PASS A40 - bias

[Braun_]

Hi,

After some reading in the part-substitution article from Mr. Pass refering to the A40, I made some simulation in the Software ARES.
Due to the fact that I couldn't be able to find any 2n5248, I tried the solution to add a 10k resistor as it's indicated on the Article. The results I obtained can be viewed on the picture below (Note that I'm using TIP 142 and TIP 147 as output transistors).

An externally hosted image should be here but it was not working when we last tested it.

http://img100.imageshack.us/img100/6048/simulation2.jpg

after I tweaked some component values in order to reach the specified bias (0,8 A per transistor) and the ddp at the specified at some resistors, the result I obtained:

An externally hosted image should be here but it was not working when we last tested it.

http://img100.imageshack.us/img100/9826/simulation1.jpg

Within this configuration, the current through the resistor R9 is 27mA.

Now my questions are:
Did I miss something? Because I thought it was weird to obtain such results (almost -32V in the output) with the substitutions Mr. Pass had recomended.

The current at R9 is not to high? If I understand right, It suposed to be near 3mA, but here I got 27mA!

Besides component values and transistors Matching, are there some other effects that could change these values?

Obs. If I made some mistake here, please, let me know.

thanks in advance for the attention to this case.

 

[Nelson Pass]

You don't want to adjust the value for R9 to bias the output stage.
Put the 10K back in and trim R12 if you need to adjust the bias.

 

[gl]

Hi Braun_,

C3 should be across Q6 too, instead of being where it is. Maybe some other eyes might see some other things.

Graeme

 

[Braun_]

You don't want to adjust the value for R9 to bias the output stage.
Put the 10K back in and trim R12 if you need to adjust the bias.

Thanks Mr. Pass for your note. I simulated the circuit with the 10k resistor placed where the Q11 should be, however, I always get the -31,7 V at the output. I will verify possible mistakes in the circuit as like the menber gl has shown for the capacitor C3.

Hi Braun_,

C3 should be across Q6 too, instead of being where it is. Maybe some other eyes might see some other things.

Graeme

Thanks for your advice. I'll made this correction.

 

[Braun_]

You don't want to adjust the value for R9 to bias the output stage.
Put the 10K back in and trim R12 if you need to adjust the bias.

Sorry to stay too long on this issue, but I'd replaced the 10k resistor in the place of Q11, however, there are no way (in the simulation) to eliminate the -31V at the output with only adjustments in R11 or R12.

After some tests, where I aded a constant current source instead of Q11, I noticed that the simulated circuit needs at least 27mA to produce the voltage parameters that Mr. Pass had published in A40 article: almost 0.6V at R7 and R10, 0.65V at R6 and 1.1V between Q7 and Q9 emiters.
Considering that I have rebild this circuit at the simulator more then 10 times, I would like to know why do I must to use a 10k resistor in the place where the 2n5248 should be? And, if possible, why my simulation-circuit is not working properly with "just" 3mA at R9?

Thanks in advance for all the help!

 

[Braun_]

I discovered why the 10k resistor has no effect at the simulated circuit. I don't know why, but the two zenner diodes from the current source were not able to hold the 1,3V ddp. I still don't know why the diodes cannot hold this voltage drop (actualy the voltage drop is about 0,7V). But this voltage drop seems to be vital for the entire circuit polarization.
I would appreciate a lot if someone could assist to make me understand why the voltage drop at d1 and d2 is not happening as it is supposed to be.

Thanks in advance for any help!

 

[gl]

Check D1 and D2 to see if one of them is shorted. Just measure the voltage across each one separately. One of them will probably measure zero. That's the one to replace.

Graeme

 

[amp_guy]

Is that a simulated shorted diode or a real one?

 

[Braun_]

Hi Graeme

This is a simulated circuit. I pretend to build one after all my doubts are gone.
I verified the diodes, but both present 360mV between each terminals.

Actually, each diode is presenting the half of the voltage drop between terminals as it should be.
I must investigate. At the 1n4148 datasheet shows that the minimal voltage drop is something about 0.6V. So I think it could be a model mistake. I'll test with another diodes models to see what happens.
Thanks Graeme for your support.

 

[nigelwright7557]

I've build the A40 with lambda BJT from german supplier (http://www.electronicpool.de/en/index.html).

While setting the bias with R11 i wondered how do one knows that the bias is correctly set up ?

Mr Nelson Pass writes
- 1.5A per PMD trail (so 0.75A per transistor), p. 70 of the a40.pdf file on passdiy
- as well as to measure 1 volt (DC?) between Q7 and Q9.
Measuring 1 volt get 0.72 (1.44 for the trail)

An other article (http://www.facstaff.bucknell.edu/esantane/movies/a40.html) states 0.8A per ransistor (so 1.6 per trail). Is there a difference (except the heat).

What shall be the target value ?

What does trigger the DC offset ? I have 40mV is that acceptable ?

The object of setting the bias is to turn a class b output inot a class AB.
What I do is set the bias to minimum then apply a 1 volt sine wave and monitor the output with a scope.
I turn the bias up slowly until the crossover distortion goes.
Turning the bias up any further is just a waste of power and pours heat into the heatsinks.

 

[gl]

For a class AB amplifier you are correct. But the A40 is a class A amplifier. So the object is to set the bias high enough to allow the amplifier to remain in class A for as wide a range of load conditions as possible and low enough to prevent it from thermally self destructing.

Regarding the efficiency? That's a personal decision. My choice would be to run class A and turn the fire down one bar.

I have built two A40's although it was long ago. Both came up without a hitch and ran hot and trouble free for years.

Cheers,
Graeme