Separate preamp power supply

[dscottguitars]

Hi,  I am building a separate preamp and I am not sure about the filtering for the power supply.  Below is what I have in mind.  Also, can anyone help me with figuring out how to calculate the voltages on the nodes?  I know the formula I=V/R but I don't know how to apply that to voltage drop.  The transformer I am getting is one used for a stand alone reverb unit, 0-260v @ 100mA.

My understanding:

365v/10K=0.0365A;   

1/.0365=27.4/100=.274;   365v*.274=100v;  365-100=265v

If I have:

365v/5K=0.073;   1/.073=13.7;   365*.137=50v;  365-50=315v

The main problem I'm having is another amp I built has these voltages in the power supply and I can't figure out how to get those values-see pic.

Thanks...

Daniel

[HotBluePlates]

Hi,  I am building a separate preamp and I am not sure about the filtering for the power supply.  Below is what I have in mind. ...

If you actually only use 3 filter caps, your ripple voltage may be higher than you think it will be. That's because you have several filter stages less than in a complete amp with an output stage.

Then again, your current draw will be very much less than a typical amp, and only about 1/50th of what the power transformer can supply, so you may get lucky. A careful sim or breadboarding will tell for sure.

... can anyone help me with figuring out how to calculate the voltages on the nodes?  ...

You decide what you want those to be, calculate the performance and current draw of the individual stages, then calculate what the dropping resistors will be to arrive at your desired node voltages. So picking resistors values (unless done on basis of calculated desired ripple reduction) is putting the cart before the horse.

365v is probably a reasonable voltage you could expect on your first filter cap, because 260vac * 1.414 = 367.6v peak, and there are 2 diode drops due to the bridge, so the filter cap will charge to 367.6v - (2 * 0.7v) = 366.2vdc.

Since 3 12AX7 stages will probably only draw 3mA or less, and your transformer will be able to deliver 100mA, sag will be non-existent. In fact, the under-loading of the transformer will probably result in higher voltage at the first filter cap than 366vdc, but how much more is uncertain until you breadboard this thing.

... is another amp I built has these voltages in the power supply and I can't figure out how to get those values-see pic.

You don't have voltage listed for the cathode resistors. Does the other amp have the indicated values for power supply nodes A & B, as well as the listed plate voltages for the preamp stages in your picture?

If so, calculate stage current draw using the supply node voltage and the preamp stage plate voltage.

Input Stage
Plate voltage shown: 225v
Supply node voltage: 265v (Node A shown)
Plate load resistor: 47kΩ
Stage current draw: (265v - 225v)/47kΩ = 0.85mA

Channel One 2nd Stage
Plate voltage shown: 135v
Supply node voltage: 265v (Node A shown)
Plate load resistor: 47kΩ
Stage current draw: (265v - 135v)/47kΩ = 2.77mA  (<- Looks suspicious; I'd verify supply node voltage for this stage and/or voltage across cathode resistor)

Channel Two 2nd Stage
Plate voltage shown: 150v
Supply node voltage: 215v (Node B shown)
Plate load resistor: 47kΩ
Stage current draw: (215v - 150v)/47kΩ = 1.38mA

NOTE: Positions of the input stage and Channel Two's 2nd stage seem backwards. The input stage is almost always supplied by the lowest voltage power supply node, which is furthest from the PT. That's because it has the most filtering and the cleanest d.c., and signal levels are small so high supply voltage is not necessary.

Now calculate the dropping resistor values. The total current for all stages will flow through the 10kΩ you have drawn. Voltage drop is the total current times the dropping resistor value.

Voltage drop to Node A
Total Current = 0.85mA + 2.77mA + 1.38mA = 4.24mA
Voltage drop = 4.24mA * 10kΩ = 42.4v
Node A Voltage (Calculated) = 366vdc - 42.4v = 323.6vdc

Note this is not the 100v difference you have indicated between the first and second filter caps. Again, I don't know where the voltages came from for the different points shown, but obviously they don't match in different places on the schematic.

For Node B, the only current through the dropping resistor will be the current drawn from Channel Two's 2nd Stage. That's because this is the only stage you have indicated as connected to this point, and the stages connected to Node A have already had their current diverted away from the dropping resistor and down through those stages.

Voltage drop to Node B
Total Current = 1.38mA (Channel Two 2nd Stage)
Voltage Drop = 1.38mA * 5kΩ = 6.9v
Node B Voltage (Calculated) = Node A Voltage - Voltage Drop = 323.6vdc - 6.9v = 316.7vdc

Again, schematic values don't match calculated values. Judging from experience, the dropping resistor values seem unrealistic and way too small. Correct values would be determined by selecting desired supply voltage level and measured/calculated current draw for that portion of the circuit (as I mentioned before) to select resistor values which will result in the desired supply voltage levels.

[HotBluePlates]

My understanding:

365v/10K=0.0365A; 1/.0365=27.4/100=.274;   365v*.274=100v;  365-100=265v

If I have:

365v/5K=0.073; 1/.073=13.7;   365*.137=50v;  365-50=315v

Your first step (bold in each) is incorrect; respectfully, as a result of the step 1 mistake the rest of the math is meaningless manipulation of numbers to try to arrive at some answer.

When you wrote 365v/10k = 0.0365A, you are finding a current through a 10kΩ resistor if all 365v is dropped across the 10k resistor. That is, if the voltage at Node A is zero, then there is 0.0365A (which is 36.5mA) through the 10kΩ resistor. This could only happen if Node A's 22uF filter had its + terminal shorted to ground, as none of the preamp stages drawn could ever draw 36mA even in fault conditions.

Similarly, your first step for the second calculation implies 365v dropped across 5kΩ, which results in 73mA of current.

So again, the correct method is to draw plate and cathode resistor loadlines for each stage with assumed, reasonable supply voltages to determine from the tube's plate curves what the idle current will be. Then, knowing current draw for each stage, calculate what power supply resistors will provide the needed voltages. Then you double-check your work by calculating the resulting ripple voltage at each point in the power supply to verify the calculated resistors provide the needed level of filtering (this last step hasn't been shown here).

[dscottguitars]

Okay, I have quite a bit to learn about how to apply the math to the stages.  To answer some of your questions:

The second schematic of the amp I have built has the actual voltages as stated on all points within a couple of volts as I rounded the values stated.  The only correction is the screen resistor values are 470, not 330.  I want to use those voltage values as targets for the preamp I want to make.  But I do not know for sure what dropping resistors to use in the power supply so I guessed based on what I have in the one amp and what I thought-obviously wrong.

The voltages on the cathode resistors were approximately 0.8 on channel 2, and 1.5 to 2.0 on the input and channel one.  I don't remember exactly but that is close to what I remember. 

The reason I have channel two on the last stage is I want less plate voltage on that stage for more distortion.  It seems to work fine with the amps I have made. 

Currently I have just switched resistor values in the power supply to get the voltages I desire.  It's easy because I didn't know the correct way to apply load lines, current draw and such.

It looks like I am going to need a bigger value for the first resistor to get the 265v I want.

Based on your examples I still can't figure out how the values I measured correspond to the math for the amp I built.  I measured 42mA across each of the 1 ohm resistors on the 6L6's.

[HotBluePlates]

The voltages on the cathode resistors were approximately 0.8 on channel 2, and 1.5 to 2.0 on the input and channel one.  I don't remember exactly but that is close to what I remember.  

Fair enough. What node is the Input stage connected to? It's not indicated on your complete-amp schematic. Node D or E?

Is the 2nd stage of Channel 1 actually a 12AV7? Do you plan to use that tube again on your new build?

Is the 2nd stage of Channel 2 really the only stage connected to Node E?

[dscottguitars]

The input tube is on node D, I don't know how it got deleted.  And, yes it's a 12AV7, I have used them before and like the price.  A 12AX7 is too much and a 12AU7 isn't enough.  A 12AY7 is more money and I read the current in a 12AV7 is higher.

[HotBluePlates]

... I still can't figure out how the values I measured correspond to the math for the amp I built. ...

Let's focus on the preamp stages, by supply node.

Node E
-  Channel 2 2nd stage only
-  Stage Current = (Supply Voltage - Plate Voltage)/Plate Load = (225v - 150v)/47kΩ = 1.6mA
-  Stage Current Check = Cathode Voltage/Cathode Resistor = 0.8v/470Ω = 1.7mA (close enough check)

Node D voltage is listed as 275v, Node E voltage as 225v, with 20k dropping resistor.
-  Channel 2 2nd stage current should equal voltage drop divided by dropping resistor value.
-  Node E Current = (Node D Voltage - Node E Voltage)/Dropping Resistor = (275v - 225v)/20kΩ = 50v/20kΩ = 2.5mA

This doesn't match stage current calculations, so you must have too much rounding of voltages or the 20kΩ resistor is actually more like 29-31kΩ. That's because current calculated through plate load and cathode resistor voltage drops are in good agreement.

Working backwards, to show how you'd select that dropping resistor value:
Assume Node D voltage is 275v and you want 225v at Node E. Also assume that the current for Channel 2's 2nd stage is 1.7mA.
-  Needed Dropping Resistor = (Prior Node Voltage - Desired Node Voltage)/Current Draw = (275v - 225v)/0.0017A = 29.4kΩ
-  Nearest standard value is 27kΩ for a commonly available resistor, so you'd select that and have slightly higher Node E voltage.

Things get more complicated as you work back closer to the power transformer. Read below...
Current through the 10kΩ resistor separating Nodes C and D is the current through all stages connected to Node D and the current through stages connected to Node E. That's because the Node E current has to be drawn through that 10kΩ resistor to get to Node E.

Node D
-  Input stage, Channel 1 2nd stage connected to this node

-  Input stage
-  Stage Current = (Supply Voltage - Plate Voltage)/Plate Load = (275v - 225v)/47kΩ = 1.06mA
-  Stage Current Check = Cathode Voltage/Cathode Resistor = 1.5v/1.5kΩΩ = 1mA (close enough check)

-  Channel 1 2nd stage
-  Stage Current = (Supply Voltage - Plate Voltage)/Plate Load = (275v - 135v)/47kΩ = 2.98mA
-  Stage Current Check = Cathode Voltage/Cathode Resistor = 2v/2.2kΩΩ = 0.91mA (does not match)

Node C voltage is listed as 375v, Node E voltage as 275v, with 10k dropping resistor.
-  Total current through the 10kΩ resistor is Node D current plus Node E current; this should match .
-  Total Current = Node E Current + Input Stage Current + Channel 1 2nd Stage Current = 1.7mA + 1.06mA + 2.98mA = 5.74mA
-  Node D Current (based on schematic) = (Node C Voltage - Node D Voltage)/Dropping Resistor = (375v - 275v)/10kΩ = 100v/10kΩ = 10mA (does not match)

At this point, I don't trust your schematic voltages because Node D current based on supply voltage measurements is calculating out to double the current based on individual stage voltages, even when using the higher of the possible 12AV7 currents. As a result, proper design procedure using plate curves is the only thing that I think will deliver a believable answer.

I'll work on a few quick loadlines and post the results.

[HotBluePlates]

Okay, loadlines done for the input stage and the two channel stages. What this does is graphically determine the operating point for each stage. Supply voltage is preselected, the red line is the 47kΩ loadline and the blue line is the cathode line for that stage's cathode resistor.

Vital stats:

Input stage (12AX7)
Supply voltage: 275v
Plate voltage: ~222v
Plate current: 1.15mA
Cathode Resistor: 1.5kΩ
Cathode voltage: ~1.73v

Channel 1 2nd Stage (12AV7)
Supply voltage: 275v
Plate voltage: ~170v
Plate current: 2.25mA
Cathode Resistor: 2.2kΩ
Cathode voltage: ~5v

Channel 2 2nd Stage (12AX7)
Supply voltage: 225v
Plate voltage: ~148v
Plate current: 1.67mA
Cathode Resistor: 470Ω
Cathode voltage: ~0.79v

Purpose of the graphs is to find what realistic current is for each stage, not to analyze stage gain, distortion or to second-guess whether each stage is optimized for plate load or cathode resistance.

Now it's time to go back to your original preamp drawing and determine dropping resistor values for the power supply. We will assume Nodes A and B in that drawing will be 275v and 225v, respectively, as found in your original amp's supply and matching the B+ voltages for the loadlines drawn. We also assume the Input and Channel 1's 2nd Stage will be connected to Node A, with Channel 2's 2nd Stage connected to Node B.

Voltage drop to Node B
Total Current = 1.67mA (Channel Two 2nd Stage only)
Desired Voltage Drop = Node A Voltage - Node B Voltage = 275v - 225v = 50v
Required Dropping Resistor = Desired Voltage Drop/Node B Current = 50v/1.67mA = 30kΩ

Voltage drop to Node A
Total Current = Node A Stages' Current + Node B Current = 1.15mA + 2.25mA + 1.67mA = 5.07mA
Desired Voltage Drop = Input Filter Cap Voltage - Node A Voltage = 366v - 275v = 91v
Required Dropping Resistor = Desired Voltage Drop/Total Current = 91v/5.07mA = 18kΩ

Note that if your transformer or winding supplies much over 5-6mA, your d.c. voltage output at the first filter cap will likely be high, as a transformer is often spec'd as "rated voltage when rated current is drawn". When you draw much less than rated current (like 5mA from a 100mA winding), winding voltage will often rise.

[dscottguitars]

Wow!!  Thanks for all of that.  Unfortunately I don't have the amp here to verify the voltages or see if I changed some resistors.  I really don't think I'd be that far off, but the math does prove things.  

On the cathode of the 12AV7, the voltage could be higher, enough to match your calculations. I only checked it once and didn't pay that close attention to how much it was, only that it wasn't too low, like on channel 2, @ 0.8v is getting close to being too low from what I've read.

Using the calculations you have for the stages-now node A and B on the new preamp I want; for the voltage at node B, I am getting:
-  (365v-265v)/(.00106A+.0017A+2.98A) = ~17K.  Does that seem correct?

I would use a 20K there instead of the 10K on the drawing.

Also, I will get the amp tomorrow and recheck everything and get back to you about the voltages and see if I can do the math now.  I'll also check the actual resistor values on my meter to see if they're far off of what they're supposed to be.

Thanks so much for the lesson, I had no idea it was that involved and easy-once it's shown...

[dscottguitars]

You added some more, it's going to take a bit to 'soak' this in...

I tried figuring out value needed for the first resistor based on your calculations from the other amp, since it's the same preamp, and got pretty close to your figure of 18K...

[dscottguitars]

One thing I should point out that may not have been noticed is that both sides of the preamp tubes are wired as one big triode in parallel.  I'm thinking it makes a difference to know since my plate and cathode resistors are cut in half.   

[HotBluePlates]

Using the calculations you have for the stages-now node A and B on the new preamp I want; for the voltage at node B, I am getting:
-  (365v-265v)/(.00106A+.0017A+2.98A) = ~17K.  Does that seem correct?

Your third current is shown as 2.98A instead of 2.98mA, but I know what you meant.

Your calculated currents are now closer to the currents predicted by the loadlines, and therefore your 17(.4)kΩ is close-enough to my 18kΩ for the same resistor.

So yes, now the math looks like its correctly specifying the dropping resistor.

... I had no idea it was that involved and easy-once it's shown...

It is easy. The hard part is that because the power supply in this amp is one long series string you have to know/realize that the currents for stages closer to the input jack get pulled through the dropping resistors closer to the power transformer.

For example, if you now tried to calculate the dropping resistor needed between the node feeding your phase inverter and next output tube screen voltage node, You'd have to know the current drawn by the phase inverter and add it to all of the preamp currents. They all get drawn through the resistor in question.

Which is why you'll typically see large-valued resistors feeding the stage closest to the input jack, with progressivly lower resistances as you move along the B+ supply back toward the PT. At each point working back to the PT, more current is being pulled through each dropping resistor.

One thing I should point out that may not have been noticed is that both sides of the preamp tubes are wired as one big triode in parallel.  I'm thinking it makes a difference to know since my plate and cathode resistors are cut in half.  

Yep, you should have said that up front, as none of your drawings indicate this. Is this the case for all stages?

Well that invalidates all the loadlines, because tube current won't necessarily be double. Or at least, all loadlines will have to be drawn as though the plate and cathode resistors are double the values shown on the schematic to find a new idle current, and then those currents could be doubled to get an accurate answer. They very likely will not be the same currents I obtained from the loadlines I drew, and there's no such thing as a set of plate curves for paralleled triodes.

[HotBluePlates]

After a quick check one one stage with doubled plate load, doubled cathode resistor and same supply voltage, idle current is not 1/2 of the previous value, but might be close enough to call it a wash.

So calculated dropping resistor values will probably work, especially if a few volts in either direction is acceptable.

[dscottguitars]

Sorry for that confusion, I thought that most drawing use only 1/2 of a tube symbol for just the one triode.  I won't make that mistake again and I'm glad the loadlines are close enough...

Thank you very much!

[HotBluePlates]

Sorry for that confusion, I thought that most drawing use only 1/2 of a tube symbol for just the one triode. 

Yeah they do, but when 2 triodes are used in parallel, both are shown wired in parallel.