diyAB Amp The "Honey Badger" build thread

Finnised the top lid to day. 5x35mm + 3x9mm vents pr heatsink.
The wiremesh came from a paper trashbin :)
I also got around to finnish the front panel. Just missing the LED's.
 

Attachments

  • IMG_20191002_235427.jpg
    IMG_20191002_235427.jpg
    593.5 KB · Views: 455
  • IMG_20191003_110858.jpg
    IMG_20191003_110858.jpg
    404.8 KB · Views: 460
  • IMG_20191003_125544.jpg
    IMG_20191003_125544.jpg
    387.4 KB · Views: 458
  • IMG_20191003_125907.jpg
    IMG_20191003_125907.jpg
    404.5 KB · Views: 448
  • IMG_20191003_152846.jpg
    IMG_20191003_152846.jpg
    382.4 KB · Views: 451
  • IMG_20191003_152936.jpg
    IMG_20191003_152936.jpg
    365.1 KB · Views: 144
Member
Joined 2016
Paid Member
Yes, it is correct
Wow. I did not realise this at first. This then means that the emitter degeneration is only 7:1. With the LTP current source set to 3.75mA as recommended by the build guide.
I calculate the following. Maybe someone can check these numbers for me to.
re' = Vt / Ic
Ic = 3.75 / 2 => 1.875
re' = 26 / 1.875
re' = 13.87 ohms
Emitter degeneration = (Re + re' / re')
Emitter degeneration = (83.33 + 13.87) / 13.87
Emitter degeneration = 7

Isn't 10:1 usually recommended as it provides better linearization of the input differential pair and better local negative feedback to Q1 and Q2 admittedly at the cost of some IPS gain.

I was looking at this because I wanted to have some idea what the gain crossover frequency of this amplifier was.

I have not seen any specifications published for this amplifier.

So maybe someone can check these numbers for me to.

Gain crossover frequency (Fc) = 1 / 2 x pi x Rltp x CLG x CM

Rltp = (83.33 + 13.87) x 2
Rltp = 194.4 ohms

Closed Loop Gain (CLG) = (R6 + R5) / R5
Closed Loop Gain (CLG) = (33000 + 820) / 820
Closed Loop Gain (CLG) = 41.24 or 32.3db

Miller Capacitor (CM) = 100pf

Gain crossover frequency (Fc) = 198,520 hz

And finally Slew rate (SR) equals
I = C * dv/dt
SR = 37.5V / uS
So slew rate would equal 37.5V /us
 
Last edited by a moderator:
Member
Joined 2016
Paid Member
ah, the rule of 10, if the Re is greater than re by 10 or more......i agree here...since the Re dominates....

Leach used 270 ohms Re with tail current of 4ma.....

I guess that we can do it more accurately to work with the current emitter resistor that we have.

We can transpose and rearrange the following two formulas to caculate Ic.

1. re' = Vt / Ic
2. Emitter degeneration (Ed) = Re + re' / re')

So the result using the current resistor values and a emitter degeneration of 10 becomes
Ic = Vt x (Ed -1) / Re
Ic = 2.8mA

This will then required a new calculation for the appropriate values for C7 + C8

I guess that the point of my last post was to check that the Gain crossover frequency is caculated correctly.
 
Last edited:
Got some painting done. 1 layer of primer on both sides. And 1 layer of paint on the inside, and 2 layers of paint on the outside. Just missing the bottom panel and brackets.
 

Attachments

  • IMG_20191008_155319.jpg
    IMG_20191008_155319.jpg
    289.7 KB · Views: 89
  • IMG_20191008_155201.jpg
    IMG_20191008_155201.jpg
    305.2 KB · Views: 100
  • IMG_20191008_155102.jpg
    IMG_20191008_155102.jpg
    369.6 KB · Views: 95
  • IMG_20191007_135111.jpg
    IMG_20191007_135111.jpg
    375 KB · Views: 81
  • IMG_20191007_135104.jpg
    IMG_20191007_135104.jpg
    251.8 KB · Views: 80
  • IMG_20191007_135051.jpg
    IMG_20191007_135051.jpg
    275.5 KB · Views: 93
i can not speak for the honey badger, i haven't read any about its open loop gain crossover frequency, i suspect they would be higher given the higher Ft devices used in the design compared to the 1980's leach amps than used the 2n3440...

i am sure this was discussed in the development of the design by Pete V. just don't have the time dig up...

a cornering frequency, say in the 20khz region would be best as far as Leach wrote...
 
The chassis is comming together. Don't mind the glitch on the backpanel. That will be gone as sone as the heatsinks are in place.
 

Attachments

  • IMG_20191011_183542.jpg
    IMG_20191011_183542.jpg
    331 KB · Views: 113
  • IMG_20191011_183511.jpg
    IMG_20191011_183511.jpg
    354.2 KB · Views: 341
  • IMG_20191011_183425.jpg
    IMG_20191011_183425.jpg
    306.6 KB · Views: 348
  • IMG_20191011_183358.jpg
    IMG_20191011_183358.jpg
    284.7 KB · Views: 349
  • IMG_20191011_183326.jpg
    IMG_20191011_183326.jpg
    313 KB · Views: 364
  • IMG_20191011_174302.jpg
    IMG_20191011_174302.jpg
    402.1 KB · Views: 367
What supply voltage would i have to run to hit 200-210W into 8 ohms? ive been looking at these boards for a couple years now and now im getting back into it


the calculation for power is as follows....p=(2x power xR)..square root the total...then divide the answer by the square root of 2(1.414)to get the ACv of the secondary.... so 2x200 x8=3200..sq root of 3200=56.5vdc then divide by 1.414=40v AC...