Gaincard pictures revisited

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Hi all,
my name is Grigor and this is my first post in this forum although I'm watching it for a long time.
There are two things that are still unclear to me and, AFAIK, nobody mentioned anything about them.
1. Why is pin 11 connected to negative PS rail? I can assume that soldering this pin to a board makes the whole thing more stiff, but why is it connected to PS?
2. In an article "Op-amps are More Fun Than Tubes" Junji Kimura says "In my case, the op-amp has eight feet...". If this is not a mistake then at least chip which was initially used was neither 1875 nor 3875. Which one then it was?

Please, could somebody explain this.

Kind regards,
Grigor
 
Welcome to this forum, Grigor.

I dont know what the others do, but I used Pin 11 for V- connection in order to have a more symmetric layout of the amp. For that, I made a connection from pin 4 to pin 11 under the chip.

GCC1b.jpg


I dont have an answer to your second question....

Mick
 
I have built a copy of the Gaincard (report is here ) , with some delibarate deviations from the original design. After a week testing on my second system, I have now connected the amp to my main system (Source: Sony Playstation 1, Speakers Audio Physic Tempo 3i) and made first serious listening tests.

Particularly interesting for me was the comparison with the Audiodigit (Autocostruire) amp. Here are my findings:

The first thing that struck me in a direct comparison is the much stronger presence of the music. I had the very strong feeling that with the GC the music came about two metres closer to me than with the Audidigit. While with the latter, the soundstage extends from about 1m in front of my speakers and about 2m behind it. With the GC the stage was extending about the same in the back, but in the front it was about 3m deep. I was very close to sitting in the middle of the music. Also it was very direct, lively and it was a little bit more sordid than the Audidigit. Hope you know what I am trying to say. This was just the feeling I had when I heard the deviation from the extreme cleanness of the Audiodigit.

On the other hand, the Audiodigit has qualities which the GC lacks - it is more transparent, analytic and clear. In fact, knowing the sound of the Audiodigit, I was missing some sparkling high treble when the GC played. Maybe I should use some more metalfilm resistors with the next approach, for this one I used only carbons (only for the mute I used a metalfilm).

Finally, the GC is less holographic. Before I had the GC, I found the holographic staging of the Audiodigit a very positive thing. The GC, however, still locates every feature in the music into a clearly defined position in space, but not as extreme as the Audidigit. Eventually I found this less tiring, and more pleasant how the GC does it, in the sense that I much easier found my way into the music. With the Audiodigit I was rather listening to staging and imaging, with the GC I was captured by the music itself.

To summarize everything in a few words: The Audiodigit rather addressed my intellect, the GC addressed my emotion.

Mick
 
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Christer said:
Tony,

The size of the caps decide how much the voltage over them drop depending on the current you draw from them. Take a charged cap and draw a constant current I from it, then the voltage will drop linearly and the voltage drop dV after time dt will be

dV = I * dt / C

If the load is resistive or varying the relationship will not be linear, so you have to solve a diff equation instead, but the above formlula is usually a good approximation if you use the average current drawn. For 50 Hz and fullwave rectification, we get a charge cycle every 10 ms so that also limits how much the voltage can drop. Actually dt is less than 10 ms, since the recharging also takes some time, but using 10 ms is a safe approximation. For 60 Hz we get a slightly shorter dt and thus slightly less voltage drop for the same current and capacitance.

Example:
50 Hz, fullwave rect., average load current 2 A, C = 22000 uF

dV = 2 * 0.01 / 0.022 = 0.9 V

If we instead use 2200 uF we get dV = 9 V


Hi Christer,

I did some tests. I think that the theory looks sound, but in practice something else is happening. but one point first which should (I think) halve the voltage drop that you were calculating. were you working out the current required in total from the ps, or the current from each rail?? if total then I guess divide by two to get the figures per rail :)

Now I have done these calcs based on current being drawn by the load, so by rights the amp should be drawing even more.

I have 1000uF per rail per channel. so the calcs are really easy. basically for a 1A current should have 10V drop for your equation.

I measured the power output of the amp (27.5V) rails. It managed 21W into 8 Ohms before clipping. the sag on the PS was 4V (ripple was about 2V)..... If I take 800ma draw per rail (to make up the 1.6A at the speaker) then I should have a rail drop of 8V :xeye: so the voltage sag appears to be roughly half what your calculation suggests. maybe this is because I used a pure sine wave, so only roughly half duty cycle???... perhaps with multipe sine waves the current draw would be higher. Might try that later :)

Tony.
 
Tony,

The calculations I did were approximate, which I also said. Typically they will give a larger voltage drop than in an actual circuit, which is a good thing if we use such calculations for choosing capacitors. The major source of error is probably that charging was assumed instantaneous, which is not the case. For some part of each half cycle, the transformer will supply both charge current and load current, so the capacitors only need to supply load current during the rest of the half cycle. The length of the charging period will depend on the total resistance in the charging circuit (secondary winding R, cap ESR, wire R etc.). I did a few simple sims in Spice using 50 Hz, 1000 uF cap, 900mA load current. With no resistance, the charging period is 2.5 ms and the voltage drop is 7.1 V. With 1 Ohm, the charging period is 3.4 ms and the voltage drop is 6.3 V.

After writing this I realized that maybe we mean different things with "voltage drop"? I should perhaps have made it clear that I meant the total p-p voltage variation, ie. I don't separate ripple and average voltage sag. Maybe you measured something else?
 
Tony,

If you use a multimeter only, you will not take the ripple into account, but that is important too since it is the lowest voltage ever that limits the output power, not the average voltage. If you read 4 V on the multimeter, that probably means you have about 8 V pp ripple, which correlates well with my calculation. Do you have access to a scope? Then you could get a better picture of this.
 
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Hi Christer,

I have a pc based scope which is fine up to about 10Khz so it is good for looking at ripple on the PS :) one other thing, do you measure ripple on just one rail at a time or do you measure across both rails??

I've only ever looked at a single rail which gives asymetric ripple (understandable with a sine wave and class AB amp), might try across both rails :) (but not till Monday as I'm not home).

Tony.
 
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