Output Capacitor Value

[dr._sleep]

I was trouble shooting some noise problems in my Aikido Headphone/Preamp (hissing and crackling). Found a couple of suspect, cold solder joints at the 20uF caps out to trannys and transformerless outs.



I had a couple of .47uF Auricaps and put them in place of the 20uF Solens fulling expecting to lose the bottom end. This did not happen and the frequency balance seamed to actually improve, certainly the back ground became much darker and silent, better highs.

This smaller value means that I can put some serious paper in oil, copper or silver caps which I prefer at a reasonable price.

Anyone know the math behind selecting this cap value. Does this cap form a high pass filter with the load. And what is the calculated load? Thanks in advance for an answer.

dr._sleep

[sreten]

Hi,

the capacitor appears to be inside a feedback loop, which will reduce
its effect on the frequency response depending on the open loop gain.

Inside the feedback loop the main effect would be to reduce voltage swing
capability in to lower impedance loads without affecting frequency balance.

You probably need a circuit simulator and a model of your loads.

/sreten.

[cerrem]

The calculations are straight forward....
You need to know or provide the transformer's primary inductance and it's turns ratio... I assume you are driving 600 ohm headphones????

Chris

[dr._sleep]

The transformer's primary inductance is 137H, the turns ratio are 12.1:1:1, Primary Z is 10K, 150 Ohm Headphones.

dr._sleep

[cerrem]

Originally posted by dr._sleep
The transformer's primary inductance is 137H, the turns ratio are 12.1:1:1, Primary Z is 10K, 150 Ohm Headphones.

dr._sleep

Terrific!!!!! I am almost ready to crunch the numbers...
Just one thing I am not understanding... The OPT turns ratio is 12.1 to 1 ????? WIth the Headphone being 150 ohms????
If thats the case then you get roughly 22K Primary Z ....... Is that your intention??? Rather than the 10K load????
One last thing.... What is the output tube you are using in your pre-amp??? I can't read the tube# on the schematic, my eyes are not terrific....

Chris

[dr._sleep]

Yes max ratio is 12.1:1, and the output tubes are 5687's.
RATIO@ 30 to 100 ohms 12 : 1
@ 100 to 250 ohms 6 : 1
@ 250 to 600 ohms 4 : 1
I am using them at a ratio of 6:1 @130 ohms.
dr._sleep

[cerrem]

OK...I still confused on exactly how this is going...
Based on the info I have... You are using the 6:1 ratio of the transformer and terminating this to 130 ohms....I thought it was 150 ohms for the headphones???
Either way assuming 130 ohms for secondary termination....leaves you with roughly 4.7K plate load....
For low frequency response, first thing you calculate the L needed...You take the source impedance and parallel it with the transformer primary impedance...in this case the source is roughly 100 ohms output ....so 100 ohms wins out since it is grossly dominant... Now I divide this by (2*pi*fr).... I use 3.3Hz and the result is roughly 5 Henries needed ...so you are overkill on the L..
Next you can look at two things... The RC POLE produced by the plate load and the coupling cap....which is a -20dB per decade roll-off below the POLE...which is = 1/( 2*pi*RC) ..... Then you have your LC DOUBLE-POLE = 1/(2*pi*(LC)^.5) , this is where your resonant peak occurs and then you have a -40dB roll-off....
The Q of the this peak is dictated by the damping effect of the plate load, which is a fixed , so not much to do...BUT, by carefully positioning the resonant peak just below the RC POLE, you can null the effect of this resonat peaking....As of now you have a +5dB "hump" at about 3Hz with the 20uF cap.... By using a 4.7uF cap and moving the resonant peak closer into the RC POLE, you get a flat response at the low end without the "hump"....
Keep in mind that this is just "open-loop" analysis...With proper feedback compensation you could also flatten this response...
As for your high frequency response....thats a mess...since this can be dictated by the ESR and ESL of the caps you use in combination with the leakage inductance and interwinding shunt capacitance of the windings, this is easily modeled and derived if you knew all these specs ... I would suggest using a 4.7uF output cap then "bypass" by smaller cap of your choice.....

Chris

[dr._sleep]

Thanks for taking the time for these calculations.

The Sowter website states: "The ratio of the transformer and the impedance of the headphones will determine the reflected load seen by the valve. It does not matter if the headphones see a reflected impedance of the valve which is lower than the headphone impedance. The lower the impedance seen by the headphones the better. The anode (in this case cathode) will see the headphone impedance multiplied by the ratio squared."

130 x 6x6 = 4680 then

f=1/2piRC
f=1/2pi(4680)x(.0000047)
f=7.232 Hz

Is this not my low frequency cutoff?
dr._sleep

[cerrem]

Hi...
When it comes to this type of analysis, you need to refer all secondary termination to the primary....
I am sorry for the confusion, since I was trying to make an analogy to the RC vs LC... Actually I should just should have kept it straight forward.... This circuit is refered to as a LCR tank circuit... In this particular case, you can ignore the "voltage" source resistance of 100 ohms, for simplicity.....
The corner frequency for your low-frequency roll-off is a LC DOUBLE-POLE = 1/(2*pi*(LC)^.5) , this is where your resonant peak occurs and then you have a -40dB roll-off....
You have a damping Resistance which dampens the resonant frequency hump.... This Resistance is the primary load and in your particular case happens to be 4680 ohms..... Since your 4680 ohms is somewhat fixed....you need another means to flatten this resonant hump....you can apply negative feedback as the case of the circuit you show.... Or as mentioned before...If you move the resonant frequency up a bit..then the 4680 ohms becomes more effective at damping.... When the Cap is moved to 4.7uF, the damping response by the 4680 ohms is perfect and the response is flat right down to the roll-off freq.....this is all prior to closing the feedback loop...
Keep in mind you can't always move the resonant frequency up to get a better Q, since it can monkey up the audio band...but in this case it is doable... Now if you make the capacitor to small, then you can make a RC POLE over-ride the LC pole and you get roll-off happening in a much higher frequency...may not be good for audio...

Chris

[sreten]

Originally posted by dr._sleep


130 x 6x6 = 4680 then

f=1/2piRC
f=1/2pi(4680)x(.0000047)
f=7.232 Hz

Is this not my low frequency cutoff?
dr._sleep

Hi,

you are assuming the capacitor is outside the fedback loop, it isn't.

/sreten.