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I am making an analysis for this circuit and I want to make a KVL at mesh 1 (in red), but should I take the current source as it is or does it have another voltage equivalent since I am working with voltages.

This is the equation I got: KVL @ mesh 1: (RB x IB) + (VBE) + I + VEE = 0

Note that I want to find the current IB and I am solving using expressions and not values.

Thank you in advance.

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This what I am asked to do. I guess it is necessary to find the expressions in DC mode so I use these in the AC mode.

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    \$\begingroup\$ Do you know the relation between the emitter current and the base current in a BJT? \$\endgroup\$ – G36 May 3 at 8:47
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    \$\begingroup\$ Because you don't need to do a KVL to solve for an IB current if you know the emitter current \$\endgroup\$ – G36 May 3 at 8:49
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    \$\begingroup\$ For a DC analysis, you need to include the current source I_E = I and IB = I_E/(beta +1). But for an AC small-signal analysis the current source is shorted via CE capacitor thus, for AC signal emitter is shorted to GND via CE capacitor, so you do not have to worry about the current source. \$\endgroup\$ – G36 May 3 at 9:22
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    \$\begingroup\$ If you want the small-signal input resistance of this circuit, you have to replace the transistor symbol by its hybrid-\$\pi\$ model showing \$r_\pi\$ and the collector current source \$\beta i_b\$. \$\endgroup\$ – Verbal Kint May 3 at 12:30
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    \$\begingroup\$ You can't have the I term in a KVL equation. All of the terms must be voltages. \$\endgroup\$ – Elliot Alderson May 3 at 13:28
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So many good points made in the comments, P_M. And there are so many mental concepts you need to gather up, I fear. Let's look at the schematic for a moment. (By the way, you should have used the schematic editor available here. Your writing is nice, but the schematic editor is nicer.)

schematic

simulate this circuit – Schematic created using CircuitLab

For problems expressed in this form, as I'm sure you already know (but for the benefit of those who do not), the circuit is inherently non-linear because of the BJT. When powered on, the circuit first must establish for itself a "DC operating point."

(This is like releasing a rock to roll down an embankment, which takes some time but eventually the rock stops somewhere. You need to know where the rock finally rests at before you can work out what happens when you go down there and try to roll it around a bit, this way or that.)

I suspect your "KVL attempt" is a matter of trying to work that out. Finding the DC operating point is just a matter of treating capacitors as open circuits (inductors as wires) and seeing where that leads. Stripped away, you have:

schematic

simulate this circuit

Before I go on to the KVL error you made, you need to know that the dimensional units must always match up when you are making sums. As the phrase goes in English, "You can't mix apples and oranges." So if you are using KVL, each term must be a voltage. You cannot add a current to a voltage and get anything meaningful. You can convert a current to a voltage by multiplying in by a resistance, for example. But you cannot just add a current to a voltage. Doesn't work. But that's what you did. And you need to learn to apply dimensional analysis to any expression you develop to make sure that the dimensional units are consistent when adding or when comparing or setting equal to each other. It's important that you learn this, soon, and practice it continually through life.

(Sidebar: Even then, there are subtleties. For example, torque is Newton-meters. But this is also Joules. However, torque is never expressed in Joules. Part of the reason here is that torque is really Joules per radian, where a radian is itself unitless. Worse, torque is a vector. But the dot-product of a force through a distance is a scalar in Joules. So while the boiled-down dimensional units of torque may "look like" the dimensional units of work, they really aren't. [Torque is more like the work done if you rotate through one radian.] So dimensional analysis isn't a brain-dead application of learned, rote rules. You still do have to also think for yourself, as well.)

Okay. So your addition of \$I\$ is a mistake. Here's how you can repair it:

$$0\:\text{V} -I_\text{B}\cdot R_\text{V}-V_\text{BE}-V_I=V_\text{EE}$$

All I've done is substituted some unknown voltage across that current source (the voltage across it isn't necessarily \$0\:\text{V}\$, as it may be anything.) Now the dimensional units are, at least, appearing to be consistent.

Note that on the left I started with ground, or \$0\:\text{V}\$, less the voltage drop through \$R_\text{B}\$, less the base-emitter junction voltage, and less the voltage drop across the current sink, before arriving at what then must simply be \$V_\text{EE}\$.

Some problems still exist. Unless \$V_\text{BE}\$ is given, that's an unknown (and there are equations you can use to calculate it if it isn't specified.) But let's say that one is given. You still don't know what \$I_\text{B}\$ is. And without knowing the value of \$\beta\$ for the DC operating point (which may be device specific, ambient temperature specific, collector current specific, and so on), you really cannot work out that value. So, you'll either require \$\beta\$ or else you will need to solve this symbolically and substitute that in at a later time. But once you know \$\beta\$, you can work out the collector and base currents from the emitter current, which is known because \$I_1\$ establishes it.

This won't protect you, completely, because it is possible that the value of \$I_1\$, together with other circuit values when you have them available, may result in a saturated BJT situation. In that case, \$V_\text{CE}\$ acts a lot more like a voltage source than a current source and you cannot just apply some value of \$\beta\$, arbitrarily, and get the right base and collector currents, anymore. So you still need to think. But assuming the circuit can be treated validly as active mode, then you can make the usual calculations to get the base and collector currents.

  • G36 immediately asked you about \$\beta\$, because you need that if you assume the BJT is in active mode in order to compute the base current (and collector current.) And without the base current, you can't compute the voltage drop across \$R_\text{B}\$. And without that voltage drop and \$V_\text{BE}\$ you can't work out \$V_I\$, either.
  • Verbal Kint addressed finding the small signal input resistance (which you also appear to want) and this requires, again, knowing the DC operating point in order to work out the dynamic emitter resistance, \$r_e\$, which you'll need in order to get \$r_\pi\$. And you'll need that, together with \$R_\text{B}\$ and possibly other details, to work out the input resistance of the circuit (ignoring \$R_\text{SIG}\$.)
  • Elliot Anderson, of course, addresses your failure to handle the dimensional analysis part when working out the KVL equation.

So far, I've avoided dealing with the small signal analysis. For this, once you have the DC operating point and you have confirmed that the BJT is operating in active mode, you can then and only then proceed on to the AC analysis part of the problem. Here, there are many possible complexities (the emitter is AC-grounded, there is an Early Effect you may or may not need to deal with, etc.) But you need to get the DC operating point first. So I'll leave that other part alone, since you can't even develop a single mesh equation without tripping over a dimensional analysis problem.

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