just built this cct of the CJ-PV12 based on 12AU7s. sounds good and was trying to understand the cct. as i saw the first stage is basically a common cathode and the second stage seems to be of cathode follower.
the first stage will invert the signal input as its a common cathode and this inverted signal would be fed into second stage cathode follower which will have the same inverted signal appearing on the output. hence the signal at the output would be 180deg inverted from the original. must this be corrected or im i missing something which does this already?
thks
the first stage will invert the signal input as its a common cathode and this inverted signal would be fed into second stage cathode follower which will have the same inverted signal appearing on the output. hence the signal at the output would be 180deg inverted from the original. must this be corrected or im i missing something which does this already?
thks
Attachments
there are certain parts of design in the PASS amp design as well which tends to be inverted but some changes are done to keep them (the in and out signal) in phase. at least the output is an exact replication somewhat. impairment of sound? ..im not sure.
another thing which is bothersome is the heater supply im using is derived from the LM7812 which is suppossed to regulate at 12Vdc at least. im reading 11.88Vdc (loaded). would this be impactous? what can i do to get spot on 12.6Vdc?.....have swapped 3 LMs already and they all the same!
thks
another thing which is bothersome is the heater supply im using is derived from the LM7812 which is suppossed to regulate at 12Vdc at least. im reading 11.88Vdc (loaded). would this be impactous? what can i do to get spot on 12.6Vdc?.....have swapped 3 LMs already and they all the same!
thks
puginfo said:
Yes.
Yes.
"Corrected" how? So long as it's stable, it doesn't need any correction. As for the phase reversal, that means nothing. How many times did that occur between the mic in the recording studio and your speeks?
The only time that it matters is in a feedback situation where the difference between 0deg and 180deg is positive feedabck and negative feedback.
Total sound polarity is generally more important if you have more than one sound reproduction device and source going on at the same time (like on stage)...generally. If it really bothers you, and some people are pretty serious about this type of thing, you can swap your speaker leads so they're out of polarity on the power amp. Ta da!
Regarding your 12V supply. If you want to turn your 12V into 12.6V then simply put a Small sig diode between your regulators gnd reference. If your using the 7812 regulator then it will raise its output voltage by 0.6V or whatever the voltage the diode rops across it. The regulator will still work fine with no new side effects caused by adding the diode. On the LM7805 this will be the middle leg that normally goes to ground. Place the diode between this leg SO IT CONDUCTS DOWN TO GROUND. If your unsure i suggest you check the output befor re-connecting the heaters.
Leigh
Leigh
thks for the inputs guys. it has been very informative. the LM7812 with the diode in the middle leg does work like a treat now!.....up by 0.6volts.
the sound of the 12AU7 seems to be a bit too bright to my taste and hence i wish to tone it down a little. i'm trying to understand this from some analysis ie. ac/dc loadline shift but have some problems.
1 - the stage 2 valve does not have a Ra, so what would be the Ia?
2 - can the Vgrid be calculated fr the cct? how pls?
3 - there are 2 coupling caps from the stage 2 to output . why is this and can they be replaced buy just 1?
4 - there are 3 resistors parallel in cathode cct stage 2. why is this so?
5 - 10pF between grid and cathode of stage 1. why is this and can this be removed?
thks folks.
the sound of the 12AU7 seems to be a bit too bright to my taste and hence i wish to tone it down a little. i'm trying to understand this from some analysis ie. ac/dc loadline shift but have some problems.
1 - the stage 2 valve does not have a Ra, so what would be the Ia?
2 - can the Vgrid be calculated fr the cct? how pls?
3 - there are 2 coupling caps from the stage 2 to output . why is this and can they be replaced buy just 1?
4 - there are 3 resistors parallel in cathode cct stage 2. why is this so?
5 - 10pF between grid and cathode of stage 1. why is this and can this be removed?
thks folks.
Attachments
puginfo said:
#1, #2) Since you already have it built, just measure.
#3) Big capacitors like that 4u7 have a good deal of internal inductance. The smaller one bypasses that for improved high frequency performance.
4) They wanted a 40K load.
5) Probably for RF rejection. You can probably remove it if RFI isn't going to be a problem.
12AU7s are considered less than premium for audio. (I've seen way more RF and quasi-digital circuits using these puppies than audio, and most of that has been gee-tah amps.) You're probably getting some h3 in there, causing a bright, edgy, aggressive sound. About the only thing you could do is to try to linearize it with some gNFB. Otherwise, try a more linear type such as the 6SN7 next time.
Tutorial - Long
puginfo,
See what Miles wrote above, here is teh "full rave".
A bit of "TLAR" (That looks about right) analysis on your posted schematic - a Tutorial if you like.
Lets look at the 1st stage (Common Cathode)
Its bit of an iterative process:
In general that first stage should have the anode voltage at about half the +250V rail.
For that to be the case you need 125V drop across the 82K Ra
125/82K => 1.5mA tube current.
Download the Anode Characteristics graph of the 12AU7 data sheet.
Thats the graph which has Plate Current vs Plate Voltage for various Grid Voltage.
If the tube has no current through it then there will be no voltage drop in Ra so the plate Voltage will be 250V. Place "dot" on the bottom (Plate Voltage) axis at the 250V point.
If the tube is "Full ON" then there will be no voltage dropped across the tube and the current would be set by the 82K (Plus the 3K3 but ignore that) and would be 250/82K = 3mA
Place "dot" on the vertical Plate Current axis at 3mA.
Draw a line through the 2 "dots" - This is your "load line" and the tube MUST be operating at some point along that line.
From that "load line" you can see that for Plate Voltage of 125 (half the rail) then the grid bias voltage would be approx -7 volts, you need to interpolate between the Eg = -6 and Eg = -8 curves.
Actually you don't have -7V on the grid. The grid is at 0V and the cathode is at +7V to give you Vgk = -7V
Now that 7V is dropped across the cathode resistor
So 7V/1.5mA => Cathode resistor should be 4.6K
BUT you have Cathode Resistor of 3.3K so our assumptions above are a bit off. Go around the loop again and again till you get to:
Assume Tube Current = 1.65mA
Va then is 250 - 82K x 1.65mA = 115V
Bias from load line = -6.0V
6V/1.65mA => 3K6 for cathode resistor. Thats close enough to what you have.
So your tube current in the 1st stage is about 1.65mA
Your voltage at the anode is approx 115V
Your voltage on the cathode is approx 6V
At this point if you want to get really fancy you can look at the data sheet "Mutual Characteristics" graph and read off the following values.
For a Plate Current of 1.65mA
mu (u) = 14.5
gm = 0.65 mA/V (650 umhos "micro-mho"s)
These values are useful to calculate gain, output impedances etc., we wont go there.
You asked "What does the 10pF from Grid to Cathode do?"
Well that depends - if that 3K3 cathode resistor really has a 47pF across it then it will give some Radio Frequency pickup supression. It MIGHT stop you from hearing the local taxis talking to their base. I'd leave it in.
You asked "Why no Ra in the 2nd stage?".
Well in effect it does have an Ra - Its those 3 x 120K in parallel (=40K). Its just been shifted to the cathode side to make that stage into a cathode follower.
As far as analysis goes, the power supply, tube, Ra series connected loop remains the same.
They have just use 3 x 120K resistors instead of 1 x 40K to average out the power dissipation across 3 devices (More on this later).
Now lets analyse the 2nd stage.
Using Ra = 40K draw a load line as before between 250V at 0mA and 250/40K = 6.25mA at 0V.
We know that the Grid Voltage = 115V because its connected to the anode of the 1st stage.
The cathode volatge will be higher than this (to give us our vgk = -bias volts).
A squizz (technical term) at the load line suggests that the cathode voltage will be about 120V giving an effective Vgk bias voltage of -5V and Plate Current Ia = 3mA.
Revisit above Power = I squared x R = (0.003)^2 x 40K = 0.36 Watts.
Today I would use a 39K 2W metal film resistor and be done with it.
When this design was done they obviously decided to use 3 x 120K 1/2 watt Carbon Comp resistors in parallel so each one dissipated 0.12 Watts.
You asked "Why 2 caps in parallel?" at the output.
When this was designed that 4.7uF cap would have been an electrolytic (with the +ve end to the tube cathode). Electrolytics in those days were seriously compromised by available technology and as frequency went up they stopped behaving like capacitors and started behaving like inductors (chokes) instead. It was (and still is) a usefull trick to parallel the big cap with a film cap of about 1/100th (in this case 1/30th) of the value. At higher frequencies the little cap takes over. If you are using a 4.7uF film cap the the 0.15uF is'nt strictly required. If you are using an electrolytic it IS. For maximum performance at higher frequencies I've found it is normally better to do this parallel "trick" regardless of whether the big cap is an electrlytic or a film cap. Some of the guys who spend $30 to $50 on audio "jewellery" capacitors disagree.
There - I think I've answered all your questions and maybe (hopefully) made things somewhat clearer to you.
As far as mods to the circuit go:
The input stage is running at a very low current. A lot lower than usually sounds best for a 12AU7.
You want some homework?
To get the best sound always try to set things up so that there is no net AC current draw from the power supply. The 2 stages have current draws which are 180 degrees out of phase with each other.
To see how much of the above you have absorbed.
As a 1st approximation to no net AC current draw try to calculate new RA and RK for that first stage such that it operates at 3mA (the same as the 2nd stage).
If you come up with values of Ra = 47K and RK = 1K2 or 1K5 you get 95%, If you decide that RK should be 2 x 2K7 (=1K35) in parallel you get full marks.
Cheers,
Ian
puginfo,
See what Miles wrote above, here is teh "full rave".
A bit of "TLAR" (That looks about right) analysis on your posted schematic - a Tutorial if you like.
Lets look at the 1st stage (Common Cathode)
Its bit of an iterative process:
In general that first stage should have the anode voltage at about half the +250V rail.
For that to be the case you need 125V drop across the 82K Ra
125/82K => 1.5mA tube current.
Download the Anode Characteristics graph of the 12AU7 data sheet.
Thats the graph which has Plate Current vs Plate Voltage for various Grid Voltage.
If the tube has no current through it then there will be no voltage drop in Ra so the plate Voltage will be 250V. Place "dot" on the bottom (Plate Voltage) axis at the 250V point.
If the tube is "Full ON" then there will be no voltage dropped across the tube and the current would be set by the 82K (Plus the 3K3 but ignore that) and would be 250/82K = 3mA
Place "dot" on the vertical Plate Current axis at 3mA.
Draw a line through the 2 "dots" - This is your "load line" and the tube MUST be operating at some point along that line.
From that "load line" you can see that for Plate Voltage of 125 (half the rail) then the grid bias voltage would be approx -7 volts, you need to interpolate between the Eg = -6 and Eg = -8 curves.
Actually you don't have -7V on the grid. The grid is at 0V and the cathode is at +7V to give you Vgk = -7V
Now that 7V is dropped across the cathode resistor
So 7V/1.5mA => Cathode resistor should be 4.6K
BUT you have Cathode Resistor of 3.3K so our assumptions above are a bit off. Go around the loop again and again till you get to:
Assume Tube Current = 1.65mA
Va then is 250 - 82K x 1.65mA = 115V
Bias from load line = -6.0V
6V/1.65mA => 3K6 for cathode resistor. Thats close enough to what you have.
So your tube current in the 1st stage is about 1.65mA
Your voltage at the anode is approx 115V
Your voltage on the cathode is approx 6V
At this point if you want to get really fancy you can look at the data sheet "Mutual Characteristics" graph and read off the following values.
For a Plate Current of 1.65mA
mu (u) = 14.5
gm = 0.65 mA/V (650 umhos "micro-mho"s)
These values are useful to calculate gain, output impedances etc., we wont go there.
You asked "What does the 10pF from Grid to Cathode do?"
Well that depends - if that 3K3 cathode resistor really has a 47pF across it then it will give some Radio Frequency pickup supression. It MIGHT stop you from hearing the local taxis talking to their base. I'd leave it in.
You asked "Why no Ra in the 2nd stage?".
Well in effect it does have an Ra - Its those 3 x 120K in parallel (=40K). Its just been shifted to the cathode side to make that stage into a cathode follower.
As far as analysis goes, the power supply, tube, Ra series connected loop remains the same.
They have just use 3 x 120K resistors instead of 1 x 40K to average out the power dissipation across 3 devices (More on this later).
Now lets analyse the 2nd stage.
Using Ra = 40K draw a load line as before between 250V at 0mA and 250/40K = 6.25mA at 0V.
We know that the Grid Voltage = 115V because its connected to the anode of the 1st stage.
The cathode volatge will be higher than this (to give us our vgk = -bias volts).
A squizz (technical term) at the load line suggests that the cathode voltage will be about 120V giving an effective Vgk bias voltage of -5V and Plate Current Ia = 3mA.
Revisit above Power = I squared x R = (0.003)^2 x 40K = 0.36 Watts.
Today I would use a 39K 2W metal film resistor and be done with it.
When this design was done they obviously decided to use 3 x 120K 1/2 watt Carbon Comp resistors in parallel so each one dissipated 0.12 Watts.
You asked "Why 2 caps in parallel?" at the output.
When this was designed that 4.7uF cap would have been an electrolytic (with the +ve end to the tube cathode). Electrolytics in those days were seriously compromised by available technology and as frequency went up they stopped behaving like capacitors and started behaving like inductors (chokes) instead. It was (and still is) a usefull trick to parallel the big cap with a film cap of about 1/100th (in this case 1/30th) of the value. At higher frequencies the little cap takes over. If you are using a 4.7uF film cap the the 0.15uF is'nt strictly required. If you are using an electrolytic it IS. For maximum performance at higher frequencies I've found it is normally better to do this parallel "trick" regardless of whether the big cap is an electrlytic or a film cap. Some of the guys who spend $30 to $50 on audio "jewellery" capacitors disagree.
There - I think I've answered all your questions and maybe (hopefully) made things somewhat clearer to you.
As far as mods to the circuit go:
The input stage is running at a very low current. A lot lower than usually sounds best for a 12AU7.
You want some homework?
To get the best sound always try to set things up so that there is no net AC current draw from the power supply. The 2 stages have current draws which are 180 degrees out of phase with each other.
To see how much of the above you have absorbed.
As a 1st approximation to no net AC current draw try to calculate new RA and RK for that first stage such that it operates at 3mA (the same as the 2nd stage).
If you come up with values of Ra = 47K and RK = 1K2 or 1K5 you get 95%, If you decide that RK should be 2 x 2K7 (=1K35) in parallel you get full marks.
Cheers,
Ian
Miles Prower said:
thks for the input Miles. i was under the impression that the more linear the valve was, the more h3 would be there (ie. solid state). would it be possible to shift the Q point on the loadline towards more negative grid to induce warmth-h2?
yes i could try some NFB to see but where does it connect? stage 1 or 2? not sure of the 6SN7 but i have also been testing the CV4010 in cascade with this cct.......the bass seems to be deeper, mid staging improved and high trimble decay longer.
"To see how much of the above you have absorbed"
wow, Ian, thats exactly what i needed to be able to sit down and work out. Thks so much.
cheers
puginfo said:
That's not right. The more linear your active devices, the less harmonic and IMD you get. Solid state is the absolute worst when it comes to linearity, and that's why it requires such high levels of NFB to make it sound something like right. I'm not familiar with 12AU7s since I've been avoiding them. I'm not so sure you can run them as thin as you are here. If anything, you need to get the plate current of that input stage up, not down. Even then, the linearity won't be anything special.
You'd have to connect it to the input stage, and connect it like an inverting op-amp.
The CV4010 is probably a good deal more linear than the 12AU7. It's getting late here, so I'll have to get back to you on that.
Re: Tutorial - Long
Yes Ian, things are lot more clearer now.
As for the homework : 3mA of Ia stage 1
Ra = 125Vdc/3mA = 41.7k
For Vgk -7V (as the original cct)
Rk = 7V/3mA = 2.3k
As mentioned above, for Rk = 1.35k & 3mA, the Vg is actually 4V vs the 7V earlier.
Are we saying that the stage 1 should be better with 3mA = Ia and Vg = -4V?
Also, would be good if you can explain further on the net AC current draw as both the B+ and heater are DC for now.
cheers
gingertube said:
Yes Ian, things are lot more clearer now.
As for the homework : 3mA of Ia stage 1
Ra = 125Vdc/3mA = 41.7k
For Vgk -7V (as the original cct)
Rk = 7V/3mA = 2.3k
As mentioned above, for Rk = 1.35k & 3mA, the Vg is actually 4V vs the 7V earlier.
Are we saying that the stage 1 should be better with 3mA = Ia and Vg = -4V?
Also, would be good if you can explain further on the net AC current draw as both the B+ and heater are DC for now.
cheers
Here's something that should improve the performance greatly, and help get rid of the excessive brightness. You can run both sections of the 12AU7 hotter for much lower estimated THDs.
Here's the loadline for the input stage: First Stage Loadline.
This one doesn't have a too excessive Vpk, and can still be direct coupled to the cathode follower.
Cathode Follower Loadline.
This should give a Zo ~= 426R. Since it's a follower, there is minimal THD coming from this stage.
Here's the loadline for the input stage: First Stage Loadline.
This one doesn't have a too excessive Vpk, and can still be direct coupled to the cathode follower.
Cathode Follower Loadline.
This should give a Zo ~= 426R. Since it's a follower, there is minimal THD coming from this stage.
have set this cct up now with stage 1 : Rk=1.47k and Ra=41.0k. stage 2 same as before. details:
B+=230Vdc
stage 1:
Rk=1.47k Rk=1.47k
Ra=41.0k Ra=40.9k
Va=121V Va=119V
Vg=0v Vg=0V
Vk=4.0V Vk=4.0V
Ia=2.7mA Ia=2.7mA
Vgk=4.0V Vgk=4.0V
stage 2:
Va=230V Va=230V
Vg=121V Vg=119V
Vk=124V Vk=122V
Ia=3mA Ia=3mA
Vgk=3.0V Vgk=3.0V
not sure if reading are ok.
1 - B+ had dropped from 250V to 230V. should this be riased again?
2 - stage 1 Va is not at half of the B+. is this an issue?
3 - both sides of the valves do not have the same voltages. is this an issue? how can this be balanced?...can a trim pot be used?
4 - Vgk for both tsage are diff. is 4.0V and 3.0V .... an issues here.
the valves are running quite hot now and sounds different with sufficient brightness. just nice but still lacks warmess and bass. how do we go about now? thks
cheers
B+=230Vdc
stage 1:
Rk=1.47k Rk=1.47k
Ra=41.0k Ra=40.9k
Va=121V Va=119V
Vg=0v Vg=0V
Vk=4.0V Vk=4.0V
Ia=2.7mA Ia=2.7mA
Vgk=4.0V Vgk=4.0V
stage 2:
Va=230V Va=230V
Vg=121V Vg=119V
Vk=124V Vk=122V
Ia=3mA Ia=3mA
Vgk=3.0V Vgk=3.0V
not sure if reading are ok.
1 - B+ had dropped from 250V to 230V. should this be riased again?
2 - stage 1 Va is not at half of the B+. is this an issue?
3 - both sides of the valves do not have the same voltages. is this an issue? how can this be balanced?...can a trim pot be used?
4 - Vgk for both tsage are diff. is 4.0V and 3.0V .... an issues here.
the valves are running quite hot now and sounds different with sufficient brightness. just nice but still lacks warmess and bass. how do we go about now? thks
cheers
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