Hello all. We've been given this circuit and asked to derive the transfer function using Mesh analysis as well as Nodal Analysis. I'm just curious if there's any considerations/alterations I should make before I start to hammer out my mesh/nodal equations.

Here's the circuit that we've been given. As you can see, we should be able to obtain 1 mesh current by inspection, namely \(\beta i_{bl}\). After doing so I'm still left with 3 mesh currents to solve for. (Not the nicest scenario, but doable)

Is there a way I can further simplify this? I was thinking of doing a parallel combination of the impedances of the capicator and the resistor to eliminate that 1 mesh, but I'm curious to see how we would simplify this anymore than that.

Sort of the same scenarion with nodal, we have one node voltage by inspection. (Namely, the node connected to the positive terminal of the voltage source) Is there a way we can make our lives easier when applying nodal analysis to this circuit to obtain the transfer function.

Remember our main goal from all this is to obtain the Transfer Function by using these two methods.

I'd love to hear what you guys think.

Thanks again!

 

I would think that one thing you could do would be to remove Rb1 & Rb2 from the analysis as both are simply in parallel with the the source voltage Vin. If the source included a series resistance you could not do this.

 

I would think that one thing you could do would be to remove Rb1 & Rb2 from the analysis as both are simply in parallel with the the source voltage Vin. If the source included a series resistance you could not do this.

So I can remove these without any consequence?

I'm trying to picture what my resulting circuit would look like.

Is it something like this?

Is this correct?

EDIT: There should be another ground at the bottom of my circuit as well.

After thinking about this some more, I've got another question. After I've removed the 2 resistors, can I do a parallel combination of the two impeadances between Re2 and Cb? This would further simplify the circuit giving me two loops, which can easily be solved with either Mesh or Nodal analysis. Is this doable?

 

Which would be 'simplified' as this .....

 

Which would be 'simplified' as this .....

As mentioned in the EDIT in my post above, if I choose to do so, can I combine the parallel combination of impeadances of the capciator and the resistor in parallel?

Or will that effect my transfer function?

Basically what I'm trying to understand is, what alterations can I NOT make to the circuit. (because it would result in an incorrect transfer function)

EDIT: I've done work some work if I can infact make that parallel combination, followed by another series combination. (i.e. \(z = (R_{e2} // \frac{1}{sC_{b}}) + R_{e1})\) )

See figures attached.

If I can't do this I'd love to know why. Thanks again!

 

Yes you can combine the parallel Re2 & Cb and represent as a single impedance - for the purposes of determining the TF.

 

Are you supposed to derive a transfer function in symbolic form, or perhaps a numeric value at high frequencies where Cb looks like a short?

 

u could change vin and rb2 to a current src in series vd a resistance.. it'll help in mesh analysis.
plus u could use laplace transforms on the impedances and parallel, series combine them

 

Are you supposed to derive a transfer function in symbolic form, or perhaps a numeric value at high frequencies where Cb looks like a short?

I'm assuming its symbolic form for this portion of the questions, because it doesn't give us any values to determine any voltages. Without these voltages, how are we supposed to obtain the currents?

In another question it asks to find \(V_{o}(t)\) if \(V_{in} = 0.001V u(t)\). Here we must determine a numeric value I'm assuming.

 

The reason I asked is because your original schematic shows values for all the components, as well as symbolic designators.

I'm assuming its symbolic form for this portion of the questions, because it doesn't give us any values to determine any voltages. Without these voltages, how are we supposed to obtain the currents?

The transfer function is a RATIO of voltages (Vo/Vi), so you don't need to know what the input voltage is.

You don't need the currents in the final answer; they are only involved in intermediate steps in finding the transfer function. The currents shouldn't even appear in the transfer function.

 

The reason I asked is because your original schematic shows values for all the components, as well as symbolic designators.

I'm assuming its symbolic form for this portion of the questions, because it doesn't give us any values to determine any voltages. Without these voltages, how are we supposed to obtain the currents?

The transfer function is a RATIO of voltages (Vo/Vi), so you don't need to know what the input voltage is.

You don't need the currents in the final answer; they are only involved in intermediate steps in finding the transfer function. The currents shouldn't even appear in the transfer function.

So did I solve for the transfer function incorrectly then? (See figures attached)

EDIT: I've got 2 elements in my transfer function that it shouldn't rquire (i.e. \(i_{bl}, \quad V_{in}\)) which I'm not sure how to get rid of. (According to the previous post(s), I should be able to obtain it without these values) Did I do something wrong?

 

Your stated TF includes current terms - which is incorrect. Recall the Electrician's last post.

The TF should only include the designated circuit values - Rc, β, Re1, etc.

 

Your stated TF includes current terms - which is incorrect. Recall the Electrician's last post.

The TF should only include the designated circuit values - Rc, β, Re1, etc.

I'm trying to figure out what I did wrong...

Is the circuit attached to this post correct? (See figure)

Should I be able to derive the transfer function from this representation of the circuit?

What did I do wrong previously?

 

Your 'Z' value looks correct - perhaps its simply a case of completing the necessary algebraic simplifications/cancellations to come up with a result that doesn't include the current terms.

 

Your 'Z' value looks correct - perhaps its simply a case of completing the necessary algebraic simplifications/cancellations to come up with a result that doesn't include the current terms.

I thought this might have been the case as well, but when I look at my mesh equation for \(i_{1}\), how am I ever going to solve for it without containing either \(V_{in}\) or \(i_{bl}\)? Either one is going to cause problems.

Can I really work for this simplified version of the circuit or will this cause my transfer function to contain currents and voltages instead of solely containing designated values?

EDIT: If that was the case I'd be working from the expression I have given in previous posts,\(\frac{-R_{c}\beta i_{bl}}{(z+r_{p1})i_{1} - zi_{2}}\),

Now I could replace \( i_{1} \) with what I obtain from my Mesh equation for that loop but then it will contain a voltage source \(V_{in}\), so no matter what I do, that term is going to cause problems. (Unless it turns out that \(i_{1}\) is proportional to some factor times \(i_{bl}\) and the two somehow cancel eachother...?

Any ideas?

 

FYI I have

\(\frac{V_o}{V_{in}}=-{\frac{\beta R_c}{\[r_{p}1+(1+\beta)Z\]}}\)

 

FYI I have

\(\frac{V_o}{V_{in}}=-{\frac{\beta R_c}{\[r_{p}1+(1+\beta)Z\]}}\)

Okay, did you obtain that from the same simplified circuit I had posted?

What did you do? Note that we are required to solve it using both mesh analysis and nodal analysis.

EDIT: Working with your expression and my expression. I might draw a conclusion soon!

AHA!

\(i_{1} = i_{bl}\),

\(i_{1} - i_{2} = (1 + \beta)i_{bl}\)

Then my expression converts into yours after a little bit of simplification, but does this count as solving using "mesh analysis"? We almost made it too easy to solve it with mesh... ?

 

Here are some pointers ....

Call the common center node Vb

Then

\(V_b=V_{in}-i_{b1}r_{p1}\)

Also

\(V_b=(1+\beta)i_{b1}Z\)

Eliminate Vb from these two equations to give Vin purely in terms of ib1.

Also

\(V_o=-\beta i_{b1}Rc\)

So you would then have Vo and Vin purely in terms of ib1 which vanishes in the ratio Vo/Vin

 

Here are some pointers ....

Call the common center node Vb

Then

\(V_b=V_{in}-i_{b1}r_{p1}\)

Also

\(V_b=(1+\beta)i_{b1}Z\)

Eliminate Vb from these two equations to give Vin purely in terms of ib1.

Also

\(V_o=-\beta i_{b1}Rc\)

So you would then have Vo and Vin purely in terms of ib1 which vanishes in the ratio Vo/Vin

I obtained the same results in the EDIT of my post above, but are these conclusions still valid? Because after all, the question asks me to solve it specifically using mesh analysis, and then specifically using nodal analysis.

We simply simplified the circuit and solved it using KCL and KVL really... But that's all mesh analysis is! KCL & KVL!

Should I be okay leaving my work like so?

 

I obtained the same results in the EDIT of my post above, but are these conclusions still valid? Because after all, the question asks me to solve it specifically using mesh analysis, and then specifically using nodal analysis.

We simply simplified the circuit and solved it using KCL and KVL really... But that's all mesh analysis is! KCL & KVL!

Should I be okay leaving my work like so?

Your original post stated that you were required to find the transfer function using both methods. To me the process of using either mesh or nodal analysis is an intermediate (and ultimately transparent) step in finding the required solution. In this regard, I would anticipate having exactly the same result for the relationship Vo/Vin irrespective of the method used. Therefore (by implication) the derived relationship would make no reference to the nodal or mesh parameters used during the intermediate solution process. To include any reference to those parameters in the relationship would mean that the solution (albeit correct thus far) is still incomplete.

This point was essentially made by the Electrician in one of his earlier posts.

Clearly you need to be able to demonstrate to the examiner that you are familiar with the nodal and mesh techniques but you also need to distill the technique into something of use at the end - which in this case is the Transfer Function.

I'm not sure if you are aware that this particular problem is closely related to BJT common emitter amplifier analysis.