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#1
12-18-2006, 01:43 PM
 star Junior Member Join Date: Dec 2006 Posts: 19
BJT Simulation in PSpice

Hi,

I have to simulate a bipolar junction transistor in PSpice and investigate its behaviour. One of the things I have to investigate is the BJT with a base current bias. The circuit I am simulating in PSpice is attached. What I'm trying to do is plot the base current (Ic) vs the voltage across the collector and the emitter (Vce). The graph I get below is when I simulate using a DC sweep through the voltage source from 0V to 10V with a 1V increment. However, this is different to those I see in textbooks, which is a steep increasing curve, which then starts to flatten out past around 0.7V. Can someone please help me figure out what I'm doing wrong?

Thanks a lot!
Attached Images
 bjt_circuit.jpg (9.4 KB, 279 views) bjt_graph.jpg (37.5 KB, 197 views)
#2
12-18-2006, 03:01 PM
 JoeJester Senior Member Join Date: Apr 2005 Location: Grand Prairie, TX, USA Posts: 2,361

Star,

Look at your scaling, the linear scale looks much different than the log scale.

See attachment.
Attached Images
 graph.jpg (20.7 KB, 175 views)
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Joe
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#3
12-18-2006, 03:45 PM
 n9352527 Senior Member Join Date: Oct 2005 Posts: 1,198

Is it Ic or Ib? You wrote base current but also wrote Ic.

If you are looking for something like transfer curves, then you need to revise your test circuit a little bit.
#4
12-18-2006, 04:03 PM
 star Junior Member Join Date: Dec 2006 Posts: 19

Thanks for the replies.

Joe, I noticed the difference with the log scale too, but the other graphs from textbooks I've been comparing against have linear scales. One example of such graphs is attached, taken from 'Solid State Electronic Devices' by B.Streetman. I'm wondering if there's something wrong with my circuit or simulation profile.

n9352527, sorry about the mistake. I meant collector current (Ic). What are transfer curves?
Attached Images
 bjt_graph2.GIF (13.0 KB, 126 views)
#5
12-18-2006, 04:07 PM
 n9352527 Senior Member Join Date: Oct 2005 Posts: 1,198

Sorry, I meant a characteristic curves, as illustrated on this page http://www.allaboutcircuits.com/vol_3/chpt_4/4.html, (Ic vs Vce for different Ib).
#6
12-18-2006, 05:37 PM
 n9352527 Senior Member Join Date: Oct 2005 Posts: 1,198

I've seen the example curves that you attached. We are talking about the same thing. To measure the Ic, you have to set the Ib to a constant value (use a current source) and then sweep the Vce (DC source). No resistors are needed in series with base and collector. Then, read the Ic value for each sweep.
#7
12-18-2006, 06:22 PM
 star Junior Member Join Date: Dec 2006 Posts: 19

Yes! That is what I'm trying to get. When I tried your circuit with the current source, the characteristic curves are what I expect them to be. However, my problem is that the circuit I have been given involves the two resistors Rb and Rc, both connected across the top to Vcc, so I have to use the resistors. The original circuit doesn't actually show the Vcc grounded at the other end - I'm just assuming that. I've just realised that if I sweep the DC source, and keep the resistor value constant, then I effectively vary Ib, which perhaps explains why I'm not getting the curve I want at the moment. How would I get the same characteristic curve with the resistors?

Btw, thanks a lot for your help so far.
#8
12-19-2006, 10:36 AM
 n9352527 Senior Member Join Date: Oct 2005 Posts: 1,198

To get a constant Ib, you could either use a current source (easier, since the current value is fixed to whatever you set it to), or you could use a voltage source with a series resistance Rb (you need to calculate Ib as (Vsrc_b - Vbe)/Rb).

Setting up a constant Vce, which is necessary to obtain the characteristic curves, is not going to be possible, I think, with a series resistance Rc. Because the voltage drop across Rc will vary with Ic and Vce = Vsrc_c - IcRc.

Are you sure you were asked to obtain the characteristic curves with that given circuit?
#9
12-19-2006, 08:02 PM
 star Junior Member Join Date: Dec 2006 Posts: 19

I think I have managed to crack this thing by using the fixed voltage source in series with a resistor (see attached Fig1). In the simulation, V1 is kept constand and a DC sweep is performed through Vcc. A parametric sweep through Rb is performed to observe the characteristic curves of the BJT with different Ib. Do you think that's close enough to the circuit I was given originally in Fig2?
Attached Images
 Fig1.jpg (12.8 KB, 139 views) Fig2.GIF (3.6 KB, 97 views)
#10
12-09-2007, 08:58 PM
 VVS Member Join Date: Jul 2007 Location: London Posts: 66

Star,
Are you by any chance a EEE first year in Imperial College London???? i am doing exactly the same thing
If you are looking for Ib vs Vce characteristic do the following:
sweep Rc from 5 to 80k and use secondary sweep and enter a value list for Rb using 10k,20k,30k,40k if u want more go ahead.
Then add Ic as a trace and change your axis variable to V(Q:c). You will get a really nice graph!

Last edited by VVS; 12-09-2007 at 09:18 PM.

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