hi,
What do you think the V output would be?

E

hi,
What do you think the V output would be?

E

well in A.C what i got is a signal that is increasing from 0 to 9V, until it reaches 0.7V nothing appears in output, when i got more than 0.7V i can see it in output, so from 0 to 8.3V ( i know its a bit contradictory with what i said about d.c component first , but i didnt fully understand it yet..)

 

the basic thing to understand is that most of the characteristics of transistor are not linear (there is buch of exponential terms for example). this makes calculations and predictions hard. so when dealing with the problem of this nature one uses two approaches:
1. do a rough estimate to get quick results even if not perfect.
2. do a detailed calculation that produces better (more accurate) results.

the 0.7V is used in first case. crude estimates...

so based on that all input signals are treated based only if the input value at some point is above or below that 0.7V level.
if it is below, it is approximated that transistor is off and no collector current is present. if the input signal is above 0.7V then further check is needed if transistor is in linear or saturation region. when in linear, then crude estimate is applied. this still produces remarkably useful results even if not very accurate.

according to estimate, result of next two inputs are not distinguishable

 

Hi,

well in A.C what i got is a signal that is increasing from 0 to 9V,

No, An AC signal with no DC offset swings above and below zero, so 1Vac swings-1v > 0v +1v > 0v > -1v
e

 

Hi,


No, An AC signal with no DC offset swings above and below zero, so 1Vac swings-1v > 0v +1v > 0v > -1v
e

ah ye my bad, then in that case if it swings from -9v to 9v for example, what i will see in output is a sine wave that goes from 0 to 8.3V. Because what happens is: until 0.7V it wont conduct ( so negative voltage wont make the diode conduct ) above that voltage it starts conducting, and the signal will have a drop of 0.7v) or im wrong?

 

the basic thing to understand is that most of the characteristics of transistor are not linear (there is buch of exponential terms for example). this makes calculations and predictions hard. so when dealing with the problem of this nature one uses two approaches:
1. do a rough estimate to get quick results even if not perfect.
2. do a detailed calculation that produces better (more accurate) results.

the 0.7V is used in first case. crude estimates...

so based on that all input signals are treated based only if the input value at some point is above or below that 0.7V level.
if it is below, it is approximated that transistor is off and no collector current is present. if the input signal is above 0.7V then further check is needed if transistor is in linear or saturation region. when in linear, then crude estimate is applied. this still produces remarkably useful results even if not very accurate.

according to estimate, result of next two inputs are not distinguishable
View attachment 347049

ye i know its incorrect to say 0.7V but its not the problem here, im trying to understand how things works first, then i will look about real diode drop etc. but thanks

 

hi,
0V to +9V
E

 

hi,
0V to +9V
E
View attachment 347051

yeah good, as i expected, from 0 to 8.3V, i guess i understood now how things really work

 

yeah good, as i expected, from 0 to 8.3V, i guess i understood now how things really work

and during the negative cycle of the sine wave, the diode doesnt conduct cus its acting like an open circuit no? (semplification clearly)

 

correct... from post #24 you seem to be on the right track.
but keep in mind that there are other concerns too: base voltage Vbe cannot be negative too much or transistor will be destroyed.

 

correct... from post #24 you seem to be on the right track.
but keep in mind that there are other concerns too: base voltage Vbe cannot be negative too much or transistor will be destroyed.

ye i gotta look for the maximum VEB (cus its reversed) that my transistor can withstand before popping right? ( this i look i nthe datasheet). thanks btw

 

that would be job for a diode:

 

that would be job for a diode:
View attachment 347053

what do u mean? why did u add this diode in reverse polarity? ( like this when there is -9V the diode conducts and current flows from output to input no? ( from anode to cathode ) but why so? can u explain me please?

 

that would be job for a diode:
View attachment 347053

from what i understand here is that, basically the current goes to signal source, which sinks the current ( but this wont cause any problem to the input generator?)

 

diode is used to protect transistor from negative voltage, regardless of input voltage polarity, Vbe is always within +/-0.7V and this is safe since considerably smaller than base breakdown voltage. most transistor have Vce breakdown of 45V or or higher, and Vc=9V, input could be not just -9V, it could be -36V
9V-45V=-36V

if you connect -9V to base, current will be (-9V-(-0.7V))/1k = -8.3mA

if you connect -36V, then current will be -35.3mA

not sure if your signal generator can handle this but ... transistor will be safe

 

Here's a plot that shows how the Vbe (red trace) changes versus the input voltage and output current:
Note that it doesn't reach 0.7V until the output current (blue trace) is near 9mA.

 

f you connect -36V, then current will be -35.3mA

not sure if your signal generator can handle this but .

You can add a resistor in series with the generator to limit that current, if needed.

 

that would be job for a diode:
View attachment 347053

Adding that diode changes the transfer characteristic radically.

Ignoring breakdown considerations (which can't be ignored in real life), without the diode, the output is essentially 0 V for any input voltage below about 0.7 V. But, with the diode, the output follows the input for negative voltages below about -0.7 V.

The transistor provides buffering between the input and the output, but the diode removes that buffering for negative input voltages.

 

If you connect the diode from base to ground with a series resistor to the input generator, then the base will be protected and the output won't see any negative voltages from the input.

 

Here's a plot that shows how the Vbe (red trace) changes versus the input voltage and output current:
Note that it doesn't reach 0.7V until the output current (blue trace) is near 9mA.

View attachment 347061

Hi, Crutschow, what is the simulator app I see you experienced guys using here?

 

Adding that diode changes the transfer characteristic radically.

how so? even without the diode, output of shown circuit still cannot reach negative voltage. transfer characteristic is not changed, all the diode does is protect transistor. in case AC input is applied