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I basically have a transistor circuit which works as a switch. I want to ask, why is the output voltage looks like that? a bit explanations please

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Nashhyyy

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1595099251239.png
 
Your circuit has a output which goes form +5 to almost -5V because of V2, V3.
The input turns on when V1 is larger than -4.7V (or 0.7V above V3). The only time the circuit is off is when the input is below -4.7V.
A transistor responds to current in the Base. Current flows into the Base when B-E voltage is above 0.7V.
 
can you explain a bit more what's happening inside the circuit? I know it works as a switch. What's the different with the circuit above and the one below? I really understood the simplest circuit below. But I just don't know how to analyse the circuit above like why must it be at -4.7V and the graph looks so different from this one.
1595100342855.png
 
Because you understand picture #3 then you should understand this:
Using +10 not +5 but other wise the same.
1595101689328.png

Now this circuit is the same as in post #1. I renames -5V to 0V, renames 0V to +5, and renamed +5 to +10, but it is the same.
1595101746785.png
 
Many people, new to transistors, read that if you apply 0.7V to a Base or 10V to a Gate of a MOSFET the part turns on. That is not true. You need to apply this voltage from Base to Emitter (Gate to Source).
In this circuit the transistor can not know where ground is. The emitter is at -5V so the turn on voltage is -4.3V. (Base-Emitter=0.7V)
1595122273442.png

To be more accurate the transistor Base responds to current not voltage.
 
In my opinion, if I wanted to build an amplifier circuit, I'd consider a higher current gain BJT so that my signal can be amplified at its best. This gives me with T2 and T3. I'm not so sure with the rest parameters, thus it leaves me here wihtout a definite answer.

for switch circuit, I'd consider the least power losses. The fact that using BJT will give us power loss due to switching from cut off and saturation region, this thing is something that we can't surely avoid. But of course, a best switch is when there's no power losses on it. Thus, I will choose T1? I hope you could help me
 
In my opinion, if I wanted to build an amplifier circuit, I'd consider a higher current gain BJT so that my signal can be amplified at its best.
A good amplifier has a high voltage gain so that plenty of negative feedback can be used to reduce distortion.
Current gain (beta) is not voltage gain.
 
ah okay, i mixed things up. so in your opinion, which BJT will you choose as amplifier purpose and switch purpose?
 
Which BJT will you choose as amplifier purpose and switch purpose?
A transistor has a maximum allowed voltage, current and amount of heating.
A transistor used as an amplifier has some distortion that is reduced by adding negative feedback that also reduces the voltage gain.
A transistor used as an amplifier needs to have its input biased with R3, R4 and C1 in my circuit so that its input and output can swing smoothly up and down.
The hFE (beta) of the transistor determines its input bias current from 3 and R4.

I simulated with a 2N3904 transistor;
1) No input bias, 1.25V peak sinewave input and 2.5V output. The output is severely clipping into a rectangular wave.
2) No input bias, 0.73V peak input (so that the transistor barely turns on) and about 2.2V peak output. Again, severe distortion.
3) Simple input bias, 0.03V peak input producing 1.3V peak output. High voltage gain of 144.3 times but fairly high distortion.
 

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