor. Hope this helps.

 

The -10 volts you see is due to the fact that during that period, the diode is reverse biased; therefore no current flow (except into the measuring device) and no voltage drop through the current limiting resistor. During the positive half cycle of your source, the diode is forward biased and is allowing current flow into the battery. The remainder of the positive 10 volt half cycle is dropped across the current limit resistor. Hope this helps.

very helpful thank you