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schematic

simulate this circuit – Schematic created using CircuitLab

\$V_{cc}=20V\\V_{BE1}=V_{BE2}=0.5V\\V_{BE3}=0.6V\\\beta_1=\beta_2=100\\\beta_3=50\$

I need to find V in this circuit. I tried to use the path for KVL with the least unknowns but a lot of times I end up with linear dependent equations that leads to nowhere. It's been 3 hours since I'm trying and I didn't get anything close to a result on this. I think my major issue is (1) figuring out what path to take on KVL's equations and (2) figuring out when the KVL equation will be linear dependent of another one. What's a good approach to analyze circuits like this?

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Normally, a circuit like this really would take careful consideration. But less so, provided some of the givens. I'm a little concerned about your value for \$R_4\$, only because the schematic editor uses that value as a default and I'm not sure you intended that value. \$R_3\$ seems a bit odd, as well. But let's go with it:

schematic

simulate this circuit – Schematic created using CircuitLab

I think you can see that pretty much everything turns on the value of \$V_x\$. So let's just do nodal analysis and shoot for its value:

$$\begin{align*} \frac{V_x}{R_2} + \frac{V_x}{R_3} + \frac{V_x}{R_4} + I_{B_1} &= \frac{600\:\textrm{mV}}{R_2} + \frac{20\:\textrm{V}}{R_3} + \frac{V_x-1\:\textrm{V}}{R_4} \\ \\ V_x\cdot\left(\frac{1}{R_2} + \frac{1}{R_3}\right) + I_{B_1} &= \frac{600\:\textrm{mV}}{R_2} + \frac{20\:\textrm{V}}{R_3} - \frac{1\:\textrm{V}}{R_4} \\ \\ V_x &\approx 12\:\textrm{V} - 755\cdot I_{B_1} \end{align*}$$

Hmm. We don't know \$I_{B_1}\$. So let's re-group a bit.

You know that \$Q_3\$ has \$\beta=50\$ and that it's collector current must include \$I_{R_4}\$ and the base current of \$Q_2\$. This must be more than \$10\:\textrm{mA}\$. It follows that \$I_{B_3}\ge 200\:\mu\textrm{A}\$. So you know that \$I_{R_2}\ge 260\:\mu\textrm{A}\$ and therefore also that \$V_x\ge 9.18\:\textrm{V}\$.

You also know that \$R_3\$ must include \$I_{R_2}\$, \$I_{R_4}\$, and \$Q_1\$'s base current. So \$I_{R_3}\ge 10.26\:\textrm{mA}\$ and therefore \$V_x\le 12.07\:\textrm{V}\$.

So we now can at least say this:

$$\begin{align*} 9.18\:\textrm{V} \le \left(V_x \approx 12\:\textrm{V} - 755\cdot I_x\right) \le 12.07\:\textrm{V} \end{align*}$$


Thanks for asking questions about this. Here's my additions. I've added the current \$I_{B_1}=I_x\$ to the schematic. (This must be returned to the collector of \$Q_3\$ via the base of \$Q_2\$, since both \$Q_1\$ and \$Q_2\$ share the same \$\beta=100\$ value.)

So we have the above equation. But what is missing is the value for \$I_x\$.

\$I_x\$ adds to the \$10\:\textrm{mA}\$ from \$R_4\$, so that the collector current of \$Q_3\$ increases by that amount. This means that the base current for \$Q_3\$ increases by \$\tfrac{1}{50}\$ of that. So we can set this up:

$$\begin{align*} V_x = 12 - 755\cdot I_x &= 600\:\textrm{mV}+\left(60\:\mu\textrm{A}+\frac{10\:\textrm{mA}}{50}+\frac{I_x}{50}\right)\cdot R_2 \\ \\ 12 - 755\cdot I_x &= 600\:\textrm{mV}+\left(260\:\mu\textrm{A}+\frac{I_x}{50}\right)\cdot R_2 \\ \\ I_x &\approx 2\:\textrm{mA} \end{align*}$$

From this, we can now estimate \$V_x\approx 10.5\:\textrm{V}\$ and then that \$V\approx 10\:\textrm{V}\$.

Thanks for asking additional questions. It helped me to add my own additional thinking to this and to provide what I think is now a more complete answer to your question.

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  • \$\begingroup\$ Jonk, thank you so much. Interesting the way you set the nodal equations, going to consider it for sure. I found to be wierd that I couldn't get the values of all variables by Kirchhoff, but I'm a rookie so I thought it was my mistake. Are you saying that there's no way to find all currents and voltages in all transistors without eliminating the basis currents, or with some sort of approximation? I was holding the idea that as in a pure resistive circuit, if you know the values of the resistances and sources you could always find the value of currents and voltages. Thank you again sir. \$\endgroup\$ – João Pedro Nov 18 '16 at 0:11
  • \$\begingroup\$ @JoãoPedro No, the collector currents for Q1 and Q2 are model dependent and you don't have the full models laid out there. You can see this problem in Tony's write-up as well, where he just lays out a pair of diodes for you with 1 V across them. How much current in a pair of diodes like that?? It's model-dependent and you don't have the model info to know. But given the low voltage there, I'd say the currents are small and so the base currents are negligible and can be ignored. \$\endgroup\$ – jonk Nov 18 '16 at 0:15
  • \$\begingroup\$ I don't know if what I'm about to say is related to the model you talk about. But the only thing we learned so far, is that Vbe operates with approximate 0.7V, the value of Vce and Vbc we analyze depends on the given data, we can't correlate those two with any other parameter. Other than that, the fact that the voltages Vbe, Vbc and Vce sum to zero, and Ie=Ib(1+Beta) considering Ic=Beta*Ib. Are those equations sufficient for finding the currents and voltages in the circuit? \$\endgroup\$ – João Pedro Nov 18 '16 at 0:36
  • \$\begingroup\$ @JoãoPedro Let me reconsider what I wrote from a different perspective in the "re-group" section. I'll add a note or two. \$\endgroup\$ – jonk Nov 18 '16 at 0:46
  • \$\begingroup\$ @JoãoPedro Done. Please check the results! \$\endgroup\$ – jonk Nov 18 '16 at 1:25
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schematic

simulate this circuit – Schematic created using CircuitLab

This is a simple way with 2 steps

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  • \$\begingroup\$ I didn't know I could ignore the base currents, that makes sense. I'll look carefully to your solution right now to understand it! \$\endgroup\$ – João Pedro Nov 17 '16 at 21:56
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    \$\begingroup\$ check last edit \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 17 '16 at 21:57
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You could start by ignoring the base currents of T1/T2 and see whether they matter.

First find the voltage Vx at the junction of the three resistors (which is V+0.5V assuming a load to ground). You write the equation for Ib3 as a function of Vx by inspection.

Ic3 = \$\beta_3\$ * Ib3, which gives you Ic3 as a function of Vx.

Now you can find Vx as a function of IC3 using KVL, which gives you two equations in two unknowns, then solve for Vx/Ic. If Ic <= 10mA then you don't need to worry about base currents.

If it is more than 10mA then you just replace the transistor bases T1/T2 with a 1V voltage source and repeat (or do this in reverse, depending on what you think is likely).

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  • \$\begingroup\$ Thanks Spehro. What you mean by "base currents T2/T2"? Do you mean T1/T2? The reason we ignore the base currents of T1/T2 is because \$\beta\$ is much larger than 1, correct? I'll try your solution as soon as I can, thanks again. \$\endgroup\$ – João Pedro Nov 17 '16 at 21:58
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    \$\begingroup\$ Yes, T1/T2 (typo). If the bases are not conducting (more than 1V across the 100R resistor then you may be able to ignore them. If they are large enough to matter a lot, as a practical matter, the transistors will destroy themselves. \$\endgroup\$ – Spehro Pefhany Nov 17 '16 at 22:02

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