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Consider the following circuit: enter image description here enter image description here

Now I'm asking to impose the base current as zero and $$R_C=452.35 \Omega $$

Now I have a lot of issues with this circuit because I can't find a systematic way to approach this.

Here is the work I've done and problems I found:

1) ASSUME ACTIVE ZONE ONLY USE KIRCHOFF'S LAWS

We are told to neglect the junction voltages when the junction is directly polarized so I should neglect \$U_E\$

Now from Kirchoff laws we will obtain: $$I_C=I_E$$ $$U_C=-E_C +R_CI_C$$ $$0=E_B +R_EI_E$$

which leads us to $$I_C=60 mA$$ $$I_E= 60 mA$$ $$U_C=-2.859 V$$

which makes me accept my hypothesis... However if I try now to apply Ebers-Molls equations I get to

$$I_C=\beta I_B + I_{CE0}(e^{\frac{U_C}{u_T}}-1)$$

I obtain $$I_C= 10 \mu A$$

So something is terribly wrong with this approach. Is anything wrong with my equations. My suspect is that I'm wrongly substituting the resistor as a short-circuit to apply Kirchoff laws. Since current is zero I should instead substitute it by using an open-circuit? I never quite understood this resistor paradox from circuit analysis:

Zero current is equivalent to an open-circuit ´ However zero current means zero voltage across resistor from Ohm's law which means a short-circuit.

I think that might be what I'm messing up here. Can someone clarify me please?

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    \$\begingroup\$ wrong formula Vc should be Vbe en.wikipedia.org/wiki/… \$\endgroup\$ – Sunnyskyguy EE75 Jan 2 at 22:49
  • \$\begingroup\$ Dont be confused but Uc=Ec=Vc are international differences in standard terms \$\endgroup\$ – Sunnyskyguy EE75 Jan 2 at 22:59
  • \$\begingroup\$ I'm using an equation given by my book, which yes is a little different from what we usually see but it is what my professor uses. It's from a semiconductor textbook I think. \$\endgroup\$ – Granger Obliviate Jan 2 at 23:00
  • \$\begingroup\$ I suspect that when you're told to set base current to zero you're actually being told to ignore base current. Meaning, don't worry about it, just pretend that it's zero, and use KCL. Because realistically, the transistor is operating in reverse mode, which means that the base current is 1/5 of the collector current. If you can ignore that, you can ignore the Ebers-Molls equations. \$\endgroup\$ – TimWescott Jan 2 at 23:38
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I don't recognize your Prof's textbook reference for \$U_{Cdisr}\$ which has been translated for something like \$V_A\$ the early voltage or \$V_{CE}\$max.

In North America and Asia, we normally use Vbe instead of Ube. etc.

Normally you can ignore \$I_{CEO}=10uA\$ This equiv. leakage current with "base open cct." has an equivalent resistance is called \$ r_o\$ that changes with Vce and is defined by the Early Voltage.

The "Early Voltage", \$V_A\$ is an asymptote that defines the leakage current slope for any opposite voltage/current of normal biased BJTs not in saturation.

Below for an NPN would be reversed about I axis for a PNP. enter image description here

Not shown above is the Iceo with Ib=0 which would appear as an asymptote close to the Ic=0 or 10uA at Vce max. ≡ Uce max = \$U_{Cdisr}\$ ?? which is negative for PNP and positive on graph above for NPN.

In any case the equivalent CE leakage will be something like \$Vce=(-50V/10uA = 5MΩ\$ or more. So this may start to become significant when Rc + Re are more than 10% of this value.

So ignoring this leakage, we get;

Let's call the junction between split supplies as GND which by def'n is 0V @ 0Ω

The collector current is only controlled by Vbe and Emitter current includes base current. Also we know Ie = Ic+Ib as long as CE is not saturated and in linear mode.

The important part to know is that the collector is a current source and it is the base voltage and emitter current that controls everything. As long as the collector is not saturated, KVL rules are simple. When saturated β rapidly reduces to β/10 or 10 as used in most datasheets (Ic/Ib=10)

Using KVL \$ E_B = 0V (gnd) + I_B*R_B + U_{BE} + I_E * Re = 30V\$

  • At this current range \$U_{BE}\$ will be something just under 0.7V which may be computed exactly from the EB model .

  • We know \$I_E= (β+1)I_B\$ thus you can solve for everything for any value of Rc and Uc.

  • You will note that none of this dependent on \$R_C ~or ~ E_C\$ as long as Vce is not saturated and proper polarity.

    • That means as long as our original assumption is satisfied.

    \$R_E+R_C<< r_o ~or~ < 0.1 r_o ~= r_o~approx= -50V/10uA=5MΩ \$

schematic

simulate this circuit – Schematic created using CircuitLab

Now I'm asking to impose the base current as zero and Rc=452Ω

Answer

ignore β * Ib and use only Iceo

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  • \$\begingroup\$ For a simulation run this tinyurl.com/ycrvu9g4 and drag Rb into position, then for giggles remove Uc and drag wire from collector to 0V Gnd. You get the same current. \$\endgroup\$ – Sunnyskyguy EE75 Jan 3 at 2:22
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Kirchoff law is correct. As mentioned above ignore the base of the transistor, because \$I_b\$ is insignificant compared to the \$I_c\$ current, so it is safe to say that $$I_c = I_e$$ and total resistance is close to 1K ohm (\$R_c\$ + \$R_e\$), factoring in the internal resistance of the transistor.

Now you have \$E_c\$ and \$E_b\$ in series to produce 60 volts across essentially 1K ohm which is 60 mA of current flow. Bring it down to basics and ignore that which is defined as 'zero', and compute the rest.

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I'm wondering if everyone is overthinking this. The question states (paraphrasing)

the base current is zero amps (open circuit)

By definition \$I_{CEO}\$ means "collector-to-emitter current with the base open" at some defined junction temperature and some defined collector-to-emitter test voltage \$V_{CE}\$. For a PNP transistor, if the emitter voltage is more positive than the collector voltage, then \$I_{CEO}\$ is negative—i.e., \$I_{CEO}=-10\,\mu A\$.

It seems to me that for this problem statement \$I_{CEO}\$ is a "given value" for the stated circuit conditions, and no further calculations are expected to adjust \$I_{CEO}\$'s value.

(n.b. \$I_{ECO} = -I_{CEO} = 10\,\mu A\$)

(n.b. \$U_{C_{disr}}\$ seems to be \$V_{CBO}\$. See fig. 1 in THIS DOCUMENT. It would not make sense for \$U_{C_{disr}}\$ to be \$V_{CEO}\$ because with the transistor in cutoff, the observed value \$V_{CEO} \approx -60\,V < -50\, = U_{C_{dist}}\$ which means the transistor would be at risk of breakdown in the collector-emitter path.)

Assuming the GROUND node is the junction of \$E_{C+}\$, \$E_{B-}\$, and \$R_B\$, and noting that \$I_{CEO}=I_C=I_E\$, then by KVL the voltage \$U_E\$ at the emitter (not the voltage \$U_{EB}\$ across the emitter-base diode) is given by

$$ U_E = E_B - I_E\,R_E = E_B - I_{ECO}\,R_E = +30V - (10\mu A)(500\,\Omega) = 29.995\,V $$

Likewise, the voltage \$U_C\$ at the collector (not the voltage \$U_{CB}\$ across the collector-base diode) is given by

$$ U_C = E_C + I_C\,R_C = E_C + I_{ECO}\,R_C = -30V + (10\mu A)(452.35\,\Omega) \approx -29.995\,V $$

And the theoretical voltage \$U_B\$ at the base is given by Ohm's law:

$$ U_B = I_B\,R_B = (0\,A)(10\,k\Omega) = 0\,V $$

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