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Old 31st October 2008, 01:14 PM   #1
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Default Help modifying circuit

Hi Folks,

I am looking at replacing the parallel 12ax7 arrangement with a single side of a 6sn7 to lower the gain. Here is the schematic -

http://www.audiodesignguide.com/New845/New845v2.html

Am I correct in thinking that all I really need to do (aside from wiring the 6sn7 appropriately) is to replace the 75K resistor with 47K.

Thanks for the advice,

Rob
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Old 31st October 2008, 01:57 PM   #2
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That should work.

The valve will run at approximately 140V/4.5mA, -4.5Vgk and the 220F cathode bypass capacitor will give a LF -3db cutoff at around 0.3Hz.

It should work, but you will get more linear operation with a higher current and a CCS as a load.
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Old 31st October 2008, 04:25 PM   #3
kmaier is offline kmaier  United States
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That's a pretty big drop in gain... have you considered using either a 12AY7 or 6072? Probably a direct drop-in replacement with no changes and a mu of 40-44 versus 100. Just a thought.

Regards, KM
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Old 31st October 2008, 09:42 PM   #4
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Hi KM,

I think that you are on to something...

I am just getting the hang of playing around with circuits - up to this point I have just been a copier.

The 12AY7 looks like it has some good characteristics. As far as I can see, it is a drop in replacement in parallel configuration with an anode resistance of about 30K.

What I don't know how to do, is to work out if it will fully drive the EL 34. Do you have a beginners explanation for working this out?

Cheers,

Rob
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Old 31st October 2008, 11:21 PM   #5
kmaier is offline kmaier  United States
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Well, the 12AY7 and 6072 are long time favorites... I've used these in several designs over the years. As for a simple explanation for how they work, no. Some basic tube theory goes a long way to understanding how to work with a set of plate curve charts, picking bias operating points and more. I would recommend finding an older ARRL manual, mid to late 50's and do some reading. These are excellent reference manuals and I still keep mine around (1958).

In this particular design, if you examine the fixed bias on the EL34, it's set at -29 volts. To drive the EL34 at maximum class A1, you would require an input voltage swing of double that, or 58 volts peak-to-peak. This would drive the EL34 input grid to zero on positive swings and down to -58 volts on negative swings.

If you examine the biasing on the 12AX7 (or 12AY7), you need to ensure proper biasing so that the required output voltage swing (around 60 volts), versus the actual gain of the stage, does not require driving the grid of the input tube more positive than the cathode (determined by the cathode resistor value and the idling current). As an example, if the cathode is at 1 volt, and the stage gain is only 25 volts/volt, you won't be able to develop more than 50 volts peak-to-peak swing before pushing the grid positive. This results in class A2 operation and high distortion and non-linearity results.

So, back to your chosen circuit... as you are only looking to swap the input tube for a lower gain, your plate supply is already defined. So in this case, I would be inclined to "back into" the bias points. With 320 volts, I would be inclined to split that between the tube and the plate load resistor, or 160 volts each (less the cathode bias voltage). From previous experience, I've used the 12AY7 as a paralleled input tube with 1.4ma on each triode section, but at a lower plate voltage. Looking at the plate curves, 160 volts and 2ma of current per section would be around 2.5 volts bias. In this config, you should expect a voltage gain ~30 depending on the tube. This should work and give you plenty of headroom for overload on the input grid. Now to calculate values.

Cathode Bias resistor = 2.5 volts @ 2ma, or 2.5 divided by 0.002 = 1250 ohms.

Plate Load resistor = 320 - 160 divided by 0.002 = 80K

Then half those values for paralleling the two triode stages = 40K plate resistor and 625 ohms bias resistor.

This is where I would start, do some scoping/measuring and fine tune your bias points if needed. What you're trying to do is ensure low distortion in each stage plus wide bandwidth and adequate headroom to ensure proper operation as the tubes age and be flexible to handle normal variations in tubes of the same type.

Again, this is backing into an operating point... you could design the stage from scratch and arrive at a more optimal operating point which could require a different supply voltage. Hope this helps.

Regards, KM
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Old 1st November 2008, 02:43 AM   #6
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There are two stages cascaded here.

The 845 will need 120V(peak) for full output in without control grid current, but the inductively loaded EL34 stage will provide a gain of about 8.5. Therefore, the 6SN7 only has to swing 120/8.5 = 14.1V(peak) into the next stage. With a 47K load, the 6SN7 will provide a gain of around 17, so the input sensitivity of the amplifier will be 0.83V(peak) or around 0.59Vrms for full output.

Arguably, you have too much gain already. Modern amplifiers are usually around 2Vrms sensitive for full output. There's enough gain there with the 6SN7 input stage for around 10dB of negative feedback...

To the above poster: we don't need to drive the EL34 to full output, it is the 845 that one should be interested in as it is the valve providing useful power output to the speaker. The numbers above presume class A1 operation.
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Old 1st November 2008, 05:56 AM   #7
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Hi KM,

I thought I was getting the hang of things until I read your post! I am going to work through it slowly. Thanks for taking the time to provide such a detailed answer and just as importantly real useful practical advice.

I am leaning more towards your suggestion of the 12AY7. Firstly it is a classic easily available tube. Secondly, gain wise it is somewhere between the 12AX7 and the 6SN7 and finally, the 6SN7 looks like it has been attacked with the ugly stick, so I am not keen on it from this point of view alone.

Hey Jason, thanks for weighing into this discussion. As far as I can tell, the 12AY7 has about double the gain of the 6SN7 and so it should give me (roughly) about 1.2Vrms for full output which should be just about perfect as I am running it straight from a CD player. I am running it in class A so I think that KM's calculations should be OK.

Now the next question is, given the new 12AY7 driver, can I do better than the EL34, keeping in mind that I have already sourced the power transformer and all of the caps (although I could probably sort out a 5v filament winding if needed as they have not started winding the transformer yet). I was actually looking at the 300B and the 2A3 (I have always wanted to use one of these) but I could not see a simple way of making it work without re-designing the whole circuit, which is beyond my skill at this stage - although I am improving!

Rob
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Old 1st November 2008, 07:13 AM   #8
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Quote:
Originally posted by Rob11966
[B]Hey Jason, thanks for weighing into this discussion. As far as I can tell, the 12AY7 has about double the gain of the 6SN7 and so it should give me (roughly) about 1.2Vrms for full output which should be just about perfect as I am running it straight from a CD player. I am running it in class A so I think that KM's calculations should be OK.
The 6SN7 already has too much gain. The 12AY7 will be even worse. The 6SN7 would give a stage gain of about 17, while the 12AY7 with KM's values will give a stage gain of around 33.

When designing a power amplifier, work backwards fro the output, because ultimately it's the output that we're interested in. This is the load line that the designer has chosen. The red spot is 900Va and 90mA Ia, which is the idle operating condition shown in the schematic. The orange line is the load line - as the grid voltage swings up and down, the anode voltage and current will change. If we plot horizontal lines for anode current and vertical lines for anode voltage at regular intervals while the circuit is operating, the lines would intersect on this orange line. This is called the load line and its slope is equal to the load impedance placed on the valve - in this case 11.5k (derived from the specified turns ratio of 38:1 and an 8 ohm loudspeaker).

Click the image to open in full size.

For class A1 operation, the grid voltage is always less than zero (= always negative). To keep the valve at 900V / 90mA, as the designer specified, we can see that we need a grid voltage of around -130V. (The designer specifies -120V, but may have used a different set of curves). Therefore, for full class A1 sine wave output, the grid has to swing from its idle voltage at -130V, up to 0V, back down to -130V and on to -260V, then up to -130V. An easier way to say this is that we need 260V peak to peak, or 130V peak. Incidentally, we can also estimate power output from this information. P=(Vrms^2)/R, so 17W in this case. (Note, we could actually improve the efficiency and increase power output by designing to allow the grid voltage to swing positive, but this would be class A2 operation, but the designer specifies that the amplifier should be class A1.)

Now, the interstage transformer has a step up ratio from the EL34 to the 845 of 1:1.125. Therefore, since we need 130V peak at the grid of the 845, we need 130/1.125 = 116V peak at the EL34 anode. The EL34 is triode strapped and inductively loaded, so its gain is approximately equal to the EL34's in triode mode. This is 8.5, so for 116V peak at the EL34 anode, we need 13.6V peak at the EL34's grid, which will be provided by the 12AY7's anode.

Right at the beginning, I said that the 12AY7's stage gain will be about 33. So, for 13.6V peak at the 12AY7 anode, we will only need 0.41V peak at its grid. For a sine wave, peak voltage is a factor of the square root of two greater than the RMS voltage, which gives an input sensitivity of 0.41/1.414 = 0.292Vrms for full output.

This is way too sensitive. You only need a stage gain of 4.8 for the amplifier to be driven to full output by a CD player.
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Old 1st November 2008, 11:00 AM   #9
kmaier is offline kmaier  United States
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Quote:
Originally posted by audiousername
There are two stages cascaded here.

The 845 will need 120V(peak) for full output in without control grid current, but the inductively loaded EL34 stage will provide a gain of about 8.5. Therefore, the 6SN7 only has to swing 120/8.5 = 14.1V(peak) into the next stage. With a 47K load, the 6SN7 will provide a gain of around 17, so the input sensitivity of the amplifier will be 0.83V(peak) or around 0.59Vrms for full output.

Arguably, you have too much gain already. Modern amplifiers are usually around 2Vrms sensitive for full output. There's enough gain there with the 6SN7 input stage for around 10dB of negative feedback...

To the above poster: we don't need to drive the EL34 to full output, it is the 845 that one should be interested in as it is the valve providing useful power output to the speaker. The numbers above presume class A1 operation.

Hi Jason, I guess I'm the "above poster".... don't know who "we" is, unless you have a frog in your pocket.

The amount of overall gain (of the original design) is not the main point to this discussion, Rob is simply looking for less gain, but no discussion of how much less. The designer (who is not green at this) used a 12AX7 and that would easily have the highest gain. Referencing the charts in the back of the RCA RC-30 manual for Resistance coupled amplifiers, typical gains would be:

12AX7 - 57, or about 35dB
12AY7 - 28, or about 29dB
6SN7 - 16, or about 24dB

As you can see, the difference is only 11dB total from highest to lowest with the 12AY7 around the middle. Again, it depends on what the individual builder is looking for in overall gain. There was no discussion around using negative feedback but again, an easy personal choice... just not mine.

The 845 with a negative grid bias of 120 volts requires a voltage swing of +/- 120 volts, or 240 volts peak-to-peak for full class A1 drive. The EL34 will easily drive this, but I would still optimize all stages for best linearity. Focusing only the output stage is not sufficient, IMHO. Optimizing all stages provides for additional headroom and allows each stage to operate in a more linear region.

Not sure how you are calculating gain on the EL34, but from the data I have, in triode mode it would have a mu of ~8 and an actual (in circuit) gain closer to ~5. I also can't see the 20ma of current in the EL34 at 29 volts of negative fixed bias... works out closer to 50ma. To get down to 20ma, bias voltage increases to -36 volts, but we don't the brand or condition of the tubes used by the designer. For argument's sake, I'll use the 20ma bias point which is -36 volts of bias. Looking at the required voltage swing required from the EL34 assuming the 1:1.125 driver ratio, that would be ~215 volts peak-to-peak, divided by the gain of 5 results in 43 volts peak-to-peak.

The 12AY7 input amp with it's about gain of 28, equates to ~1.5 volts peak-to-peak input. This is ~530mv RMS input. Not much of a difference from your calculations above. Again, the original design was intended to not require a line level preamp. Of course, if you decide to build one, do it your way.

Rob, the 6SN7 is an excellent tube, so much so, that it's pricing has gone up considerably. So if you want less gain, you can use it or a 12AU7 (another favorite of mine). For the driver, I would tend to stick with the EL34 for now... going to a DHT driver is fine but the added complexity around a DC filament for quiet operation and a large increase in required voltage swing might be more work than you're looking to take on right now. Probably a good idea to approach the build and any modifications slowly... decide what it is you're really trying to accomplish.

Regards, KM
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Old 1st November 2008, 12:43 PM   #10
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Quote:
Originally posted by kmaier
Not sure how you are calculating gain on the EL34, but from the data I have, in triode mode it would have a mu of ~8 and an actual (in circuit) gain closer to ~5.
On the 845's side, in AC terms at useful frequencies of interest, one end of the transformer is grounded and the other end is connected to the control grid. The control grid in negative grid bias presents a large impedance. Taking into account Miller capacitance, the grid still presents an impedance of around 115k at 20kHz.

This is reflected across to the EL34's side at the square of the turns ratio. The load is much greater than the EL34's Ra, so the stage's gain tends to .

Or have I done something incorrectly?
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