Hi Guys,

Could anyone help to understand the DC Operation point analysis of transistor in LT Spice ?
Thanks in Advance !

 

The DC operating point establishes the initial conditions of the simulation. It asks what happens to the circuit if nothing is changing.
It establishes voltages:
Vc -- The voltage on the collector
Vb -- The voltage on the base
Ve -- The voltage on the emitter

The spice command .op replaces the .tran command, and the results are in the error log.

 

hi mishra,
Check out this LTS video.
You build a simulation circuit and post your asc file.
E

 

See

 

That circuit is not a good example for real-life transistor biasing since it depends upon the Beta (current-gain) of the transistor, which often varies over a 3:1 range, even for transistors of the same model.
Normally some negative feedback is used to stabilize the bias point.

 

You need to learn how to read a datasheet and to learn how a transistor works.
The transistor is severely saturated. I was taught to NEVER bias a transistor like that.

 

Depends on what the TS is trying to do. If he wants to see the operating point of a saturating switch that is controlling 9mA at 9V, then it is quite correct.

Bob

 

Could anyone help to understand the DC Operation point analysis of transistor in LT Spice ?

That's kind of like using a calculator to add 200+200.

Base current will be about half a mA and will be sufficient to saturate the transistor with a 9mA collector current. Collector voltage will be close to 0V. Did all of the calculations in my head.

 

hi Mishra,
You asked about BJT , DC operating point using LTSpice.
This sim will give a good indication of the Q point.
The sim of your circuit with the fixed 3.3v DC input is not a viable design for an amplifier.
E

 

I have a different simulator another approach to the same thing step by step.
looking at section III. Turning Your Oscilloscope into a Curve Tracer:
https://www.keysight.com/upload/cmc_upload/All/BJT.pdf

 

hi Mishra,
You asked about BJT , DC operating point using LTSpice.
This sim will give a good indication of the Q point.
The sim of your circuit with the fixed 3.3v DC input is not a viable design for an amplifier.
E

Thanks for simulation reply !
How can i see/decide Q point in simulation as below wiki.

 

hi m,
Look at this marked up image, the colour plots are synchronised in time.

Tell me why I have chosen the ringed operating point for the circuit.?
E

 

hi m,
Look at this marked up image, the colour plots are synchronised in time.

Tell me why I have chosen the ringed operating point for the circuit.?
E

Thanks for pointing out the question and sorry for delay reply !

My understanding for attached circuit is as per below :
1. The operating V(vc) is at Vcc/2
2. Ic(Q1) is at Ic(max)/2
3. V(vb) is at max Vbe.

So my understanding is now to choose Q point at Vcc/2 to have faithful amplification.

Regards,
Shiv Mishra

 

hi mishra,
The Collector load line is normally drawn from the maximum possible Collector current [ determined mostly by the value of the load resistor] to the minimum load current.
As you say the 'Q' point is usually approx Vcc/2, but you should also consider the effect of the Emitter voltage, if there is an Emitter to 0V resistor. (Vcc-Ve]/2
You want the maximum possible Collector voltage swing without any signal clipping.
E

 

A single transistor with high voltage gain (no negative feedback) has such a high amount of distortion that it causes half of its output swing to be squashed then symmetrical clipping is impossible.

 

hi agu,
My post says: Collector voltage swing without any signal clipping.
No mention of: symmetrical clipping

E

 

Usually the transistor is biased so that the collector DC voltage is set at half the supply voltage for maximum output voltage swing without clipping.
But my simulations show that it cannot be done when the transistor has a high voltage gain caused by no negative feedback.