2
\$\begingroup\$

I'm playing around with BJT amplifiers and have some questions regarding the gain rolloff you see at higher frequencies, I'm using the 2N2222 for my tests and my simulations and measured results match up pretty well, example below:

enter image description here

So at 3MHz or about I see my rolloff start affecting my gain, which is described in another answer, at around 18MHz the gain drops to 0dB. My questions are:

  • Is this rolloff caused by the Miller effect?
  • If not caused by the Miller effect what is the cause?
  • Can you compensate for the gain rolloff to some extent or do you need to choose a faster BJT?. I've played around with the input impedance which I thought would help but this made little difference.
\$\endgroup\$
2
\$\begingroup\$

You've got two lots of miller capacitance to factor in here. The output stage miller cap is shunting the collector resistor thus reducing the gain at higher frequencies. Also the first transistor's miller capacitance is acting as a shunt to the collector resistor too.

You've got the tools right there - try seeing what the performance is with the 2nd transistor removed - does this improve the frequency response (maybe it will double it to about 6 MHz). Plenty of things you can generally do with a simulator that allows you to focus on a problem.

It's not worth trying to compensate for the gain roll-off - better to choose an improved BJT.

EDIT - just so there's no confusion. I'm saying "no" to miller effect but "yes" to miller capacitance affecting the high frequencies by shunting the collector resistor. AFAIK, miller effect is due to the miller capacitor causing feedback to the base but, because the base driving impedance is low (probably close to zero in the OP's simulation), the miller effect is minimal.

\$\endgroup\$
2
  • \$\begingroup\$ It did help somewhat but only made the gain loss less, the general shape and location of the output remained the same. Any other ways to improve this? \$\endgroup\$ – s3c May 23 '14 at 10:24
  • 1
    \$\begingroup\$ try simulating a different transistor or, go into the data for the transistor and reduce the parasitic capacitances. Maybe also trying increasing the supply voltage to see what happens - a bit more collector current always helps! And do make sure your input source impedance is low <100 ohms \$\endgroup\$ – Andy aka May 23 '14 at 10:58
2
\$\begingroup\$

You're not actually getting any significant Miller effect showing up with this circuit.

The output stage has no voltage gain, so there's just regular old capacitance, and the input stage is driven with a zero-impedance source in series with a 1\$\mu\$F capacitor (almost a short too). Even with a more realistic 50\$\Omega\$ source, the modest gain of -2.4 and ~5.5pF \$C_{cb}\$ doesn't lead to much roll off due to Mr. Miller at those frequencies.

The output capacitance of 5.5pF (times two because there is a \$C_{cb}\$ load from the emitter follower transistor) is significant in comparison to the 2K collector load above a few MHz. If you halve the collector and emitter resistors, you should see about a doubling in the -3dB cutoff frequency.

Alternatively, use a pair of RF transistors and you can probably get 5:1 improvement in cutoff frequency at the same current drain (or make a nice 500MHz oscillator, but you won't see that in simulation).

\$\endgroup\$
2
  • \$\begingroup\$ How did you get to the ~5.5pF Ccb? \$\endgroup\$ – s3c May 23 '14 at 12:49
  • \$\begingroup\$ From a graph in a datasheet and the voltage across the CB junction. I may not have used the datasheet you linked. \$\endgroup\$ – Spehro Pefhany May 23 '14 at 12:59
0
\$\begingroup\$

I addition to the capacitive effect as mentioned before, there is another effect which causes a reduction in current gain for high frequenvcies: The limited mobility of the charged carriers which form the collector current.

This mobility aspect is closely related to the inertial mass each charged carrier has.

\$\endgroup\$
2
  • \$\begingroup\$ Do you know what this is called? \$\endgroup\$ – s3c May 23 '14 at 10:19
  • \$\begingroup\$ Instead of "mobility" it is perhaps better to say: Transit time (through the E-C path). Such transit time effects lead to a delay which - for each frequency - is equivalent to a phase shift. Hence, you can compare such an effect with a lowpass characteristic (in addition to other capactice effects mentioned already). \$\endgroup\$ – LvW May 23 '14 at 12:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.