Help needed to upgrade the capacitor in OTL tube amp - Page 8

Quote:

Originally Posted by xmdkq 

 http://www.diybuy.net/thread-461889-1-1.html

WOW.... thats some intense amp building!!! thanks for sharing.

Quote:

Originally Posted by StalkerAssassin 

Hello people.
How you these capacitors Jantzen Cross Cap as output separation? I want to use them for DarkVoice 336SE. Will there be enough 100uF for channel?

I don't remember now, what is the value of output coupling cap in DV336se? What headphones will be plugged into it?

Quote:

Originally Posted by StalkerAssassin 


In stock Darkvoice 336SE - 20uF per channel, La Figaro 336C - 40uF per channel. But I will only use the circuit from 336SE, I will do all of the new in large chassis.
Sennheiser HD 650.

F = 1/(2 * pi * C * R)
F= 1 / (2 * 3.14 * 0.0001 * 300) = 5.3 (Hz) This means that up to 5 Hz, frequency response is linear, with no recession on the low frequency?

Well bigger enclosure, it is understood so u have a lot space to put good but large film caps, the mkp jantzen will do this job pretty well, with 100uf and 300 ohm headphones you'll get -3bB at 5Hz and the optimal frequency response is about 53Hz It means the roll off start from 53Hz to reach the point at 5Hz, but if u really want flat frequency response 20Hz at 0dB u need 265uF cap

This is the frequency response graph which my measurements, the white plot show the 130uf cap with the 300 ohm load and the yellow & blue 265uf on to 300 ohm, so as you see only with the bigger cap the FR is flat up to 20Hz 

Don't forget too the impedance (or resistance... I forget which) of the power tubes.  It gets added to the load impedance of the headphone in the high-pass cutoff frequency equation.

So in a TS-5998, dual triode, paralleled output configuration its ~70ohm + ~70ohm.  together in parallel result in 35ohms... that gets added in series to the headphone load impedance.  32ohm grado thus is really 32 + 35 total in the f-Hz calcs.

Quote:

Originally Posted by kramer5150 

Don't forget too the impedance (or resistance... I forget which) of the power tubes.  It gets added to the load impedance of the headphone in the high-pass cutoff frequency equation.

 

So in a TS-5998, dual triode, paralleled output configuration its ~70ohm + ~70ohm.  together in parallel result in 35ohms... that gets added in series to the headphone load impedance.  32ohm grado thus is really 32 + 35 total in the f-Hz calcs.

good point Kramer!  have a look on FR graph, it is good example, the plots red and purple tung-sol 7236(mu-4.8) higher output impedance vs TS5998(mu-5.4) slightly lower, 265uf output coupling cap into 50 ohm dummy load resistor. We can see how the tube impedance affect the FR, small difference because 7236/5998 are similar but there are.

I have another good example even more interesting, this is FR graph(ARTA software) taken out from my current setup LF339, tung-sol 5998, gec1067 and 325uf output coupling capacitors.

5998, 325uf 5% tolerance on to 33 ohm dummy load, the -3dB point is about 9.5Hz, ok let's mats, for the perfect model high-pass cut-off frequency for 325uf and 33ohm load will be 14.84Hz at -3dB so we have 14.84 (calculated) - 9.5 (measured)  = 5.34

5.34Hz is the tube impedance effect on frequency  response.

Ok, let’s see for an amp impedance, to get the 9.5Hz at -3bB with the 325uf cap we will need load about 51.5 ohm.

51.5(calculated) - 33(measured) = 18.5

Does this mean the 18.5 ohm is a output impedance? hmmm...slightly too low for a amp with 5998 as a output tube. It can be concluded, that impedance has a partial (some) influence on the high-pass cut-off frequency equation IMO.

DV337 & LF339 are cathode follower amplifiers, in this case, I think the output impedance for TS-5998 will be a little bit different.

A cathode follower never has a voltage gain more than 1, Vin = Vout, what it does offer is current gain, that is why it is used to drive hungry current cans.

The output from the CF is taken from the cathode, which has an impedance of approximately 1/gm(higher gm a lower output impedance) the transconductance of the tube, is also in parallel with the cathode resistor which primarily sets the idle current through the tube.

Knowing that the (Zo) formula for a cathode follower should be, Zo = rp/(mu+1), where (rp) is the plate resistance, (mu) is the amplification factor.

TS-5998:

               - (rp) 350 ohm approx                                                                                                

               - (mu) 5.4 some datasheets gives 5.5

350/(5.4 +1) =54.68 / 2 (because parallel triode section) = 27.34 ohm.

Does this mean the 27.34 ohm is a output impedance? hmmm...close enough, but not exactly. Data sheet shows only limited information, what will be the plate resistance for example 140V?, the (mu),(gm) can varies depending on the current flow, the 5998 wit max plate current amplification factors can rise to 6.2 which has a large impact on the final impedance of amplifier.

Only by measuring the impedance on working amp(driving the load) can give reliable and correct number of output amp impedance!