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"Info"

The bias point (without the 623mv input) is calculated as

\$I_b = \frac{5-0.6}{5.5 M\Omega} = 800nA\$

\$Ic = \beta I_b = 290*800nA = 233\mu A\$

These calculations checks out in the simulator. But when I connect, the Input signal, whose peek is at 623mV, the transistor saturates. Why?

Because, if I do the calculations again with \$V_{BE} = 0.62318V\$, the results do not change much from the previous calculation.

\$I_b = 795.78nA \$ which should give the \$I_c = \beta I_b = 290*795.78nA = 230.8 \mu A\$, and this is \$ << I_C(sat)\$. Then why is the transistor saturated?

I know the transister is saturated because the \$V_{CE} = 67.4mV\$, when it should have been \$5 - (230.8\mu A * 10k) = 2.7V\$

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You write that the peak base current, with the signal source connected is given by

$$i_B = \frac{5V - 0.62318V}{5.5M\Omega} = 795.78nA $$

But this isn't true (which should be obvious as it's less than the bias current!). What's true is

$$i_{R2} = \frac{5V - 0.62318V}{5.5M\Omega} = 795.78nA \ne i_B$$

The resistor current and base current are not equal. According to KCL at the base node:

$$i_B = i_{R2} + i_S $$

where \$i_S\$ is the current out of the signal voltage source. But you don't know what this current is.

In fact, the base current depends exponentially on the base-emitter voltage. We can estimate the change in base current as follows

$$\frac{i_{B2}}{i_{B1}} = \frac{e^{\frac{v_{BE2}}{V_T}}}{e^{\frac{v_{BE1}}{V_T}}} = e^{\frac{v_{BE2}- v_{BE1}}{V_T}} = e^{\frac{0.62318V - 0.6V}{25mV}} \approx 2.53$$

Thus, the peak base current should be larger than the DC base current by a factor of about 2.53 or

$$i_{B_{peak}} = 2.53 \cdot 800nA = 2.02\mu A$$

This gives a collector current of

$$i_{C_{peak}} = 2.53 \cdot 233\mu A = 589\mu A$$

If this were the actual collector current, the collector voltage would be

$$5V - 589\mu A \cdot 10k\Omega = -0.895V $$

So, yes, the transistor will saturate first.

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  • \$\begingroup\$ The circuit is a simplified version, of what I was doing. The original circuit, has a sine wave input signal, with capacitor coupling. Now, with that input sine wave, when the Vb gets to 0.623v the transistor gets into saturation. In order to understand what was happening, I replaced the sin wave source and put just the peek voltage through the voltage source. \$\endgroup\$ – Arjob Mukherjee Mar 12 '14 at 21:04
  • \$\begingroup\$ @ArjobMukherjee, I think I follow you but you're still not calculating the base current correctly. If the signal source raises the base voltage to 0.62318V peak, the base current is not equal to \$(5V - 0.62318V)/5.5M\$. This calculation is invalid. You're calculating the current through the 5.5M resistor, not the current through the base. \$\endgroup\$ – Alfred Centauri Mar 12 '14 at 21:16
  • \$\begingroup\$ @ArjobMukherjee, I've updated my answer to show why the transistor is saturated. \$\endgroup\$ – Alfred Centauri Mar 12 '14 at 21:38
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Your calculation of base current at base voltages of 0.6000 V and 0.6312 V is fine - both results indicate that the current is about 800 nA. However, you haven't considered that the base will not be at either of those voltage but somewhat less - it might actually be 0.4V - what does this do to the calculation of base current?

It doesn't do much on the face of it - it increases it from about 800 nA to 836 nA - not a major difference but, the point I'm trying to make is that you don't really know what the base voltage will be when it connects to a 5 V source via 5.5Mohm.

That's the first point and the second point (and more important one) is that the base-emitter junction is a forward biased diode and above 0.4 volts (a bit of hand waving going on) a small increase in base voltage results in a large increase in base current and by the time you get to 0.6 V there will be milli amps flowing into the base. That's why the transistor is saturated.

Conclusion - use your simulator to give you the actual base current or base voltage and see for yourself.

This is quite a good link for explaining stuff. It covers the ebers moll equation which allows you to predict collector current for a given base-emitter voltage at a given temperature.

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  • \$\begingroup\$ !Andy Without the signal, the simulator is showing the same results. Vbe at 0.599v and Ib, Ic are as per the calculations. But not after the 0.62318V signal was added. Also can you explain why the 'voltage might actually be 0.4V'; I do not understand that. \$\endgroup\$ – Arjob Mukherjee Mar 12 '14 at 19:41
  • \$\begingroup\$ If you force 0.599V onto the base what happens? Look at the forward voltage curve for a diode and see what the voltage is with an input current of 800 nA - it might be 0.3v or it might be 0.5v or somewhere in between. 0.4v is just my kind of rule of thumb for "nothing much happens below this level". \$\endgroup\$ – Andy aka Mar 12 '14 at 19:44
  • \$\begingroup\$ Ok, the 800nA or so current is causes the voltage drop to be 0.4v or so. But when I force that 0.623v onto the diode, the current increases exponentially. Is my understanding correct? Why do the transistor spec does not include the BE junction characteristics? \$\endgroup\$ – Arjob Mukherjee Mar 12 '14 at 20:01
  • \$\begingroup\$ @ArjobMukherjee I guess pretty much all BJTs have the same sort of characteristic and that is a diode biased forward. Plus, when biasing a transistor there are better and more reliable ways. For instance using (say) a 1Mohm resistor from the collector tends to ensure that the collector is about mid-point - if the collector voltage rises, more current feeds the base which forces the collector lower. You barely need to worry about b-e forward volt drop when you do this and the result is much more stable with temperature and between different devices. \$\endgroup\$ – Andy aka Mar 12 '14 at 20:53

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