Break frequencies for directly coupled transistors

Click For Summary
SUMMARY

The discussion focuses on determining the break frequency at the base of transistor Q2 in a circuit involving directly coupled transistors. The equation used is f(break) = 1/(2*pi*Ceq*Rth), where Ceq includes the Miller capacitances. The user initially misinterprets the Miller effect capacitance and its interaction with other capacitances, leading to confusion about whether to treat them separately or in parallel. Ultimately, the conclusion is that combining the Miller effect with parasitic capacitances yields similar transfer function results, demonstrating a common approximation in electronics.

PREREQUISITES
  • Understanding of transistor operation and configurations
  • Familiarity with Bode plots and frequency response analysis
  • Knowledge of the Miller effect in capacitance
  • Proficiency in using PSpice for circuit simulation
NEXT STEPS
  • Study the Miller effect in detail and its implications on circuit design
  • Learn how to create and interpret Bode plots for various circuits
  • Explore advanced PSpice techniques for analyzing frequency response
  • Research the impact of parasitic capacitances in transistor circuits
USEFUL FOR

Electrical engineers, circuit designers, students studying electronics, and anyone involved in analyzing transistor frequency response and gain characteristics.

programmer
Messages
13
Reaction score
0

Homework Statement


The problem asks to determine the break frequency at the base of Q2 (which is fine I have no problem with that), but the whole assignment never asks for the break frequency at the COLLECTOR of Q1, then later asks to produce a bode plot of the transfer function. See the image I've linked (pardon the sloppy paint job).

directcoupledtransistorsACmodel.jpg


Homework Equations


f(break) = 1/(2*pi*Ceq*Rth)


The Attempt at a Solution



Well I thought since the capacitor equivalent at collector1 (Cmiller(output) + Cce) wasn't actually in parallel (due to the non-bypassed emitter resistors) with the capacitor equivalent at base2 (Cmiller(input) + Cbe), I would have to treat them separately and produce two different break frequencies. Is this correct?
 
Physics news on Phys.org
still nothing?? Well, I've consulted with my colleague, and he told me that my interpretation of the miller-effect capacitance was incorrect. He told me that they always goes to ground (can anyone verify this??)...and he also told me to ignore the capacitances Cbe and Cce (ignore? never! mwhahaha)...

So looking at this NEW image:

directcoupledtransistorsACmodelpt2.jpg


You can see now that the "break frequency due to the base of Q2" probably means the two miller capacitances added together since now they are truly in parallel. BUT, to be 100% correct, wouldn't I now need a break frequency due to Cmiller(combined), Cce, AND Cbe(3 total)...not to mention the capacitors due to the rest of the circuit.

If I've confused anyone..I'm talking about break frequencies on Bode Plots of the gain vs. frequency response where the slope changes(20dB/decade...40dB/decade..so on), or the -3dB point on the actual transfer function (at least the first break frequency on the midband gain).

Can anyone help me understand this?
 
Well after extensive research (PSpice and a graphing calculator), I've concluded that adding the Miller effect to the parasitic capacitances (treating them as "parallel") gives nearly the same transfer function results as treating them separately. I guess it's just another one of those electronic approximations that are far too common.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
3
Views
3K