My Dac must produce greater voltage than my amp's input sensitivity (1.5v) because I use a passive pre TVC. Dac currently puts out 2vRMS. I'm tinkering with the value of the I/V resistor and tube analog stage of the Dac, and this changes the voltage. My question is, beyond striving for voltage "around 2vRMS" to allow for headroom, is there some math around this subject...that I can use to calc the minimum voltage out of the Dac for optimal synergy with the amp (i.e. sufficent headroom)?
The schematic for the Welborne DRD300B amp is half way down this link. 1.5V input sensitivity. 100K ohm.
I use a TDA1541 Dac. With an I/V resistor value of 30ohm it puts out only .093 volts, and hence needs a gain stage. I'm considering building 5695 in SRPP with 5687 cathode follower. I don't have a schematic yet. It's my understanding the the gain is half the mu of the 5695 or 47/2=23.5. So 0.093x23.5=little over 2 volts. Yet, the TDA1541 likes to work into zero ohms, so I could lower the I/V resistor value but this also lowers the output voltage. I'm wondering how low I can go on the Dac's output voltage but still provide adequate headroom?
http://welbornelabs.com/etmreview.htm
I use a TDA1541 Dac. With an I/V resistor value of 30ohm it puts out only .093 volts, and hence needs a gain stage. I'm considering building 5695 in SRPP with 5687 cathode follower. I don't have a schematic yet. It's my understanding the the gain is half the mu of the 5695 or 47/2=23.5. So 0.093x23.5=little over 2 volts. Yet, the TDA1541 likes to work into zero ohms, so I could lower the I/V resistor value but this also lowers the output voltage. I'm wondering how low I can go on the Dac's output voltage but still provide adequate headroom?
http://welbornelabs.com/etmreview.htm
OK, when you said “I'm tinkering with the value of the I/V resistor and tube analog stage of the Dac”, I thought you meant an I/V resistor in the feedback path of a tube I/V output stage. Instead you meant the “audiophile” approach using a simple shunt R of low value followed by a lot of voltage gain, which is not what the DAC designers had in mind. As you know, you then have the dilemma of needing that resistor to have a very low value to keep the current output pin of the DAC happy (linear), but the lower you go, the more voltage gain (with its added noise and distortion) you will need. In don’t recall the exact current swing on the TDA1541 outputs, but let’s assume it’s +/- 2mA (4mA peak-to-peak or 1.41mA rms). Since you have no line stage gain, you might want to give yourself 6dB of headroom over your amp’s 1.5Vrms sensitivity, which would then require +/-4.2 V (on peaks) out of your tube stage. If you loaded the DAC with only a 10 ohm resistor, that would generate +/-20mV at the input, full-scale peaks, to the tube stage. So you’d need a gain of 210, or about 46dB. If you take that path, you could adapt many the RIAA stages posted around here to use a gain stage. They typically have about 40dB of gain at 1 kHz, but 60dB in the bass. So you’d have to adjust the gain and remove the RIAA filter to get about 46dB across the range. Noise would be an issue, so I’d pick a gain stage with a high-gm input tube. You could bump the resistor up to 30 ohms or so, so you’d need about 10dB less gain (36dB), but you’re allowing a larger voltage swing on the DAC output (with undocumented ramifications to the DAC’s linearity when that happens). Your 5695 is OK, but it will need more gain than as used in the SRPP if you only follow it with a CF. Maybe just one stage of the 5695 with resistor loading and perhaps an un-bypassed cathode resistor tweaked to set the gain, followed by the 5687 as a grounded cathode stage with output taken off its plate (low rp tube).
Another approach is to use a tube stage as an opamp with the DAC directly driving the inverting node, and the I/V resistor becomes a feedback resistor (about 2100 ohms). I am working on such a feedback design that has an input resistance to the DAC of only 4 ohms due to feedback. If you’re interested in that approach, I can share more details although it is a project in work.
Another approach is to use a tube stage as an opamp with the DAC directly driving the inverting node, and the I/V resistor becomes a feedback resistor (about 2100 ohms). I am working on such a feedback design that has an input resistance to the DAC of only 4 ohms due to feedback. If you’re interested in that approach, I can share more details although it is a project in work.
Brian, this is a very helpful post.
You said: "Since you have no line stage gain, you might want to give yourself 6dB of headroom over your amp’s 1.5Vrms sensitivity, which would then require +/-4.2 V (on peaks) out of your tube stage"
Does 4.2V on peaks mean 2.1v rms? If I follow you, this is what I want to shoot for if I want 6db headroom.
You said: "If you loaded the DAC with only a 10 ohm resistor, that would generate +/-20mV at the input, full-scale peaks, to the tube stage".I would like to learn how to calc to 20mV here. What is the math?
It's actually 2 tda1541 chips so its 8mA. I'm pursuing this analog circuit at the suggestion of Thorsten. Please confirm if my math is correct: 8mA / 2 x .707 x 30ohm I/V = .0848 vrms. If I used 5965 in SRPP (mu=47) then voltage output would be .0848 x 47/2 = 1.99...which is not quite 2.1v...if I have this right, then I'd need to either raise the I/V value or find a tube with a higher mu if I wanted to stick with srpp+CF circuit. Like you said, the chip wants to work into zero ohms, so maybe a higher mu tube with high gm would be the ticket.
You said: "Since you have no line stage gain, you might want to give yourself 6dB of headroom over your amp’s 1.5Vrms sensitivity, which would then require +/-4.2 V (on peaks) out of your tube stage"
Does 4.2V on peaks mean 2.1v rms? If I follow you, this is what I want to shoot for if I want 6db headroom.
You said: "If you loaded the DAC with only a 10 ohm resistor, that would generate +/-20mV at the input, full-scale peaks, to the tube stage".I would like to learn how to calc to 20mV here. What is the math?
It's actually 2 tda1541 chips so its 8mA. I'm pursuing this analog circuit at the suggestion of Thorsten. Please confirm if my math is correct: 8mA / 2 x .707 x 30ohm I/V = .0848 vrms. If I used 5965 in SRPP (mu=47) then voltage output would be .0848 x 47/2 = 1.99...which is not quite 2.1v...if I have this right, then I'd need to either raise the I/V value or find a tube with a higher mu if I wanted to stick with srpp+CF circuit. Like you said, the chip wants to work into zero ohms, so maybe a higher mu tube with high gm would be the ticket.
No, 4.2/1.414 = 3 volts rms approx. 3 Vrms is twice the 1.5 vrms spec, so that's 6 dB. I picked 6dB arbitrarily - you could go higher or a bit lower.
I'm not sure that I've got a proper grasp on the TDA1541 spec sheet, but I think it says that it puts out 4mA at full-scale (all digital 1's), and zero mA at digital zero. There is a 2mA "DC" bias at midpoint therefore (a digital 1 followed by 15 zeros). So from this midpoint, the maximum signal current peaks are + and - 2mA. Across a 10 ohm resistor, you'd see a steady 20mV DC at midpoint (2mA*10), 40mV at full-scale (4mA*10), and 0 mV at digital zero. So the swing is +/- 20mV from the midpoint. Peak-to-peak is 40mV. Rms input (ignoring the DC component) is 20mV/1.414 = 14.1mV. And then double everything with two TDA1541s in parallel. If I read the spec sheet right...
I'm not a huge fan of SRPPs (mu-followers I like), so my approach would be to give the tube a higher plate load, giving higher gain. There is no reason for an SRPP in my book if you have a CF after it to buffer the first stage and to provide a low output resistance.
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