Hello,
I have some questions regarding the AD1955 I/V stage presented in the datasheet here , page 18. I'm intersted in learning the design principles of this stage in order to try and design (adapt in fact) my own vacuum stage. However I think the AD1955 has some particularities that need do be adressed. I hope the questions are not too dumb as I have no studies in electronics and little experience with digital and signal conversion.
The first question is regarding the Vbias pin of the I/V stage. The datasheet indicates this is to be conected to the Vref (2.39V) from the dac or to the Vrefa (2.90) obtained with a voltage divider from the AVDD (the analogic voltage line). It seems to me that this vbias conected to the non-inverting pin of the op-amps acts in creating a voltage reference diferent from the ground with 2.39-2.9V. My question is why is this needed, what are the tehnical considerations for this?
Other questions will come for sure. Regarding the output stage I'm tinking to adapt: I'm intersted in the advantages of transformer-coupled, differential circuits using triodes and parallel feed. Some examples are here on Mr. Dave's Davenport site here for the general topology and here for the parafeed. Another great source are Lynn Olson's articles.
What to do in order to adapt this circuit (in my opinion):
1. find a way to rise the ground potential to the 2.39-2.90 specified in the datasheet. I would probably use a Jung regulator for this.
2. design a current source for the ad1955 as it has a current sink output. The same superregulator will be used for the voltage suply, as for current conversion I would probably try first a simple resistor just as in the datasheet.
3. I have no clue if a triode can act as an inverting amplifier as in the datasheet output stage either.
Well, I'm still not sure if this isn't just a bunch of pseudo-tehnical non-sense I'm thinking of, but that's why I wait your opinions.
Thank you
I have some questions regarding the AD1955 I/V stage presented in the datasheet here , page 18. I'm intersted in learning the design principles of this stage in order to try and design (adapt in fact) my own vacuum stage. However I think the AD1955 has some particularities that need do be adressed. I hope the questions are not too dumb as I have no studies in electronics and little experience with digital and signal conversion.
The first question is regarding the Vbias pin of the I/V stage. The datasheet indicates this is to be conected to the Vref (2.39V) from the dac or to the Vrefa (2.90) obtained with a voltage divider from the AVDD (the analogic voltage line). It seems to me that this vbias conected to the non-inverting pin of the op-amps acts in creating a voltage reference diferent from the ground with 2.39-2.9V. My question is why is this needed, what are the tehnical considerations for this?
Other questions will come for sure. Regarding the output stage I'm tinking to adapt: I'm intersted in the advantages of transformer-coupled, differential circuits using triodes and parallel feed. Some examples are here on Mr. Dave's Davenport site here for the general topology and here for the parafeed. Another great source are Lynn Olson's articles.
What to do in order to adapt this circuit (in my opinion):
1. find a way to rise the ground potential to the 2.39-2.90 specified in the datasheet. I would probably use a Jung regulator for this.
2. design a current source for the ad1955 as it has a current sink output. The same superregulator will be used for the voltage suply, as for current conversion I would probably try first a simple resistor just as in the datasheet.
3. I have no clue if a triode can act as an inverting amplifier as in the datasheet output stage either.
Well, I'm still not sure if this isn't just a bunch of pseudo-tehnical non-sense I'm thinking of, but that's why I wait your opinions.
Thank you
Hello aparatusonitus,
thanks for the comments. I gave the Jung example more to ilustrate I'm after a low noise source. I didn't give too much thought to the Jung limitations. I guess I could just use a voltage divider as in the datasheet, with the jung circuit suplying the analog voltage line as a last resort.
thanks for the comments. I gave the Jung example more to ilustrate I'm after a low noise source. I didn't give too much thought to the Jung limitations. I guess I could just use a voltage divider as in the datasheet, with the jung circuit suplying the analog voltage line as a last resort.
If I ever get to build this I'm sure I'll try something simpler where I feel I can make a compromise for the begining. So the ideea with the shunt regulator is a viable option.
I think the main design guidelines are differential triode operation with active plate loading, single gain stage, parafeed operation (as in the western electric derivation)... and this should be all I guess for proof of concept..
I think the main design guidelines are differential triode operation with active plate loading, single gain stage, parafeed operation (as in the western electric derivation)... and this should be all I guess for proof of concept..
SunRa said:
Just as in the datasheet? Are we looking at the same datasheet? The one i have specifically warns agains using a "simple resistor".
No idea what is the max compliance voltage at output but it's probably really low. I don't have first hand experience with the 1955 but it doesn't seem like a great choice for resistor/tube I/V.
I was refering to the current suply of the I/V stage. The AD1955 dac sinks current so it needs an active conversion. The datasheet states that you need the I/V stage able to supply the signal current. And I supose that the 2K R2 and R3 in series with the 12V suplly is doing just that.
Could you explain a little more? For example the PCM1794 has almost the same current capabilities of the ad1955 (in fact a little less even) and it seems it's suitable for this (the rakk dac is a very good example) . Of course the main difference between these two (AD1955 sinks current) is the one that complicates everything
...
SunRa said:
In my understanding whether it sinks or sources current is irrelevant. The datasheet recommends that instead of virtual ground the output is connected to a 2.8v potential, so the IV stage has to be biased at this voltage but as far as AC current is concerned it still has to flow in a dead short.
There may be a small ac votage at output before distortion really takes off but i have no idea of its value.
Is passive IV possible? No idea. The 1794 datasheet does not specifically warn agains it but the 1955 does. It may simply mean that a low enough iv resistor will eventually degrade the S/N.
Yes, you are right, I didn't payed atention to my explication. My english is not so good and I have some dificulties to properly express what I want.
Well I think I get it now. But in order to understand better, how exactly the op-amp circuit works? I understand it's not a simple non-inverting amplifier with gain... but then? If you have a link to some references would be great, because one reason for this thread was the fact that I can't find some resorces on the net about this..
SunRa said:
Google transimpedance amp. The textbook version has the positive input grounded, the input current applied to the negative input directly (virtual ground) and the feedback resistor develops the output voltage. Vout=-Iin*Rfb. In our case the positive input is biased to 2.8v which automatically biases the negative input to the same.
Re: Re: AD1955 I/V stage/tube version
Take a look at my regulator. It should offer performance similar to the Jung regulator and be able to work at voltages as low as 2.5V.
aparatusonitus said:
Take a look at my regulator. It should offer performance similar to the Jung regulator and be able to work at voltages as low as 2.5V.
I mean an monolith opamp, not discrete one...but someone can play with this one as well
http://www.diyaudio.com/forums/attachment.php?s=&postid=840668&stamp=1139437025
http://www.diyaudio.com/forums/attachment.php?s=&postid=840668&stamp=1139437025
Cauhtemoc, aparatusonitus,
Thanks for your reply's. In fact I was allready watching cauhtemoc's thread. It's for sure a viable option and a nice source of informations.
For now I have bigger troubles trying to figure out if I can make a transimpedance amp without an op-amp. Valve operation in this configuration is out of discusion I guess as I don't think it's possible to mentain in the same time the advantages (different signal and power loops, aka insulation between the psu and the signal, and the overall simplicity and ingeniosity of the circuit I referd to)
I guess the other option is to use an op-amp or discrete (if it's whorthy) based I/V converter followed by a tube stage to take care of the amplification. However for know I'm in a dead end, as the transimedance articles where to revelatory I guess . (thanks analog_sa)
Thanks for your reply's. In fact I was allready watching cauhtemoc's thread. It's for sure a viable option and a nice source of informations.
For now I have bigger troubles trying to figure out if I can make a transimpedance amp without an op-amp. Valve operation in this configuration is out of discusion I guess as I don't think it's possible to mentain in the same time the advantages (different signal and power loops, aka insulation between the psu and the signal, and the overall simplicity and ingeniosity of the circuit I referd to)
I guess the other option is to use an op-amp or discrete (if it's whorthy) based I/V converter followed by a tube stage to take care of the amplification. However for know I'm in a dead end, as the transimedance articles where to revelatory I guess
. (thanks analog_sa)Hi SunRa,
It is really pretty simple - don't make it more complex than it is. Start with a basic design and then refine it after the basic is working.
There are three parts to it: the current source and bias, the I/V, and the output stage. A simple current source is just a resistor. The bias point can be a voltage divider from an available voltage. The I/V is a resistor. The output stage can be something like that in my article. You need to bias the input to the output stage to allow for the 2.9V DAC bias. So your cathode bias could be something like 5V or 6V to cover the bias that the tube needs above the DAC bias.
Then you can get fancy with constant current sources and regulated bias supplies.
Dave
It is really pretty simple - don't make it more complex than it is. Start with a basic design and then refine it after the basic is working.
There are three parts to it: the current source and bias, the I/V, and the output stage. A simple current source is just a resistor. The bias point can be a voltage divider from an available voltage. The I/V is a resistor. The output stage can be something like that in my article. You need to bias the input to the output stage to allow for the 2.9V DAC bias. So your cathode bias could be something like 5V or 6V to cover the bias that the tube needs above the DAC bias.
Then you can get fancy with constant current sources and regulated bias supplies.
Dave
I supose you ask this because the output of the dac needs to see an impendance as close to 0 possible. Mr.'s Davenport active stage has the advantage of transformer coupling and so the impendance seen by the dac could be reduced by the ratio of the transformer.
This way the resistor can be big enough to provide the voltage needed by the tube amplifier.
I am somehow skeptical however because I don't fully understand the way the transimpedance amplifier works, and if satisfies some requirements of the DAC I'm not aware about. I'll just have to take a more in-depth look at those articles before asking some more questins
Mr. Davenport, thank you for the reply,
I thought the same at the begining of this thread but then it occured to me that the datasheet topology is not just a simple resistor with an op-amp amplifier. Scaning the articles regarding the transimpedance amp made me to reconsider this design.
Ok, so asuming I select a bias resistor to provide 5-6V (depending on the operation point I choose for the triodes), how should I bias to the 2.9Vof the DAC? By conecting the ground of the output stage with the 2.9V from say a voltage divider? This would be at the other end of the cathode resistor? Or at the common point of the I/V resistors? I guess the second is better as the cathode resistor should be conected to the power ground. It's a little tricky with this ground conections...
Well, I've mentioned the jung regulators just to show that I want to take care of some of these details. The first posts where directed to them although they are not my primary concern. However the contributions regarding them where welcomed as they atentioned me to some dificulties I may encounter.
Hi SunRa,
Let me first talk about the I/V. No matter what kind of fancy names that are given to output stages, the current-to-voltage conversion is always done with a resistor somewhere in the circuit. Sometimes it is hard to see which resistor is doing the I/V, but in the case of a simple resistor directly on the DAC chip or connected through a transformer, the I/V resistor is obvious.
How big is the resistor? Let's take an example of a simple tube amplifier with the I/V resistor on the grid. Start at the output of the amplifier and work backwards. First decide the maximum output level you want from the output stage, for example, 2VRMS. Second, determine the gain of the amplifier. This gives the voltage needed on the input of the amplifier. So for example, if the amplifier has a gain of ten (20dB) and you wanted 2VRMS out, you would need 0.2VRMS in. From here it is easier to figure in peak-to-peak, so that would be 0.57V p-p. Now go to the specifications for the DAC chip and determine what the maximum peak-to-peak current is. This is generally something like the 0dBFS current minis an idle, or no signal, current. Let’s say it was 10mA, to pick a round number. You now have the voltage and current to calculate the resistor value. 570mV / 10mA = 57 Ohms. This is your starting point for figuring the resistance. It never works out perfectly because of variations in the circuit so start with a 56 Ohm resistor and build the amplifier. Then use a test CD with a 0dBFS signal to see what the final output from the amplifier is. Let’s say it was 1.8VRMS instead of the 2VRMS that you wanted. Your final resistor value is this ratio times the resistor used. 2VRMS / 1.8VRMS x 56 = 62 Ohms.
Now, how to connect the bias to the amplifier. Definitely connect the cathode resistor to power ground. Then connect the bottom of the I/V resistor (common point of the two I/V resistors) to the bias. Then connect the reference for the bias voltage to the same ground that the cathode resistor is connected to. See the article here: http://www.raleighaudio.com/installation.htm Keeping in mind that the DAC chip there keeps a DC level on the I/V resistors and the REF is at ground. Your case will be different where the REF will be at 2.9V and the connection to ground will be from the reference (ground) of the 2.9V supply. It is tricky, so think it out ahead of time.
Dave
Let me first talk about the I/V. No matter what kind of fancy names that are given to output stages, the current-to-voltage conversion is always done with a resistor somewhere in the circuit. Sometimes it is hard to see which resistor is doing the I/V, but in the case of a simple resistor directly on the DAC chip or connected through a transformer, the I/V resistor is obvious.
How big is the resistor? Let's take an example of a simple tube amplifier with the I/V resistor on the grid. Start at the output of the amplifier and work backwards. First decide the maximum output level you want from the output stage, for example, 2VRMS. Second, determine the gain of the amplifier. This gives the voltage needed on the input of the amplifier. So for example, if the amplifier has a gain of ten (20dB) and you wanted 2VRMS out, you would need 0.2VRMS in. From here it is easier to figure in peak-to-peak, so that would be 0.57V p-p. Now go to the specifications for the DAC chip and determine what the maximum peak-to-peak current is. This is generally something like the 0dBFS current minis an idle, or no signal, current. Let’s say it was 10mA, to pick a round number. You now have the voltage and current to calculate the resistor value. 570mV / 10mA = 57 Ohms. This is your starting point for figuring the resistance. It never works out perfectly because of variations in the circuit so start with a 56 Ohm resistor and build the amplifier. Then use a test CD with a 0dBFS signal to see what the final output from the amplifier is. Let’s say it was 1.8VRMS instead of the 2VRMS that you wanted. Your final resistor value is this ratio times the resistor used. 2VRMS / 1.8VRMS x 56 = 62 Ohms.
Now, how to connect the bias to the amplifier. Definitely connect the cathode resistor to power ground. Then connect the bottom of the I/V resistor (common point of the two I/V resistors) to the bias. Then connect the reference for the bias voltage to the same ground that the cathode resistor is connected to. See the article here: http://www.raleighaudio.com/installation.htm Keeping in mind that the DAC chip there keeps a DC level on the I/V resistors and the REF is at ground. Your case will be different where the REF will be at 2.9V and the connection to ground will be from the reference (ground) of the 2.9V supply. It is tricky, so think it out ahead of time.
Dave
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