Hello all, i was simulating this push-pull stage, where the cross-over gets cancelled.
But i didnt understand how it works, like who's forward biasing here the diodes? the power supply of 9V and -9V? ( like the circuit i show near to it )
or the signal vin is responsible for the forward biasing? and if its the power supply then why the vin doesnt matter? like ur not applying 2 signals?

 

hi 089,
The +/-9V via the R2,R3 current creates a voltage drop across each diode of approx 0.65V
E
Post your LTS asc file.

 

hi 089,
The +/-9V via the R2,R3 current creates a voltage drop across each diode of approx 0.56V
E
Post your LTS asc file.

yeah damn, i always forget to post also LTSPice simulation file, here u go:

 

hi 089,
The +/-9V via the R2,R3 current creates a voltage drop across each diode of approx 0.56V
E
Post your LTS asc file.

Yeah, it creates a voltage drop where then we get 0.6V on each diode ( so in total 1.2V and on each resistor there is half voltage of the remaining )

 

hi 089,
The +/-9V via the R2,R3 current creates a voltage drop across each diode of approx 0.65V
E
Post your LTS asc file.

But something else isnt clear to me, like we are applying an AC signal ok? and this is also forward biasing the diodes in same time no? or the signal given by the D.C source has always "priority" over the signal u give in input? i can explain again in case i wasnt clear.

 

hi
Post your LTS asc file and will mark out the circuit voltages for you.
E

 

this is also forward biasing the diodes in same time no?

hi,
The diode bias voltage to each transistor is just enough to start the transistor conducting in it linear operation mode.
Without that bias, you get what is called 'cross over distortion'.
E

 

hi
Post your LTS asc file and will mark out the circuit voltages for you.
E

this is the voltage drop no? this is what happens also on the push-pull stage? ( like the part of diodes and resistors ) or since we are also applying another signal its different?

 

hi,
The diode bias voltage to each transistor is just enough to start the transistor conducting in it linear operation mode.
Without that bias, you get what is called 'cross over distortion'.
E

yeah this i get it, but my question is different, like i get it that without bias i would get cross over distortion and its where the vin is beetwen -0.7V and 0.7V so none of the transistor are conducting, but my question is kinda different, i can explain again in detail maybe if u didnt get me

 

hi,
I would like to help you, so please post your LTS asc file so that I can answer your question.

E

 

hi,
I would like to help you, so please post your LTS asc file so that I can answer your question.

E

Here u go, thanks

 

hi,
This shows when the diodes are shorted out, ie: no VBE bias voltage.
Note the cross-over distortion around 0V on Vout.
E

 

hi,
This shows when the diodes are shorted out, ie: no VBE bias voltage.
Note the cross-over distortion around 0V on Vout.
E
View attachment 347745View attachment 347746

who makes the diode to be shorted? and in that case what happens? the bias voltage is given from who? these diodes shouldnt be able to have always 0.7v on them, so even while the sine wave is at 0V the transistor are still conducting?

 

The signal, as you have it connected, will force its voltage at the junction of the two diodes. The bases will then be at Vin +0.65V and Vin - 0.65V. In a real circuit , there will be a series resistance in the signal path, so the voltages at the bases will not swing as far as the signal.

 

The signal, as you have it connected, will force its voltage at the junction of the two diodes. The bases will then be at Vin +0.65V and Vin - 0.65V. In a real circuit , there will be a series resistance in the signal path, so the voltages at the bases will not swing as far as the signal.

sorry but one thing? who's responsible here for the polarization of the diodes? the D.C power supply or the vin? ( sinusoidal wave ) this is what i dont get

 

lets analyze the 2 cycles of the sine wave separately ok? during first cycle, the positive half, this happens: the positive cycle is forward biasing the diode, or the d.c forces that diode to be always forward biased?

 

The signal, as you have it connected, will force its voltage at the junction of the two diodes. The bases will then be at Vin +0.65V and Vin - 0.65V. In a real circuit , there will be a series resistance in the signal path, so the voltages at the bases will not swing as far as the signal.

what i understood is:

1) during positive half cycle D2 diode is polarized directly, and on the cathode i will have VIN-0.7V
2) during negative half cycle D1 diode is polarized directly, and on the anode i will have VIN+0.7 ( since the rule says that the diode, in order to conduct, the anode needs to be at higher potentional of the cathode by 0.7 )

am i right?

but with that said, i dont have also to look at the d.c voltage? that one dont do anything to the diodes? tell me if im not clear, i can explain again and again.

 

No, the two diodes will both be always be forward biased until the input signal gets close to ±9V. The biasing current comes from the two power sources.

That does not mean the Vbe of the transistors isn’t changing, because the emitters are following the input signal.

Change your input source to a ramp from -9V to +9V and plot the voltage across each diode.

 

No, the two diodes will both be always be forward biased until the input signal gets close to ±9V. The biasing current comes from the two power sources.

That does not mean the Vbe of the transistors isn’t changing, because the emitters are following the input signal.

Change your input source to a ramp from -9V to +9V and plot the voltage across each diode.

ok, so the power supply is always forward biased cus of the voltage given by the power supply, but lets assume we are in the case where the sine wave is approaching the 9V, so it assumes that value at a certain point, at this point the 9V given by power supply isnt sufficient because the ac signal wins over the d.c voltage no?

 

Change your signal to be 9V DC. Now what is the voltage across D1? Hopefully you can do that in your head.