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I want to determine the currents in each of the branches.

I could analyse the current mirror portion and the differential amplifier portion, but facing problem thereafter.

I could find the collector current of \$ Q_2 \$.But i need to find the current through \$ R_C \$ connected to \$ Q_2\$,to find \$ v_{O2}\$, which in turn would help me in analysing \$ Q_3\$(ie finding collector and emitter current of \$ Q_3\$ ).

I dont want to assume that current through base of \$ Q_3\$ is zero.

Parameters:

\$R_1=12k\Omega\$

\$R_C=8k\Omega\$

\$R_E=3.3k\Omega\$

\$ \beta=200\$ \$\$ \$\$ \$\$ \$\$

schematic

simulate this circuit – Schematic created using CircuitLab

My attempt:

ANALYZING THE CURRENT MIRROR PART

Let current through \$ R_1=I_1\$ $$I_1=I_{C4}+I_{B4}+I_{B5}$$ (Since \$V_{BE}\$ is same for both transistors and transistors are identical, \$ I_{B4}=I_{B5}\$ and \$ I_{C4}=I_{C5}\$) $$=I_{C5}+2I_{B5}$$ $$=I_{C5}+2\frac{I_{C5}}{\beta}$$ $$=I_{C5}(1+\frac{2}{\beta})$$ $$I_{C5}=\frac{I_1}{1+\frac{2}{\beta}}$$

ANALYZING THE DIFFERENTIAL AMPLIFIER PART

Let \$I_{C5}=\frac{I_1}{1+\frac{2}{\beta}}=I_E\$

$$v_1-v_{BE1}=v_2-v_{BE2}$$ $$v_1-v_2+v_{BE2}-v_{BE1}$$ $$v_{ID}+v_{BE2}-v_{BE1}=0$$

Now, $$v_{BE1}=V_Tln(\frac{i_{C1}}{I_{S1}})$$ $$v_{BE2}=V_Tln(\frac{i_{C2}}{I_{S2}})$$ If \$I_{S1}=I_{S2}\$ $$\frac{i_{C1}}{i_{C2}}=e^{\frac{v_{BE1}-v_{BE2}}{V_T}}=e^{\frac{v_{ID}}{V_T}}\tag1$$

Nodal equations at the emitter:

$$I_E=i_{E1}+i_{E2}=\frac{1}{\alpha}(i_{C1}+i_{C2})\tag 2$$

Substituting eq (1) in (2) we get:

$$I_E=\frac{1}{\alpha}(e^{\frac{v_{ID}}{V_T}}i_{C2}+i_{C2})$$ $$=\frac{i_{C2}}{\alpha}(e^{\frac{v_{ID}}{V_T}}+1)$$

$$i_{C2}=\frac{\alpha I_E}{(e^{\frac{v_{ID}}{V_T}}+1)}$$

also

$$I_E=\frac{1}{\alpha}(i_{C1}+i_{C1}e^{-\frac{v_{ID}}{V_T}})$$ $$=\frac{i_{C1}}{\alpha}(e^{-\frac{v_{ID}}{V_T}}+1)$$ $$i_{C1}=\frac{\alpha I_E}{(e^{-\frac{v_{ID}}{V_T}}+1)}$$

My question is: How do I analyse \$ Q_3 \$.

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  • \$\begingroup\$ Are you looking for a closed solution based upon your simplified BJT model? If so, do you know about the Lambert-W function and how to apply it here? \$\endgroup\$ – jonk Aug 4 '17 at 22:08
  • \$\begingroup\$ @jonk No sir. i don't know about the Lambert-W function as of now.....I try to minimize the approximations as much as possible..hence was wondering if experts on this subject have a way to take into account the base current of \$ Q_3\$ as well(Was wondering if base current of \$ Q_3\$ can be somehow taken into account mathematically) hence decided to ask in the forum where experts might answer my question.. \$\endgroup\$ – Soumee Aug 4 '17 at 22:17
  • \$\begingroup\$ The base current is handled with \$\beta\$, which you will also need. (You can't derive it.) Are you willing to add that as an additional input? \$\endgroup\$ – jonk Aug 4 '17 at 22:22
  • \$\begingroup\$ @jonk I am in the learning phase......It would be exciting to see how this solution goes taking other things into account.... \$\endgroup\$ – Soumee Aug 4 '17 at 22:25
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    \$\begingroup\$ No problem. I have had a very lucky life that I did not deserve. I owe you (and many others) more than I can repay. And I don't mind paying it forward, especially to anyone who might appreciate it. Best wishes to you! \$\endgroup\$ – jonk Aug 6 '17 at 16:32
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I also agree with what you wrote:

I try to minimize the approximations as much as possible

But keep in mind that:

  1. Your question is simplified in the sense that you weren't discussing even the basic Ebers-Moll model for the BJT. It doesn't include the Early Effect (which appeared in later Ebers-Moll models.) You are only using the active region simplification (the Shockley diode equation, plus \$\alpha\$ to define the 3-pin device.)
  2. Your circuit stage is relatively easy to approximate by hand and, if more precision is required, one or at most two additional iterations with a calculator can achieve pretty much any rational need for precision.
  3. Your circuit stage can be well-approximated using free Spice tools, such as LTspice, and it is very easy to set up and run.
  4. Your circuit stage can well-approximated with the use of spreadsheet programs like Excel and can be solved with a variety of calculator tools and programs.
  5. It's not even hard to write your own software to handle it. Just a few lines of code, to be honest.
  6. The above only scratches the surface of what's available today without much difficulty.

For all practical purposes, there is no need for more than readily available approximations. Perhaps, as at least one part of the sweeping reasons why, most people never find a reason to develop their mathematical skills sufficiently to address your motivation here.

Regardless, all of the above does absolutely nothing whatever in answer to your motivation to avoid approximations. And I applaud your desire here.

Hence, the following analytic treatment motivated by your request and offered for your consideration and interest.


I'll call your collector resistor \$R_B\$ for the purposes of analyzing \$Q_3\$, since it appears in its base circuit. I'll call the other two resistors for \$Q_3\$, \$R_E\$ and \$R_C\$. Your positive source voltage will just be \$V\$ for now. Your value of \$I_{c_2}\$ is taken as an input. Then using the standard nodal approach, you get:

$$\frac{V_{B_3}}{R_B}+I_{C_2}+I_{E_3}\cdot\left(1-\alpha_3\right)=\frac{V}{R_B}$$

This leaves us with two unknowns, \$V_{B_3}\$ and \$I_{E_3}\$. But:

$$I_{E_3}=\frac{V_{B_3}-V_{BE_3}}{R_E}$$

So,

$$I_{C_2}+\frac{V_{B_3}}{R_E}\cdot\left(1-\alpha_3\right)-\frac{V_{BE_3}}{R_E}\cdot\left(1-\alpha_3\right)=\frac{V-V_{B_3}}{R_B}$$

Or,

$$V_{B_3}=\frac{V\: R_E+V_{BE_3}\: R_B\left(1-\alpha_3\right)-I_{C_2}\:R_B\:R_E}{R_E+R_B\:\left(1-\alpha_3\right)}$$

Which doesn't help yet, because now we have a different unknown and that means we still have two unknowns.


However, we do know something else:

$$V_{BE_3}\approx V_T\cdot\operatorname{ln}\left(\frac{I_{C_3}}{I_{SAT_3}}\right)=V_T\cdot\operatorname{ln}\left(\frac{\alpha_3\cdot I_{E_3}}{I_{SAT_3}}\right)$$

This finally provides us with a segue:

$$\begin{align*} V_{BE_3}&=V_T\cdot\operatorname{ln}\left(\frac{\alpha_3\cdot I_{E_3}}{I_{SAT_3}}\right)\\\\ &=V_T\cdot\operatorname{ln}\left(\frac{\alpha_3\cdot \frac{V_{B_3}-V_{BE_3}}{R_E}}{I_{SAT_3}}\right)\\\\ &=V_T\cdot\operatorname{ln}\left(\frac{\alpha_3\cdot \left(V_{B_3}-V_{BE_3}\right)}{R_E\cdot I_{SAT_3}}\right) \end{align*}$$


Now we have two equations in two unknowns:

$$\begin{align*} V_{B_3}&=\frac{V\: R_E+V_{BE_3}\: R_B\left(1-\alpha_3\right)-I_{C_2}\:R_B\:R_E}{R_E+R_B\:\left(1-\alpha_3\right)}\\\\ V_{BE_3}&=V_T\:\operatorname{ln}\left(\frac{\alpha_3\: \left(V_{B_3}-V_{BE_3}\right)}{R_E\: I_{SAT_3}}\right) \end{align*}$$

Let's expose the above two equations in a somewhat simpler form by making the following assignments:

$$\begin{align*} A&=\frac{V\: R_E-I_{C_2}\:R_B\:R_E}{R_E+R_B\:\left(1-\alpha_3\right)}\\\\ B&=\frac{R_B\left(1-\alpha_3\right)}{R_E+R_B\:\left(1-\alpha_3\right)}\\\\ C&=\frac{\alpha_3}{R_E\: I_{SAT_3}} \end{align*}$$

Then we can re-express the pair of simultaneous equations and then solve by substitution:

$$\begin{align*} V_{B_3}&=A+B\:V_{BE_3}\\\\ V_{BE_3}&=V_T\:\operatorname{ln}\left[C\:\left(V_{B_3}-V_{BE_3}\right)\right]\\\\ &\therefore \\\\ V_{BE_3}&=V_T\:\operatorname{ln}\left[C\:\left(A-\left(1-B\right)\:V_{BE_3}\right)\right],\textrm{ or,}\\\\ \left[A\: C\right]-\left[C\:\left(1-B\right)\right]\:V_{BE_3}&=e^\frac{V_{BE_3}}{V_T} \end{align*}$$


The above equation is solvable for \$V_{BE_3}\$!

$$\begin{align*} V_{BE_3}&=V - R_B\:I_{C_2} - V_T\: \operatorname{LambertW}{\left\{\frac{I_{SAT_3}\left[R_E+R_B\left(1-\alpha_3\right)\right]}{V_T\:\alpha_3} \cdot e^{\frac{V - R_B\:I_{C_2}}{V_T}}\right\}} \end{align*}$$

Or, in what I think is a little more meaningful form:

$$\begin{align*} V_{BE_3}&=\left[V - R_B\:I_{C_2}\right] - V_T\: \operatorname{LambertW}{\left\{\frac{I_{SAT_3}\left[R_B+\left(\beta_3+1\right)R_E\right]}{\beta_3\:V_T} \cdot e^{\frac{V - R_B\:I_{C_2}}{V_T}}\right\}} \end{align*}$$

Here, you can see the nominal voltage (before correction for \$Q_3\$'s base current) as the term on the left (as well as the numerator of the exponential power on the right factor within the Lambert W function), with the voltage correction term then on the right. Also, within the Lambert W function you can see the factor's numerator translates \$R_E\$ first to the base before combining it with \$R_B\$ and then translates that summed resistance over to the collector with the use of \$\beta_3\$ in the denominator.

Once \$V_{BE_3}\$ is known, you can now also solve for \$V_{B_3}\$, too. And from those, the rest all just falls out trivially.


The product-log function (Lambert W) is a very, very powerful function that isn't often taught in the undergrad mathematics required for an EE degree. It's defined this way. If you have something of the form:

$$u\:e^u=z$$

Then solving for \$u\$ gives:

$$u=\operatorname{LambertW}\left\{z\right\}$$

There remains the small problem of arranging things into the above form. Let's take our problem from above and work through the details, knowing the form we will require. I'll take the steps slowly:

$$\begin{align*} \left[A\: C\right]-\left[C\:\left(1-B\right)\right]\:V_{BE_3}&=e^\frac{V_{BE_3}}{V_T}\\\\ \big(\left[A\: C\right]-\left[C\:\left(1-B\right)\right]\:V_{BE_3}\big)\:e^\frac{-V_{BE_3}}{V_T}&=1\\\\ \big(A-\left[1-B\right]\:V_{BE_3}\big)\:e^\frac{-V_{BE_3}}{V_T}&=\frac{1}{C}\\\\ \big(\frac{A}{1-B}-V_{BE_3}\big)\:e^\frac{-V_{BE_3}}{V_T}&=\frac{1}{C\:\left(1-B\right)}\\\\ \big(\frac{A}{V_T\:\left(1-B\right)}-\frac{V_{BE_3}}{V_T}\big)\:e^\frac{-V_{BE_3}}{V_T}&=\frac{1}{V_T\:C\:\left(1-B\right)}\\\\ \big(\frac{A}{V_T\:\left(1-B\right)}-\frac{V_{BE_3}}{V_T}\big)\:e^\frac{-V_{BE_3}}{V_T}\:e^{\frac{A}{V_T\:\left(1-B\right)}}&=\frac{1}{V_T\:C\:\left(1-B\right)}\:e^{\frac{A}{V_T\:\left(1-B\right)}}\\\\ \bigg[\frac{A}{V_T\:\left(1-B\right)}-\frac{V_{BE_3}}{V_T}\bigg]\:e^{\left[\frac{A}{V_T\:\left(1-B\right)}-\frac{V_{BE_3}}{V_T}\right]}&=\frac{1}{V_T\:C\:\left(1-B\right)}\:e^{\frac{A}{V_T\:\left(1-B\right)}} \end{align*}$$

At this point, it is easy to see that:

$$\begin{align*} u &=\frac{A}{V_T\:\left(1-B\right)}-\frac{V_{BE_3}}{V_T}\\\\ z &=\frac{1}{V_T\:C\:\left(1-B\right)}\:e^{\frac{A}{V_T\:\left(1-B\right)}}\\\\ \therefore&\\\\ u &= \operatorname{LambertW}\left\{z\right\}\\\\ \frac{A}{V_T\:\left(1-B\right)}-\frac{V_{BE_3}}{V_T} &= \operatorname{LambertW}\left\{\frac{1}{V_T\:C\:\left(1-B\right)}\:e^{\frac{A}{V_T\:\left(1-B\right)}}\right\}\\\\ \frac{A}{1-B}-V_{BE_3} &= V_T\:\operatorname{LambertW}\left\{\frac{1}{V_T\:C\:\left(1-B\right)}\:e^{\frac{A}{V_T\:\left(1-B\right)}}\right\}\\\\ V_{BE_3} &=\frac{A}{1-B}- V_T\:\operatorname{LambertW}\left\{\frac{1}{V_T\:C\:\left(1-B\right)}\:e^{\frac{A}{V_T\:\left(1-B\right)}}\right\} \end{align*}$$

And from here you should be able to make all the right substitutions from the assignments I made earlier and arrive at the same answer I gave.


This kind of solution process appears quite frequently in BJT equations, where this \$C\$ is usually a function of \$V_T\$ (often just \$V_T\$.)

Many emergent phenomena, for example those that develop as a result of the statistics of the interactions of large numbers of particles (which is pretty much everything), are inherently of these kinds of mathematical relationships. Also, any series of discrete measurements (such as taking a snapshot of any process variable with an ADC) also will have this kind of relationship arrive as a direct result of the discretization process itself. (See the shift therem [which also has a similar-looking Laplace shift theorem.])

So it also appears frequently in many other places in nature. Not just in electronics.

It has very wide applications and can be used to: (a) solve differential equations (see "Using the Lambert W to express a solution of a differential equation"); and, (b) solve delay (and astrologer) differential equations; and, (c) help solve generating functions for Poisson events (and by implication also many recurrence problems); and so on.

It may be worth reading this PDF, "On the Lambert W Function", by R M Corless, et al., (1993). (Notably including D Knuth.)


I hope you find the above useful.

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  • \$\begingroup\$ Sir, if I perform the small signal equivalent circuit analysis of the above circuit, then to find \$g_m\$, I need the DC current (Quiescent current) \$ I_{CQ1}\$ and \$ I_{CQ2}\$. But in the differential amplifier stage, the AC input voltages are not connected via capacitors....So while performing DC analysis ,we cannot open circuit the AC voltage sources. Then how we are to find the quiescent collector currents?...Also even if the AC voltage sources are connected via capacitor, then while DC analysis, the base would be floating....! \$\endgroup\$ – Soumee Aug 6 '17 at 18:27
  • \$\begingroup\$ @Soumee Assume both inputs are at the same DC voltage, somewhere in between your (-) and (+) rails. If they are both the same, and since their emitters are the same, then the current will split equally (assuming identical BJT params.) It doesn't matter if their bases are both a little higher up, or a little lower down (common mode movement.) They can float in that sense while still doing their job. I want you to take a look at the MC1496's schematic -- find a datasheet or apnote on it -- look at pins 1 and 4, in particular (and assume shorting out pins 2 and 3.) \$\endgroup\$ – jonk Aug 6 '17 at 19:17
  • \$\begingroup\$ @Soumee But when there is a difference between them, then the current will split differently. \$\endgroup\$ – jonk Aug 6 '17 at 19:18
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The current through RE is $$I(R_E) =\frac{V(v_{o2}) - V_{BE}}{R_E}$$

The current into the base of Q3 is $$\frac{I(R_E)}{\beta+1}.$$

But this still depends on \$V(v_{o2})\$, and \$V_{BE}\$ depends nonlinearly on the Q3 base current. So you have a set of transcendental equations for \$V(v_{o2})\$ and \$I_B(Q_3)\$.

For hand calculation, the base current is near enough to 0 to not affect the voltage across RC noticeably.

If you want or need a more exact result, plug the circuit into a simulator (but realize that the errors due to not knowing the device parameters exactly are more than the error due to neglecting Q3's base current).

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  • \$\begingroup\$ Looking over all the equation work, I think the op wants a closed solution. That's only achievable with the Lambert-W function. A single iteration with a much simpler equation would get probably within 5 digits, which should be enough for anyone. But a closed solution requires the product-log function and there's no way around it. Then the question is more about the application of specific equations from simplified theory, than about "engineering," per se. \$\endgroup\$ – jonk Aug 4 '17 at 22:03
  • \$\begingroup\$ @jonk, Yes I agree that OP wants a closed solution. The main point of my answer is that 1. That's a PITA to get to by hand calculation and 2. The accuracy won't be better for any real circuit. \$\endgroup\$ – The Photon Aug 4 '17 at 22:05
  • \$\begingroup\$ Of course not. There is NO Early Effect in anything the OP did already. And there isn't even the simplest Ebers-Moll model here. It's just a fragment of it. But it looks like the OP wants to exercise math skills. \$\endgroup\$ – jonk Aug 4 '17 at 22:06
  • \$\begingroup\$ Oh, cripes! Look at the schematic! The voltage sources are polarized wrong! \$\endgroup\$ – jonk Aug 4 '17 at 22:09
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    \$\begingroup\$ @Soumee, so plug it in to SPICE. The the computer will both figure out what the equation should be and solve it for you in one go. But again the answer won't be any more accurate than the one you get assuming 0 base current in Q3. \$\endgroup\$ – The Photon Aug 4 '17 at 22:27
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I want to determine the currents in each of the branches.

wow, I didn't realize that simple things can be this complex.

My take:

1) Q4/Q5 is a current mirror. Q4's base sits at -12+0.7v. So the current through Q4 is (+12 - (-12+0.7v)) / 12K = 2ma.

2) that means the current through is 1ma, assuming loading from Q3 isn't much -> we will check that at the end.

3) that means Q2's collector sits at 12v-8k*1ma = 4v.

4) So the current through Q3 is (4v-0.7v)/3.3K = 1ma.

5) that means Q3's base is at 12v-4K*1ma = 8v.

6) base current for Q3 is 1ma / 200, too small vs. Q1/2's collector current -> loading is minimum.

7) done.

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