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amplifier only works half cycle. Why?

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Why is the circuit that i've rigged up in LTspice only conducting for half cycle in the output current shown in the red arrow point?
LTsamp.jpg
 
Why is your Load Resistor (R5) Only 1 Ohm?.
It is Shorting Out the DC Collector Voltage.

Put a Capacitor between the Collector and the Load Resistor.
 
Something don't jives with the circuit shown and the waveform shown.
R5 can only get current thru R1...because Q1 will take away current as it starts conducting.

But R1 is 15K, and V1 is 15 volt....a peak of only 1 milliamp can flow.....where did you get the 2.6 amps peak?
 
Something don't jives with the circuit shown and the waveform shown.
R5 can only get current thru R1...because Q1 will take away current as it starts conducting.

But R1 is 15K, and V1 is 15 volt....a peak of only 1 milliamp can flow.....where did you get the 2.6 amps peak?

I'm trying to understand the working of transistor amplifiers, so kind of testing them with LTspice.
The 2.6 amps is measured across R5. I've shorted the input capacitor, if you noticed and if that's not shorted i'm getting 2.56 mA, as opposed to 2.6 A peak with shorting that capacitor

Also I have used a transistor spice model downloaded from ON semiconductor website! If I replace it with a 2N222, then I only get milliamp range of current!
 
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I don't know what to tell you, where the mistake was made...all that I can tell you is that the circuit and values provided do not satisfy ohm's law. It is a physical impossible.
 
mA.....good.....LOL

With the capacitor missing you should only see slightly less than 1/2 of the sign wave. (just what you have)

The transistor can not amplify the input signal that is below ground. (in this case below +0.6 volts) There is a reason for the capacitor. Put it back in.
 
The type of transistor is not the problem.
The transistor can not amplify a signal below ground. The capacitor is there to block the DC and move the signal up in voltage to a range where the amplifier can work.

I don't like the value of R3 (25k). I think you should put the capacitor back and change R3 to 5k.
 
Most likely the model is either incompatible with LTSpice or just wrong...I don't know.

Anyways, before modelling a circuit, any circuit, one must do first a pencil and paper anlysis. What are the transistor's DC Quiescent points? Vcq, Veq, Iceq must be roughly calculated.
From where and to where does the current flow? What sort of voltages will be developed across the resistances the current flows thru?

Now you add a load, AC decoupled with a capacitor... how does the new AC loadline appears? After one is satisfied with aproximate values, one will then simulate the circuit.
Finally, breadboard a unit and actually tweak it further.
 
Most likely the model is either incompatible with LTSpice or just wrong...I don't know.

The model is probably fine. The circuit is F$&*!~U@.
1) That amplifier can not drive a 1 ohm load. (remove R5)
Why is your Load Resistor (R5) Only 1 Ohm?.
It is Shorting Out the DC Collector Voltage.
2) Put C1 back in.
The transistor can not amplify a signal below ground.
3) The bias is wrong.
change R3 to 5k
 
I have a simple explanation here:
 

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Thanks AG!!
I knew that when you would see the circuit, you could spot immediately the error!

The problems with simulators is that, if one does not model them correctly, totally erroneous answers will occur. For instance, all audio generators are mostly 50 ohm, others 600 ohm. But in a simulator, sources are essentially zero output impedance. They will supply 1000 amps if one is not careful with the circuit.
 
It is easy to make the source impedance anything you want.

The transistor has a high impedance output so it needs a high impedance load.
I simulated the circuit without the 1 ohm load and found that the base resistor ratio was completely wrong. I fixed it then I needed to reduce the input to 1.75V peak instead of 5V peak. My load was capacitor-coupled with a 0.22uf capacitor to a 33k ohms load.
 
Thank you all,
[/U] The capacitor is there to block the DC and move the signal up in voltage to a range where the amplifier can work.

You mean the capacitor needs to be there to block the supply dc from setting up a voltage across the AC signal supply? And by "moving up the voltage", you mean by creating a voltage drop across the collector resistor R1?
 
Since V2 is 5V peak and R5 is 1Ω, with C1 shorted out the peak theoretical current flow via the collector-base diode of Q1 will be ~ (5-0.7)/1 = 4.3A.
 
Since V2 is 5V peak and R5 is 1Ω, with C1 shorted out the peak theoretical current flow via the collector-base diode of Q1 will be ~ (5-0.7)/1 = 4.3A.
Your theory doesn't work when the base-collector diode current is extremely high.
The simulation shows that the forward voltage of the base-collector diode is 5.0V - 2.6V= 2.4V (!) which makes the current be 2.4A.
 
Your theory doesn't work when the base-collector diode current is extremely high.
It does for a theoretical transistor whose bc diode drop is independent of current (but not for a practical transistor) :D
 
A 1N4148 is a silicon diode that is similar to the base-collector diode of that tiny surface-mount transistor.
Its forward voltage is typically 0.7V when its current is only 5mA.
It can survive 1A for one second but the datasheet graph stops at 800mA when its forward voltage is typically 1.45V. It might be 2.4V or more at 2.6A.
Its maximum allowed current is 4A for a very short duration.
 

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with the short across the input capacitor, the transistor is not properly biased. the internal resistance of the LTSpice signal source is ZERO ohms. remove the short from across the capacitor.... it's not an incompatibility of LTSpice or any of it's models. when you put the short across the capacitor LTSpice is only doing what you told it to, and creates a zero ohm DC path to ground. remove the short from around the capacitor..... R E M O V E T H E S H O R T...............


remember it's only a machine and it can only outsmart you for so long......
 
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