0
\$\begingroup\$

I was dimensioning the CE capacitor. To do that I was looking at some formulas seen in class. The formula (freq_C=1/(2*piReqC)) involves the knowledge of the loop gain (open loop gain*beta). It appears to be negative in my schematic. What does it mean? Is something wrong? In other exercises I've got a positive number: I though positive numbers were to be expected. This is the audio amplifier circuit I'm working on: enter image description here

-- update --

@Tony Stewart Sunnyskyguy EE75 I have upgraded the circuit but I haven't applied the 3:1 resistors value ratio in the Vbe multiplier since the quiescent current on the output resistors was several hundreds mA... The circuit is behaving better now though. Do you think that something is wrong? enter image description here

The fact that I have good phase margin (+90.8deg) and good gain margin (-37.8dB) (do you agree?) isn't enough to assure me stability? It looks like I still need miller compensation capacitor since I've got peaking. Why do i have peaking though? enter image description here

\$\endgroup\$
  • \$\begingroup\$ That -6 degree is reasonably close to zero degrees. If it were closer to 180 or to -180 degrees, then I'd call it an "inverting amplifier" instead of the current "non-inverting". Or are you thinking the minus sign of -5 dB magnitude is involved? \$\endgroup\$ – glen_geek Jan 7 at 0:07
  • \$\begingroup\$ Lucky, When I design something like this, I sit down and make choices and then make calculations. Then I simulate and usually find out that I was right. But sometimes I also learn something new. But I don't see any calculations here or even any thought processes behind them. Could you share your thinking about the above schematic? (One of the things in the schematic that really shocks me is your use of RF1 and RF2. It just stands out like a sore thumb as "not right.") \$\endgroup\$ – jonk Jan 7 at 4:46
  • 1
    \$\begingroup\$ @jonk This design has massive open loop gain and due to lack of CC biasing is easily saturated. Closed loop gain is extreme with almost 40dB out of 90 dB but has excessive input bias current and offset sensitivity. My suggestions will patch it but rather than a bandaid it needs better DC stability.. negative gain comes from saturation. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 7 at 6:46
  • 1
    \$\begingroup\$ @jonk That was his annotation to show how open loop gain was tested and failed. I explained why and how to fix it. He wanted to do that to see what RC breakpoint was needed to give adequate gain margin. But of course, it could have been done other ways with a closed cloop. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 7 at 8:00
  • 1
    \$\begingroup\$ Perhaps you can steer him in the right direction with a more stable DC bias method. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 7 at 8:07
1
\$\begingroup\$

Your output is saturated open loop due to offset between sensitivity to R5, R9 values and hFE assumptions. I can tell by your choice of R9 = 595 Ohms and Vout = 14.3V

  • Increase R5 to say 4k7 until you get in the linear output range and test Aol with 1uVpp input. You don't have to null it.
  • Also your 4 Vbe output bias circuit needs to be more than 3:1 R ratio to make 3+1=4. Vbe=0.6V @ 1mA and 0.65V @ 10mA so the crossover output bias should be not 1.4 AMPS.

  • let R3/R4 >=3:1 consider R3/R4 = 3.0.

    • With 0.6V across R4 you expect 3x this amount across R3.
  • you might expect Aol > 90dB

\$\endgroup\$
  • \$\begingroup\$ Thanks for the detailed answer. - What does "Aol" mean? - Can you please explain this statement "Also your 4 Vbe output bias circuit needs to be more than 3:1 R ratio to make 3+1=4. Vbe=0.6V @ 1mA and 0.65V @ 10mA so the crossover output bias should be not 1.4 AMPS." a little more? \$\endgroup\$ – Lucky-Luka Jan 7 at 9:44
  • \$\begingroup\$ Which quiescent current on Q1 and Q2 you had in mind when you wrote" Also your 4 Vbe output bias circuit needs to be more than 3:1 R ratio to make 3+1=4. Vbe=0.6V @ 1mA and 0.65V @ 10mA so the crossover output bias should be not 1.4 AMPS."? \$\endgroup\$ – Lucky-Luka Jan 7 at 9:58
  • \$\begingroup\$ Using the 3:1 ratio I get 500mA and more quiescent current flowing in R1. It looks like 720ohm and 1500ohm works just fine giving Iqc=45mA or 680ohm and 1130ohm giving Iqc=18mA on R1. Don't you agree? \$\endgroup\$ – Lucky-Luka Jan 7 at 10:08
  • \$\begingroup\$ @Lucky-Luka No . To prevent crossover distortion, you need to match the voltage of 2 Darlingtons or 4 times Vbe . Which means you need 3x VBE across R3. roughly. Since R4 =Vbe, then R3/R4=3. This is called a diode voltage amplifier \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 7 at 16:54
  • \$\begingroup\$ Aol is often used to define open loop gain. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 7 at 17:02

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.