Princeton Reverb Preamp, Phase Inverter to ICEPower amp

[dbishopbliss]

I'm cleaning out the the basement and came across the ICEpower 50ASX2BTL chip amp I bought in 2022 so time to resurrect my idea of building a tube preamp for this thing. This is the same module that is used in Seymour Duncan Powerstage 170, Fender Rumble bass amp and I think a few others.


On page 27 figure 17 of the datasheet it shows how run the amp as a bridged mono amp using a balanced signal as input to the board where one of the signals is inverted. When I first floated this idea I was looking at transformers or chips to invert the signal, but then it occurred to me... Isn't that what a phase inverter does for a push-pull amp? Is there a reason why I can't just  connect the outputs of the phase inverter to the amps Vin Channel 1 and Vin Channel 2?

[HotBluePlates]

... time to resurrect my idea of building a tube preamp for this thing. ...


On page 27 figure 17 of the datasheet it shows ... using a balanced signal as input ... Is there a reason why I can't just  connect the outputs of the phase inverter to the amps Vin Channel 1 and Vin Channel 2?

You can do that, but you need a buffer of some kind between the phase inverter & the input of the ICE module.

The reason is on Page 26 of the manual you linked:
    -  The Input has a 1nF capacitor in parallel with a 270kΩ to ground.
    -  The Input Impedance will be 270kΩ at 80Hz, 130kΩ at ~590Hz, and 16kΩ at 10kHz.
    -  The manual notes a minimum impedance of 8kΩ, and you'd be wise to have a buffer able to drive a load of 8kΩ or less.

[pdf64]

Also, musicians seem to have a hard time dealing with amps that have balanced speaker outputs. Such gear isn't a good fit with the 1/4" jack connectors we use, as the sleeve and body also carries a hot signal. Speakon and similar connectors, in which all conductors are kept shrouded / insulated, seem better suited for such gear.

[tubeswell]

A perfectly balanced cathodyne has an output impedance of 1/gm at both outputs - so can be as low as 1k.  (So that boils down to not using cathode biasing for the cathodyne, in order to keep the load resistance evenly split.)

[HotBluePlates]

... you'd be wise to have a buffer able to drive a load of 8kΩ or less.

A perfectly balanced cathodyne has an output impedance of 1/gm at both outputs - so can be as low as 1k.  (So that boils down to not using cathode biasing for the cathodyne, in order to keep the load resistance evenly split.)

4 or 5 different things need to be juggled at once.

   - You're right about "1/gm," but we should consider:
   - For AC signals, the network after the coupling cap is in parallel with the tube's load, so
   - Best to keep plate & cathode resistors 1/5 or less of any impedance on the far side of the coupling cap, which means
   - Need a load resistor of 16kΩ / 5 = ~3.3kΩ or less per output.

Further, the low apparent output impedance of cathode followers, etc only happens if we draw no current from those outputs.  As soon as there's current-draw, we find 12AX7 followers can't keep up & deliver that current.


I think this stuff will be moot if the user simply picks a buffer of any-technology to place between the phase inverter & the ICEPower module input.

[pdf64]

Low output impedance is a small signal characteristic, it won't alter the fundemental AC loadline, such that the outputs may clip at too low a signal output level to be any use.

[dbishopbliss]

I think this stuff will be moot if the user simply picks a buffer of any-technology to place between the phase inverter & the ICEPower module input.

I forgot to mention that I would like this to be a pedalboard amp so I'm trying to keep it small. Perhaps including the phase inverter is not really a good idea.

Would it be "better" to use something like a TL072 chip to invert the signal? Although, I know nothing about making circuit boards and using these types of chips. Ghent Audio offers a RTX Module that is supposed to convert an unbalanced signal to a balanced signal and was designed specifically to be used with ICEPower amps. But, as I understand it, these chip based solutions will require a 24V power supply adding size and complexity to the project.

Another possible option would be to use a transformer such as the Jensen JT-11P-1 (or some more cost effective alternative). Could I simply use the output from the tube gain stages into the transformer and then connect that to the ICEPower module or would I still need some sort of buffer.


On the other hand... as I understand it the phase inverter contributes a lot to the character of a tube amp. So perhaps I would be better off running the outputs of the phase inverter into a Cathode Follower as described by The Valve Wizard. This would require an extra tube so the enclosure size is going to be bigger, but I would already have the power supply so really not that much bigger.

Any thoughts?

[dbishopbliss]

After doing some reading, I think a Princeton Reverb preamp is more what I need. I like having a separate Treble and Bass controls (plus its easy to add a mid control as well). Also, I don't need two separate channels.


I found a DIYLC layout by Rob Robinette where he removed the Reverb and Tremolo circuits - also not necessary for my needs. I modified his diagram by removing the power tubes and adding the Cathode Follower referenced in the last post. I realize that I omitted the load resistor and B+ from the diagram because I'm not really sure what values I should put there.


I also expect to simplify the power section. I don't think I have a need for that large a power transformer and all those voltage drops when I would not be using them. I would probably go with a solid state rectifier as well. Before I continue with that, am I on the right track?


Help with tweaking values is much appreciated. I don't understand the engineering part of designing an amp. I just like to build things.

[tubeswell]

I found a DIYLC layout by Rob Robinette where he removed the Reverb and Tremolo circuits - also not necessary for my needs. I modified his diagram by removing the power tubes and adding the Cathode Follower referenced in the last post.

I don’t think the CF will provide any impedance bridging benefit over and above what you’d get from having a perfectly balanced cathodyne (because the output impedance is the same in both cases).


Re HPB’s point about 16k impedance at 10kHz, or 130k at 580Hz, my back-of-envelope prediction is that you’d probably get a reasonable output impedance using a 12AU7 triode* for the cathodyne with a pair of hefty (decent wattage) load resistors of between 5k to 10k each (I.e. 10k to 20k together) and not need the CF buffers. You could always breadboard it and see


*i.e. the ‘AU’ half of a 12DW7

[dbishopbliss]

I don’t think the CF will provide any impedance bridging benefit over and above what you’d get from having a perfectly balanced cathodyne (because the output impedance is the same in both cases).


Re HPB’s point about 16k impedance at 10kHz, or 130k at 580Hz, my back-of-envelope prediction is that you’d probably get a reasonable output impedance using a 12AU7 triode* for the cathodyne with a pair of hefty (decent wattage) load resistors of between 5k to 10k each (I.e. 10k to 20k together) and not need the CF buffers. You could always breadboard it and see


*i.e. the ‘AU’ half of a 12DW7

I've updated the diagram removing the CF so it is a cleaner version of my original post except using the Princeton as opposed to the Deluxe. I also updated it to use the AU half of the 12DW7 as you suggested. I suspect I will have to update the load, grid stopper, and cathode resistor value for an AU tube.

I found a ValveWizard chapter from his online book that examines the Cathodyne in more detail. I have added the load resistors as you described. Please let me know if I have interpreted his text and your post correctly. He also suggests bypassing with some caps, etc. I might add those later.

[tubeswell]

I suggested experimenting with a 10k to 20k (combined) load for the 12AU7 (in order to improve the impedance bridging into the load). The lower plate resistance (which is different from the load resistance) will result in more current into the load under signal conditions, but it also necessitates ensuring an appropriate bias is achieved to keep the tube within dissipation limits. You probably should run some load lines to guesstimate the bias voltage needed. I also suggested running a perfectly balanced cathodyne to ensure the 1/gm output impedance condition at both PI outputs. This will probably necessitate going with a fixed bias method (given that a cathode biasing resistor will make the stage more unbalanced if a 10k plate load is adopted, which would then increase output impedance and defeat the purpose of aiming for better impedance bridging).

[dbishopbliss]

I build things but have only enough knowledge of design to be dangerous. :-D That means I need things more spelled out for me because I recognize the terms you are using when looking at an existing circuit but as soon as I have to
start putting in my own values and knowing the result I am lost.

I suggested experimenting with a 10k to 20k (combined) load for the 12AU7 (in order to improve the impedance bridging into the load). The lower plate resistance (which is different from the load resistance) will result in more current into the load under signal conditions, but it also necessitates ensuring an appropriate bias is achieved to keep the tube within dissipation limits. You probably should run some load lines to guesstimate the bias voltage needed.

I'm confused by what you mean by 10k to 20k (combined) load. Are you saying, draw out load lines for the 12AU7 where the load resistor and the plate resistance of the 12AU7 (~7K) is between 10k and 20k? So bascially, replace the 56K resistor on my diagram with something between 3k and 13k?

I also suggested running a perfectly balanced cathodyne to ensure the 1/gm output impedance condition at both PI outputs. This will probably necessitate going with a fixed bias method (given that a cathode biasing resistor will make the stage more unbalanced if a 10k plate load is adopted, which would then increase output impedance and defeat the purpose of aiming for better impedance bridging).

I need to find how to make a perfectly balanced cathodyne. Its probably in the valve wizard pages somewhere. Hopefully it has how to make it fixed bias as well.


By the way... the name of my band is Fixed Bias. :-D

[passaloutre]

Look into how Ampeg biased the cathodyne in, say, the Gemini circuit. The grid is placed in the middle of a voltage divider from B+, rather than the simple pull-down/grid-leak of the Fender cathodyne. This fixed bias offers more control over where the grid sits with respect to the cathode voltage. I think this also improves balance, as you don’t have the separate cathode bias resistor that is otherwise required.

The “Paul C mod” for the Princeton Reverb is similar.

[tubeswell]

Yes. Try 2 x 5k6 load resistors with fixed bias. You want to set the bias voltage up to aim to have the voltage drop across each resistor being 1/4 to 1/3rd HT with the remaining 1/2 to 1/3rd being taken up by the plate-to-cathode voltage. Say HT is 360V, then that’s 120v across reach load resistor, so each resistor would be seeing 120/5600 = 21.43mA on average and would need a minimum rating of 5W. And the voltage divider for the fixed bias would want to be something that puts the grid voltage at wherever the Vg curves cross the mid point between Vg=0 and the x-axis of your (11k2) load line. I haven’t bothered working it out but my hunch is somewhere around 8v below the cathode voltage, but in practice that’s somewhere between 1/3rd to 1/4 HT. The aim is 1/3rd, 1/3rd, 1/3rd (or 1/4, 1/2, 1/4)

[dbishopbliss]

Yes. Try 2 x 5k6 load resistors with fixed bias. You want to set the bias voltage up to aim to have the voltage drop across each resistor being 1/4 to 1/3rd HT with the remaining 1/2 to 1/3rd being taken up by the plate-to-cathode voltage. Say HT is 360V, then that’s 120v across reach load resistor, so each resistor would be seeing 120/5600 = 21.43mA on average and would need a minimum rating of 5W. And the voltage divider for the fixed bias would want to be something that puts the grid voltage at wherever the Vg curves cross the mid point between Vg=0 and the x-axis of your (11k2) load line. I haven’t bothered working it out but my hunch is somewhere around 8v below the cathode voltage, but in practice that’s somewhere between 1/3rd to 1/4 HT. The aim is 1/3rd, 1/3rd, 1/3rd (or 1/4, 1/2, 1/4)

Based upon the voltages in the Princeton schematic, HT will be 240V. 1/3rd will be 80v. So that means 80/5600 - 1.43mA. So, 1.43*240v = 3.43W or 5W for a common value. 

I plugged the values above into a loadline calculator and get the following:



I picked an operating point of around 4mA or Vg = -6V because the curve look more linear around there. It also happens to be almost exactly halfway. So if I am understanding this correctly, the value of Rb (fixed bias divider) would be 6V/4mA = 1.5K (1/2W).

Do I have that right?



I realize that maybe I have entered the wrong value for Ra=Rk in the calculator. I doubled it as you did, but perhaps it should be the actual value and the calculator doubles it automatically. In that case, Rb should be 7.5/5.1 which is still pretty close to 1.5K (1.47K) so I guess it doesn't matter.

[tubeswell]

I realize that maybe I have entered the wrong value for Ra=Rk in the calculator. I doubled it as you did, but perhaps it should be the actual value and the calculator doubles it automatically. In that case, Rb should be 7.5/5.1 which is still pretty close to 1.5K (1.47K) so I guess it doesn't matter.

Yes that load line is not steep enough- that box needs to be 5600.


(Quick check is HT voltage all dropped across the load gives the y-axis coordinate, in this case 240/11200 = 21.43mA is where the load line should be hitting the vertical axis)

[dbishopbliss]

Slight detour... I plugged in the values from the Princeton schematic and got the following Graph. Also plugged in the values from Sluckey's Magnatone M-10A. The load lines are very shallow compared to the recommended load of 5K6 (5600R). Is there a reason for such a change?


Princeton:



Magnatone:

[pdf64]

Consider that the next stage input impedance (ICE power module ) is 16k, not the 270k you've used in the calculator.

[dbishopbliss]

As a reminder to myself and anyone who might follow this thread in the future, here are links to the resources I have been  using:

• Rob Rob Princeton Reverb With Reverb and Tremolo Deleted
• Sluckey Magnatone M-10A
• Valve Wizard Cathodyne Phase Inverter Page
• Valve Wizard Online Book Cathodyne Chapter
• Concertina (Cathodyne) Load Line Calculator
• Voltage Divider Calculator
• Imgbb free image hosting service
• ICEpower 50ASX2BTL

[tdvt]

Can I also ask what you used to draw your schematic? ("Incomplete Draft" drawing)

[dbishopbliss]

Can I also ask what you used to draw your schematic? ("Incomplete Draft" drawing)

I have been using the DIYLC program that Hoffman uses for drawing turret boards.

[dbishopbliss]

I have updated the schematic to use a Fixed Bias Cathodyne. I'm currently using the recommended load resistors of 5K6.


Using the load line calculator it seems the quiesence point is -5.63 Vg and 6.5mA. So 6.5mA * 5.6K = 36.4V, minus the grid line voltage = ~31V. According to Valve Wizard, this value should be 1/6 of HT (240V) so it is a bit low. Not sure if that is bad so I might have to come back to this.


Then using the voltage divider calculator I found I can hit the bias voltage by using values of 10M and 1.5M. How do I determine what power rating do these resistors need to be?


I just realized the bottom of the voltage divider is going to the wrong place, but you get the idea. Will fix soon.


image upload

[passaloutre]

There is very little current associated with the bias circuit, so you're fine using standard 1/2w parts there. You could go smaller, but why bother?

More quantitatively, you've got 11.5 megohms across 240 volts, so:

240 = I  * 11,500,000

I = 0.00002 amps = 0.02 mA

P = 0.00002 * 240 = 0.01 watts

[dbishopbliss]

Playing around with some numbers... since VW says the bias point target should be 1/6 HT (not sure why), that means I should shoot for something around 40V. That's really hard (if not impossible) to get to using the 5K6 load. But, if i change the load to 15K with a Q 3.2mA/ 144Va / -6.1Vg then my bias voltage is 3.2mA * 15K = 48V minus the 6.1V ~42V, which is pretty close to the target recommendation. Then, I have a voltage divider where Vs=240, R1=10M and R2=2M to give me a voltage of 40V.


Is there a reason to go with the 5K6 load vs 15K load? As I mentioned earlier, many other designs seem to go with much larger loads.

[tubeswell]

The 10:1.5 voltage divider may leave the bias too cool but you could start with it and if it doesn’t work, keep reducing the 10M until it works. The goal is to get a bias value that enables maximum signal without clipping, gain-spike, or frequency-doubling distortion.

In terms of the load, the higher you make the load, the worse the impedance bridging will be. 16k/(5k6+16k) is a better ratio than 16k/(15k+16k) when you don’t want to reduce gain.

[tubeswell]

Also, it would be useful to increase the HT voltage to free things up a bit.

[dbishopbliss]

If I up the HT to 320V, target bias voltage is 320/6 = 53V.


Then Q = 8.4mA, 226Va, -8Vg.


So bias voltage = 8.4*5600 = 47V, minus 8V = 39V. So pretty far of the recommended 53V.


I could move the Q to 10mA, 208Va, -6.35vG, which would give me a bias voltage of 49.65V. So pretty close now. Then I could use 10M and 1.8M for the voltage divider to get 48.8V, or even 10M and 2M to land exactly on 53V.


How important is that 1/6 HT target?



Also, does that amp really need that large of a voltage swing to drive it? I don't know what to look at to determine that value.

[tubeswell]

Given the target input impedance of the power amp is 16k, the bigger the clean output signal you can get from the PI, the better (because the impedance bridging will attenuate the signal somewhat, and if the cathodyne load resistance is ‘higher’, there will be more attenuation). So increasing the HT voltage gives you more to play around with.


As to the target bias voltage, it’s more important to tweak it to get 1/4 to 1/3 division across the load resistors (in order to maximise the available clean output signal amplitude)

[dbishopbliss]

Given the target input impedance of the power amp is 16k, the bigger the clean output signal you can get from the PI, the better (because the impedance bridging will attenuate the signal somewhat, and if the cathodyne load resistance is ‘higher’, there will be more attenuation). So increasing the HT voltage gives you more to play around with.

I will increase the HT voltage. What are your thoughts on the options I shared for the Q?
Option 1: Q = 8.4mA, 226Va, -8Vg, 39V bias
Option 2: Q = 10mA, 208Va, -6.35Vg, 53V bias
Is one option "better" than the other? How would I know?

[tubeswell]

Either one will work to begin with and you can tweak the setup to optimise it once it’s running.

[dbishopbliss]

Either one will work to begin with and you can tweak the setup to optimise it once it’s running.

Will there be something obvious that will let me know it needs to be tweaked or is this something that would require a scope (which I don't have)?

[tubeswell]

Either one will work to begin with and you can tweak the setup to optimise it once it’s running.

Will there be something obvious that will let me know it needs to be tweaked or is this something that would require a scope (which I don't have)?

A simple trial & error method is set the voltage divider output to 1/3 HT to begin with and check the plate voltage and plate dissipation. A single AU triode can handle 2.75W so if it’s idling under that you’re good. Now in theory, if you aim for 1/3rd HT for cathode and 2/3rds HT for plate, the hunch is we’ll get a plate-to-cathode of 107V and with 11k2 total load about 0.022A, which is Pdiss = 2.4W; good to go. But Vgk wants to be -7ish volts for the ‘ideal’ condition, so instead of 107 (1/3rd HT), you want to try 100V, so that’s going to be 100/320ths, so 1M:2M2 ought to be about right I reckon. If it red plates then cool the bias off some more.

[passaloutre]

An easy way to tune the plate dissipation might be to make the bottom half of your bias divider with a potentiometer. A 1M pot would work, but hard to fine-tune. Maybe more like a 100k or 250k pot in series with 900k or 750k resistor.

[dbishopbliss]

I was wondering how Tubeswell came up with the 5600 load resistors value. I think this is how he came up with that number (or he just has experience and knows this stuff). 

I started with the plate characteristics chart for a 12AU7. Then I drew a line from my HT voltage of 320V to some value that looked good but was below the max plate dissipation - I chose 30mA. Then I determined the load by dividing 320V / 0.03A = 10666 Ohms. Divide that in half between the Plate and the Cathode and you get 5333 Ohms. The closest standard value resistor is 5600.


Oh, and I also measured and indeed the gridline that is the halfway point along the loadline is -8V.


Is that right?

[tubeswell]

Full marks!

You can clearly see how you’re utilising the full potential of the tube.

(As far as knowing what to do, the exercise is looking for a load resistance that gives the ‘best’ impedance bridging into the 16k@10kHz input impedance of the following device with the lowest parts-count that we can get with an off-the-shelf tube, knowing also that the 16k is the lowest input impedance of the power amp device, and so if we can get a reasonable impedance bridge for that, it might not sound too ‘muffly’)

Also, we're dealing with a cath0dyne which has unity gain, so the gain of this stage is no different whatever tube you're choosing, and it might as well be a tube that is capable of driving a low R load line - so 12AU7.

The output impedance of the driving stage and the input impedance of the following stage form a voltage divider that attenuates the signal. So 5k6 is the upper leg and 16k is the worst-case lower leg (for each output of the cathodyne). Usually (with any impedance bridge), a 1:5 ratio gives reasonable bridging without too much loss, and here we’ve got about 1:3 so there will be some loss at 10kHz (but still not too bad, as the device’s input impedance for lower frequencies that you find on a guitar fretboard is higher, so the signal attenuation won’t be as bad for what we normally expect to hear). Having noted all that, it’s all theoretical until you get it operational, so will your ears like the result? YMMV

[dbishopbliss]

A simple trial & error method is set the voltage divider output to 1/3 HT to begin with and check the plate voltage and plate dissipation. A single AU triode can handle 2.75W so if it’s idling under that you’re good. Now in theory, if you aim for 1/3rd HT for cathode and 2/3rds HT for plate, the hunch is we’ll get a plate-to-cathode of 107V and with 11k2 total load about 0.022A, which is Pdiss = 2.4W; good to go. But Vgk wants to be -7ish volts for the ‘ideal’ condition, so instead of 107 (1/3rd HT), you want to try 100V, so that’s going to be 100/320ths, so 1M:2M2 ought to be about right I reckon. If it red plates then cool the bias off some more.

I hope everyone enjoyed the long weekend. I didn't get to work on this project because I had a sick kitten at home. He's feeling better, but I'm still dealing with the less fun end of his gastric distress. :-/ 

I've been trying to determine how the 1/3HT relates to the load line I drew, or does it?

100V lands below the 0.0V grid line whereas -7.0 lands at about 215V. Is the 1/3 HT value you reference the 320V - 215V = 105V?

Referring back to Valve Wizard section 12.2.3 and substituting my values

Continuing with the earlier example, the bias voltage was −7V and the quiescent anode current was 9.5mA. This current flows in Rk, so the cathode voltage will be 9.5mA×5.6kΩ = 53.2V. We want the grid to be 7V below this, or 46.2V. This is indeed about one seventh (VW suggests 1/6) of the 320V HT and can be closely approximated with a 10MΩ/1.8MΩ divider, yielding 48.8V.

Which values should I be using?


Thanks.

[tubeswell]

If HT=320, then 1/3HT is 107 (where the cathode sits*) and so the grid wants to be ~7V below that. I.e., at 100V. The back of envelope calculation I used earlier for a voltage divider output of 100V with an HT320 is 100/320ths, and so the resistors I’d pick initially would be 2M2/1M. (because 1M/(1M + 2M2) = 1:3.2)

*Under this scenario, with a balanced cathodyne stage, we’d expect the plate to sit at 2/3HT, and so there would be 1/3HT devoted to each output and to the input, which in very simplistic terms means that we may expect to get the maximum available clean input and output all around in a unity-gain stage.

[dbishopbliss]

If HT=320, then 1/3HT is 107 (where the cathode sits*) and so the grid wants to be ~7V below that. I.e., at 100V. The back of envelope calculation I used earlier for a voltage divider output of 100V with an HT320 is 100/320ths, and so the resistors I’d pick initially would be 2M2/1M. (because 1M/(1M + 2M2) = 1:3.2)

*Under this scenario, with a balanced cathodyne stage, we’d expect the plate to sit at 2/3HT, and so there would be 1/3HT devoted to each output and to the input, which in very simplistic terms means that we may expect to get the maximum available clean input and output all around in a unity-gain stage.

Does the 100V value have any relation to the loadline I drew? (I was using -8V but its pretty easy to interpolate to -7V). Or is it simply a rule of thumb to use 1/3HT?

[tubeswell]

Does the 100V value have any relation to the loadline I drew? (I was using -8V but its pretty easy to interpolate to -7V). Or is it simply a rule of thumb to use 1/3HT?

The chart is for plate characteristics so you calculate the plate idle point and bias with that. And we combine both load resistors in the cathodyne to get the load line slope, because each resistor is half of the total load. (And the total resistive load gives us the load line gradient at a given HT voltage).
However, for a cathodyne, we split the load into plate load and cathode load, and assume that with equal load resistors, that the voltage drop across each resistor will be the same. So looking at your chart above, we see the plate idle point (for a centre-bias of about 8V) is about 2/3rds HT, (and that would be the same if we were running a normal inverting stage with an 11k2 load). So knowing that we’ve split the load for the cathodyne, it follows from this that the plate voltage will still be the same (2/3rds HT, because overall, it’s still an 11k2 load), but as the load is equally split for the plate and the cathode (I.e., into 2 x 5k6), it follows that the cathode voltage will be the same as the plate voltage (albeit the Vk will be referenced to ground). So it’ll be 1/3rd of HT. (And so we want our grid bias voltage about 7 or 8 volts below that)

[dbishopbliss]

Is this a simplistic explanation...


The loadlline on the chart is for both the Plate and Cathode. Because the values are the same (5.6K), the 214V (for -7.0V) will be evenly divided between them for 107V. Since we want the grid to be -7 volts below this we get 100V.

[dbishopbliss]

On to the power supply... I don't want (or need) the large power transformer used when there are power tubes involved. I have an Allied 6K1VF that I used for a stereo preamp that used two 12AU7 tubes. It is rated 250VCT, 25mA DC primary, 6.3V, 1A filaments so I think it could work.

I should probably know this but I forgot... how do I know how much current this circuit will draw? Do I use the values from the tube datasheet for a 12AX7 (1.2mA * 3) + 12AU7 (10.5mA) = 14.1mA. Assuming that is correct (big assumption), then using PSUD2 to model a Solid State Full Wave Rectifier I get the following. Do you see any issues?


Update: I just did some calculations looking at the circuit for the Princeton and it seems the current for the three 12AX7 stages at quiescence is 0.88mA each. Then, I know the current for the PI is 10mA, so that is a total current of 12.64mA. Should I adjust the current sink in my model to use that value?

[tubeswell]

If you’re going with a 12AU7 cathodybe HT=320V and a loadline of 11k2, that stage alone will draw 22mA. As to the other preamp stages, rule of thumb is 1mA per triode for a 12AX7. Are you wanting Reverb in your preamp? The solid state device will want a separate PS I’m assuming.

[dbishopbliss]

If you’re going with a 12AU7 cathodybe HT=320V and a loadline of 11k2, that stage alone will draw 22mA.

I was using 10mA based on where the Q of the loadline was set. How did you come up with 22mA?

Using the 22mA value, plus 3ma for the 12AX7 sections I will have 25mA - which is exactly what my transformer is rated for. Does this mean that it is OK to use or does there need to be some additional capacity?

Are you wanting Reverb in your preamp? The solid state device will want a separate PS I’m assuming.

Nope. This is (hopefully) going to be a pedalboard amp. Reverb will be a pedal. I might add an effects loop later but I want to get the simpler version going first.

[tubeswell]

If you’re going with a 12AU7 cathodybe HT=320V and a loadline of 11k2, that stage alone will draw 22mA.

I was using 10mA based on where the Q of the loadline was set. How did you come up with 22mA?

Using the 22mA value, plus 3ma for the 12AX7 sections I will have 25mA - which is exactly what my transformer is rated for. Does this mean that it is OK to use or does there need to be some additional capacity?

Are you wanting Reverb in your preamp? The solid state device will want a separate PS I’m assuming.

Nope. This is (hopefully) going to be a pedalboard amp. Reverb will be a pedal. I might add an effects loop later but I want to get the simpler version going first.

The 22mA was my lazy calculation from earlier. If you’re looking at 107v across each 5k6 resistor, that’s more like 19mA.

[dbishopbliss]

The 22mA was my lazy calculation from earlier. If you’re looking at 107v across each 5k6 resistor, that’s more like 19mA.

So the current isn't based on the Q value its based on the voltage going through the resistors.

Just to confirm I am understanding...

The Princeton Reverb Schematic shows the B+ as 240V before the load resistor in the first three gain stage. The load resistor for each is 100K and the voltage at the plate is 160V so a drop of 80V. Using Ohm's Law, I = V / R works out to 0.8mA = 80V/100K.

So my new total is 19mA + 0.8mA + 0.8mA + 0.8mA = 21.4mA.  Or, rounding up and accounting for variations 22mA + 3 * 1mA = 25mA.

Does this mean I can use the transformer I have or will I need something bigger? 

[tubeswell]

The current ratings would be fine for a 12AX7 and 12DW7. (I gather you’re going with 250-0 and FWB?)


But the ICE power amp would need a separate PS

[dbishopbliss]

I gather you’re going with 250-0 and FWB?

The transformer has a center tap so I was thinking of using a Two Phase rectifier.

But the ICE power amp would need a separate PS

ICE has the power supply built in. Just connect the mains. I will probably go with two separate power switches just in case the tubes are making noise while they are warming up. Either that or maybe I could have some sort of relay that would delay the power amp coming on for a few seconds.

[shooter]

I did a variant of this over-thought, over engineered, monster 

[tubeswell]

I gather you’re going with 250-0 and FWB?

The transformer has a center tap so I was thinking of using a Two Phase rectifier.

But the ICE power amp would need a separate PS

ICE has the power supply built in. Just connect the mains. I will probably go with two separate power switches just in case the tubes are making noise while they are warming up. Either that or maybe I could have some sort of relay that would delay the power amp coming on for a few seconds.

125-0-125 vac with 2-phase rectifier won’t make 320vDc. But if you use voltage doubler it should. (250-0 with FWB will make 350. )


If the power amp PS is switch-mode, expect noise.

[dbishopbliss]

125-0-125 vac with 2-phase rectifier won’t make 320vDc. But if you use voltage doubler it should. (250-0 with FWB will make 350. )


If the power amp PS is switch-mode, expect noise.

Ugh… I was using PSUD2 to model the power supply and it makes it seem possible.


I will look up doublers, but I would guess it might have a difference in the power rating. Either that or I adjust the circuit to use lower voltages.