You can not change the gain of the srpp stage.
Just ad a resistor devider network at the input or output.
Your schematic is great for I out dac then you can lower i/v resistor to get right output voltage.
Schematic is not so good for V out dac or you want high output voltage.
Just ad a resistor devider network at the input or output.
Your schematic is great for I out dac then you can lower i/v resistor to get right output voltage.
Schematic is not so good for V out dac or you want high output voltage.
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Thanks Koifarm,
Is there any better configuration, based on two tubes, that you would suggest with a Vout DAC?
Thanks!
Is there any better configuration, based on two tubes, that you would suggest with a Vout DAC?
Thanks!
Thanks Ray,
I've tried a few configurations including the Broskie Cathode Follower indicated on the page you linked. So far the best result was through SRPP, which I am now fine tuning.
Here is a recap of my tests:
CD Player - Tubes vs. Burson OpAmps Output
Cheers
I've tried a few configurations including the Broskie Cathode Follower indicated on the page you linked. So far the best result was through SRPP, which I am now fine tuning.
Here is a recap of my tests:
CD Player - Tubes vs. Burson OpAmps Output
Cheers
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Those circuits in post # 1 are simply to simple.
If you use current output DACs, they will not work on those schematics.
That is because the top DAC is going to try and pump current + / - into the grid.
You will find that there is no current when the signal goes negative, so the current DAC will try and increase the negative voltage. At that point, the DAC voltage will attempt to go beyond its maximum negative "compliance" voltage (the grid is not going to draw grid current, unless the voltage is enough to draw an arc).
And when the DAC goes positive, there also is no grid current. The DAC will try and go beyond its positive voltage compliance area.
With voltage DACs, one DAC is going to drive a very hight impedance load of the grid (+ / - V into no current, and no current).
And the bottom DAC is going to drive the bottom resistor load, R, in parallel with the series combination of Rk and the bottom cathode impedance.
Unequal loads for the top and bottom DAC, even for a voltage DAC out, is probably not a good thing.
Or, am I missing something?
"You should make things as simple as possible, but no simpler." A. Einstein
If you use current output DACs, they will not work on those schematics.
That is because the top DAC is going to try and pump current + / - into the grid.
You will find that there is no current when the signal goes negative, so the current DAC will try and increase the negative voltage. At that point, the DAC voltage will attempt to go beyond its maximum negative "compliance" voltage (the grid is not going to draw grid current, unless the voltage is enough to draw an arc).
And when the DAC goes positive, there also is no grid current. The DAC will try and go beyond its positive voltage compliance area.
With voltage DACs, one DAC is going to drive a very hight impedance load of the grid (+ / - V into no current, and no current).
And the bottom DAC is going to drive the bottom resistor load, R, in parallel with the series combination of Rk and the bottom cathode impedance.
Unequal loads for the top and bottom DAC, even for a voltage DAC out, is probably not a good thing.
Or, am I missing something?
"You should make things as simple as possible, but no simpler." A. Einstein
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You can reduce the voltage output of a current output DAC. Just reduce the resistance of the load resistor.
But watch out, at some point you will run out of current from the current DAC, and you will get clipping.
But watch out, at some point you will run out of current from the current DAC, and you will get clipping.
Rayma,
Post # 4.
More super triode ideas:
The very first schematic, with an op amp does the following:
One DAC sees a 10k Ohm load.
The other DAC sees a 20K Ohm load.
At that point, I stopped reading further to the other schematics.
Post # 4.
More super triode ideas:
The very first schematic, with an op amp does the following:
One DAC sees a 10k Ohm load.
The other DAC sees a 20K Ohm load.
At that point, I stopped reading further to the other schematics.
Not a problem. To equalize the load resistances, just add a 20k to ground
on the positive DAC output.
Now why did the schematic omit that fact?
Simplified schematics are a great teaching aid, just be careful.
Some of them have far worse errors if you actually build them.
Showing all the parts in a correct circuit makes it harder for the beginner,
but it makes him learn the correct way (and an explanation in the text of
why, helps too).
Simplified schematics are a great teaching aid, just be careful.
Some of them have far worse errors if you actually build them.
Showing all the parts in a correct circuit makes it harder for the beginner,
but it makes him learn the correct way (and an explanation in the text of
why, helps too).
Thanks 6A3sUMMER and Ray for your comments.
I am using a Vout DAC. So I wonder how to balance the impedance seen by both outputs.
On the + side, assuming cathode impedance in the order of 100s kOhm, the impedance would be close to R= 250Ohm.
If I wanted to equal the high impedance seen by the - side I would need to substitute R with a value in the 100s kOhm which would mess up with the tube bias.
On the - side I could put a 250Ohm resistor to ground and equal the + side.
That should work. In fact right now I am measuring Vout+= 2.34V and Vout-= 2.55 so, with a 250Ohm load on the - output, the DAC should be still working fine.
What do you think?
Normally, a balanced output is made to drive a balanced amplifier (balanced all the way through the amp). the idea is precision, and a reduction of 2nd harmonic distortion.
You got 2.34V and 2.55V
2.34/2.55 = 0.9176 (an 8.2% voltage difference)
20 x (Log of 0.9176) = -0.746.
That is -0.746 dB; the + and - are un-balanced by about 3/4 dB
With an un-balance of 3/4 dB, you might do just as well if you drive a single triode from either the + DAC or - DAC.
In other words, just use one DAC (+ or -) that has the phase that you want.
I have a CD player that has both a balanced + DAC and - DAC. It also has an unbalanced output. I have only ever used the unbalanced output.
Someday, I plan to build a balanced amp, and then I will use the balanced output from the CD player.
You got 2.34V and 2.55V
2.34/2.55 = 0.9176 (an 8.2% voltage difference)
20 x (Log of 0.9176) = -0.746.
That is -0.746 dB; the + and - are un-balanced by about 3/4 dB
With an un-balance of 3/4 dB, you might do just as well if you drive a single triode from either the + DAC or - DAC.
In other words, just use one DAC (+ or -) that has the phase that you want.
I have a CD player that has both a balanced + DAC and - DAC. It also has an unbalanced output. I have only ever used the unbalanced output.
Someday, I plan to build a balanced amp, and then I will use the balanced output from the CD player.
Yes the different voltages are due to the unequal impedances seen by the + and - outputs.
I reviewed my (initally wrong) calculations and did some tests so now I have 150k to ground on the - side so that now both Vout values are set at 2.24V
I reviewed my (initally wrong) calculations and did some tests so now I have 150k to ground on the - side so that now both Vout values are set at 2.24V
Agree, while Broskie's blog is a fantastic resource, I now realize not all the details are there ;-)
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