I always thought that diodes in forward mode had always a 0.7 V drop, I mean the transistor has it from base to emitter, but I was testing in multism, applying 1 V to the entrance of the diode and getting the value from the exit and it was acting like a short circuit. I thought that only a few diodes had no (or low) voltage drop. Is it multisim that's wrongly configured or considering the ideal case?

Because when I connect a diode to ground and conect an ac voltage source through a resistor to it, the wave gets clipped. I mean the wave is normal when it doesn't pass 0.7 V, but when it does, it will be forced to be 0.7, since there's a 0.7 V drop to ground, which has 0 V.

 

All real-world diodes drop some voltage when conducting. The amount depends on the diode material and structure and the current through the diode. If Multisim tells you there's no voltage drop, it's lying.

 

I always thought that diodes in forward mode had always a 0.7 V drop

Where did you get that?
For 1N4148:
Vf @ 1uA = 270mV, Vf @800mA = 1.43V. The Y-axis units in Fig 5 are mislabeled; should be volts (datasheet error).



Post a schematic and don't make the mistake of believing that simulators are always correct.

 

Is there a question there?
Post the circuit you are simulating.

 

Is there a question there?
Post the circuit you are simulating.

This basic circuit, I though that after the diode there would be 1.3 V.

 

All real-world diodes drop some voltage. The amount depends on the diode material and structure and the current through the diode. If Multisim tells you there's no voltage drop, it's lying.

So, imagine the basic circuit I posted in the last picture, after the diode, using a silicon diode, there would be 1.3 V, right?

 

So, imagine the basic circuit I posted in the last picture, after the diode, using a silicon diode, there would be 1.3 V, right?

Wrong.

That's not a complete circuit.
The output of the diode needs to be connected to some load to get a voltage drop.

For a true open circuit output, the diode current would be zero and thus so would the voltage drop.
No current, no voltage drop.
Thus the diode output voltage would equal the input voltage, as I believe Multisim showed.

 

Wrong.

That's not a complete circuit.
The output of the diode needs to be connected to some load to get a voltage drop.

For a true open circuit output, the diode current would be zero and thus so would the voltage drop.
No current, no voltage drop.
Thus the diode output voltage would equal the input voltage, as I believe Multisim showed.

How stupid of me, that's similiar to the functioning of transistors (when it's off, no current, therefore voltage in collector is equal to Vcc), I was studying that yesterday ahah. I connected the diode to a resistor and the resistor to ground and it worked. Thank you for your help!