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I simulated this circuit using Qucs:

With Vin = 5V, VB = 0.746V and VC = 0.024V which means that the BJT is operating in the saturation region. But I don't understand why. Vcc = 5V and Vin = 5V. RB = RC = 1k ohm. So I expect that VB = VC and the base-collector junction is reverse biased which means that the BJT is in the forward-active region. What is the wrong with my analysis? and is there a "trick" that I can use to determine the mode of operation of a BJT?

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  • \$\begingroup\$ Did you calculate (or measure) the base current? \$\endgroup\$ – Fizz Oct 30 '15 at 19:56
  • \$\begingroup\$ What does saturation mean? And how would you measure this. Would your circuit be much different if Vin was 4V, 3V...? \$\endgroup\$ – George Herold Oct 30 '15 at 19:59
  • \$\begingroup\$ @RespawnedFluff No. I don't know how but it can be calculated. IB = (5-0.764)/1K = 4.236 mA. So? \$\endgroup\$ – ammar Oct 30 '15 at 20:19
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    \$\begingroup\$ Why do you think VB = VC? \$\endgroup\$ – Austin Oct 30 '15 at 20:26
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    \$\begingroup\$ VB=Vin-RB*IB and VC=Vcc-RC*IC, so it sounds like you are assuming IB and IC are the same, which is not true. \$\endgroup\$ – Austin Oct 30 '15 at 20:35
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In saturation you'll have \$I_c < \beta I_b \$. So calculate \$I_b\$ as (5-0.7)/1k=4.3mA. Assuming beta of 100 (or even 10), you'll have an \$\beta I_b\$ in the 43-430mA range. But through the collector you can only have at most 5mA due to \$R_c\$ limiting the current. So the transistor is in saturation.

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  • \$\begingroup\$ If you somehow didn't guess this inequality was going to hold, and you wanted to show that that the transistor is in the active region, then you'd have to prove that Vce>0.7V (so that Vcb>0). If Vce>0.7, then the collector current is at most 4.3mA. So then the base current [under the active region assumption] is under 0.43mA (or even under 0.043mA) by division with beta (of 10 or 100). But this is inconsistent with the directly calculated value for Ib of 4.3mA. So this is basically the same argument we did for proving saturation, but in reverse (instead of p=>q we have shown not q => not p). \$\endgroup\$ – Fizz Oct 30 '15 at 20:59
  • \$\begingroup\$ Thanks. Can you please take a look at this question link. For Rcp=100, Rc=3.6K and RBP=1.5K, Qp cannot be in saturation. Right?. Here is my work: link. \$\endgroup\$ – ammar Oct 31 '15 at 15:52
  • \$\begingroup\$ For the previous comment, the case is when Vin=0. \$\endgroup\$ – ammar Oct 31 '15 at 16:03
  • \$\begingroup\$ @ammarx: Why not edit that data into your other question? Many more will be able to answer that way. Also, I don't have the book [and it seems my library doesn't have it either], so I won't be able to comment on much of that. \$\endgroup\$ – Fizz Oct 31 '15 at 16:25
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    \$\begingroup\$ @ammarx: The ratio of Ic/Ib in your latest example is 58. For some transistors (e.g. with beta of 100) that would be [soft/quasi] saturation, for others (higher beta) would even be hard/deep saturation, and for some power transistors it might no be; so yeah, for values like that it depends on the transistor too. Also, I've answered your other question in more detail. I do wonder if you haven't found any answers [to any of your questions, not just those where I answered] satisfactory enough to accept though. \$\endgroup\$ – Fizz Nov 1 '15 at 15:01
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To an experienced person, it is obvious at a glance that the transistor will be in heavy saturation.

I am not sure exactly where you went off track. But I do notice that in your question, you state two things that are mutually exclusive. You state that VB = VC, and you also state that VB=0.746 and VC = 0.024. That is an obvious fallacy in your question.

Conceptually, an NPN configured this way (which is very common) goes into saturation as the base current increases. Roughly speaking (assuming base supply and collector supply are the same), when the RB/RC = beta, the transistor is somewhere near the beginning of saturation. As RB gets smaller, it goes more and more into saturation.

In this case, RB/RC = 1, so you know the transistor is in very heavy saturation. You can also see this by the very low drop from collector to emitter, which is another hallmark of saturation.

When a transistor is used as a saturated switch, the ratio of Ic to Ib is lower than the datasheet beta. The ratio of the actual currents is sometimes called the "forced beta." Typical values of "forced beta" are 10 to 20. When the base and collector drive are the same voltage, you can just use a 10:1 rule where the base resistor is 10x the collector resistor.

When the system is not that simple, you can roughly calculate forced beta by assuming realistic values for Vbe and Vce and estimating the currents.

I am trying to present you with an intuitive and practical framework rather than a textbook approach. Hopefully it will help.

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  • \$\begingroup\$ @RespawnedFluff Thanks. Can you please take a look at this question link. For Rcp=100, Rc=3.6K and RBP=1.5K, Qp cannot be in saturation. Right?. Here is my work: link. \$\endgroup\$ – ammar Oct 31 '15 at 15:53
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Let's assume that the BJT is in the active regime and analyze it. Then we'll check our assumption and see what could cause it to be saturated.

Ve = 0 and if the BJT is indeed operating in the active mode, Vb = 0.7. We can then find that Ib = (5-0.7)/1 = 4.3mA.

Still assuming we're in the active regime, the current at the collector Ic = beta*Ib, which we can use to find the Vc. Vc = 5 - beta*4.3.

Now, for the active assumption to hold true, we must check that (1) The base current is positive and (2) Vc > 0.7.

Since you've already established that the BJT is operating in saturation, either (1) or (2) must not be true. (1) is true by inspection, so I can only assume that the value of beta is too small.

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  • \$\begingroup\$ If you've already established that the BJT is in saturation, then why the heck check if it's in the active region (last paragraph). -1 If you want to disprove that it's in the active region you must do this by circuit analysis, not by assuming it's already in saturation. Your proof is circular/trivial: assumes "not x" and proves "not x", where x stands for "transistor is in the active region". \$\endgroup\$ – Fizz Oct 30 '15 at 20:40

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