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NPN transistor

Sir, These are the equations I have used in my undergrad course and all these are also found on the net. I dont think there are any new equations. Still I think I will do a bit of homework and then get back to work. Please provide light if you have any ideas on what equations or what methods I should follow.


Also Sir, do you how to upconvert a VI??

I have a VI in version 5. I need to upconvert it to Version 8.5 or 10. I did post in UPCONVERT VIs group but did not get any reply yet !

Do you know how to ?


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Message 31 of 40

As GerdW has said before, we are not going to do your coursework for you - so you'll have to decide on the equations for yourself.


As for the upconvert, you'll just have to be patient i'm afraid (or find someone at your uni who has older versions of LabVIEW installed)



"If anyone needs me, I'll be in the Angry Dome!"

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Message 32 of 40

Sir, dont get me wrong. You being Labview champions, you always know a better way to solve problems or maybe you have done such problems in the past. As of me , I am certainly not an amateur but also not good as you are. So guide me the path.


Thanks GerdW, ChirssyC87

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Message 33 of 40

That's fine, people here are more than happy to offer advice and guidanceSmiley Very Happy


Have another go at your equations, but remember LabVIEW will only plot what you tell it to - so if your equations/formulas aren't up to scratch you won't get the results you're looking for!


Best of luck



"If anyone needs me, I'll be in the Angry Dome!"

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Message 34 of 40

Based on your formula, this is not the behavior I would expect using an NPN transistor in this circuit.  Your controlling Ic with Vbe.  Why are you not using Ib to do this, since BJTs are current-controlled devices.  Your treating it like a MOSFET.

First off, on average your Vbe junction will drop anywhere from .5-.9V in real world applications.  You set Ic = 0ma at Vbe of 0.6V.  This is not typical and there is not some magic number where the Ic will cut-off at a specific Vbe.

Secondly, your equation is based on the assumption that if Vbe = 0.6-0.8V your transistor is saturated, another false assumption.  Your voltage could be at 0.8V and Ic still could not be maxed, again this is based on the value of Ib, or at least should be.  I don't know what your values on your resistors are except Rc, so have know way of knowing what Ib is.  I assune you want to simulate a pot on the base of the transistor, thereby adjusting Ib, else where would the different values of Vbe come from?

I would worry about getting the proper behavior down before worrying about the graphing.

Reese, (former CLAD, future CLD)

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Some call me the gangster of love.
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Message 35 of 40

Alrite MoResse, I think I now need to do my homework properly.


Thanks 😄

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Message 36 of 40

I have tried implementing the BJT in the following VI. Also am getting approx results of output characteristics. However I now want to plot the output characteristics according to the image given below. Please suggest me how to do it. I request you seniors to help me how can I improve the attached VI for better output characteristics. Please refer only the output characteristics of the image and the attached VI.


For the VI insert Beta as 100 and Ib as in range of 0.1 - 0.4 ma and R as 10 kohm..

Vce ranging from 0-10 at 41 that Vce step is 0.25



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Message 37 of 40



Your model does not make any sense.  Your VI implements the following set of equations:


Vce[0] = Vce_Start +Vce_Step

Vce[i] = Vce[i-1] + Vce_Step,                 i > 0


Ic[i] = (Vce[i-1] + Vce_Step)/(1000*R),    Vce < 1

Ic[i] = Ib*Beta,                                      Vce >= 1


Vce is usually determined by the value of the resistor and Ohm's Law and the collector current by Ib and beta, except as limted by the external circuit.  Where is the resistor connected in your circuit?  With Ib = 0.1 mA, beta = 100, and Rc = 10000 ohms, you will have Ic = 10 mA and Vr = 100 V, where Vr is the voltage across a collector resistor.


Also you appear to be plotting Vce vs Ic rather than Ic versus Vce.



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Message 38 of 40

Hmmm.. Thanks for pointing out the mistakes. I am plotting the output characteristics ie Vce on X axis and Ic on Y axis


Ic= (Vcc-Vce )/R ie fr in linear region..


Ic=beta*ib after Vbe is 1V.. ie where Ic becomes constant

This is what we have been taught in undergrad.


However when i do Vcc-Vce I get starting values as 10-0.25 and 10-0.5 etc. This divided by R gives us something in 0.9 and 0.8. Now when I implement the other condition of Ic=Beta*ib I get constant values. So the graph goes from 0.9 0.8 0.7 0.5 0.4 0.3 to 0.5 0.5 0.5..The curve does not get proper.


So I removed Vcc- Vce and jus kept Vce/R. I hope you get it.


Vce is usually determined by the value of the resistor and Ohm's Law and the collector current by Ib and beta, except as limted by the external circuit.


How do I implement this ? and also How do I implement Vr..Please help me modify the VI



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Message 39 of 40



I looked at the GE Transistor Manual (copyright 1964) to see how they test transistors to produce those curves. They use a simple circuit with the emitter grounded, a base bias voltage connected through a series resistor and base current measurement device, and the collector connected through a series resistor to the collector current measurement and to a variable voltage source. They also have a voltmeter connected to measure Vce.


Transistor curves.png


Next I would use a simple model like the hybrid-equivalent circuit model, common emitter mode.  Put the model into the test circuit.  Develop equations for Ic and Vce from the value of Ib. Build your VI around those equations.  You can set hie = 0 since you do not care about Vbe.




Transistor models.png

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Message 40 of 40