What is the dissipation on a 12AT7 reverb driver?

[Systemloc]

Using the AA164 BF Princeton Reverb as an example, there are both sides of a 12AT7 in parallel on a 23K:8 ohm transformer. B+ is 400 volts, the cathode bias resistor is 2200 ohms, and has 8.9V on it. Thus, at idle, there's 8.9V / 2200 = 4 mA through both triodes, or 2mA through a single triode. 400V * .002A = 0.8 Watts of dissipation on a single triode at idle. The rated max is 2.5 Watts, so this about 30% of the maximum!

I'm confused about this, because it seems to be splattered on forums everywhere that the 12AT7 reverb driver in a typical fender is under a strenuous load, and this doesn't appear to be the case at all. Can someone enlighten me?

[Merlin]

It's not the power dissipation its the voltage. Because it's transformer loaded, the anode voltage can swing above 400V, theoretically to 800v peak! 12AT7s just aren't built to withstand that, and may arc.

[Fresh_Start]

At the risk of hijacking this inquiry, reading this question revealed my continuing ignorance about many aspects of vacuum tube circuits.  Despite Merlin's answer, I went to a 12AT7 data sheet thinking a load line might give me some insight.  Unfortunately, I'm not sure how to draw a load line with a transformer instead of a plate resistor.  The parallel triodes are also throwing me for a loop.  Is the graph on the top of page 3 the right place to start?

What's got me wondering is the apparent bias point chosen assuming 400 volt supply and -8 volt bias. 14ma peak across the transformer 400 / 14,500 = 28ma; 28 / 2 = 14ma per triode  Seems like an odd choice of bias point, but I must be doing something wrong.

FWIW the primary resistance on Hoffman's reverb transformer is 14.5K (4 ohm secondary) according to this chart  
http://www.el34world.com/Transformers/fentrans.htm. Hammond shows 22.8K:8 ohm ratio though.  Not a BIG difference, but possibly significant.

Cheers,

Chip

[PRR]

> The parallel triodes are also throwing me for a loop

Double all the resistances, assume one triode. (Or double all the current numbers on the curves.)

> how to draw a load line with a transformer instead of a plate resistor.

First draw the DC condition. Apparently somewhere near 400V 2mA. Strictly, you should allow for the _DC_ resistance of the OT reducing actual plate voltage. DCR is way lower than intended AC-audio impedance. If OT claims 15K, assume 1K-3K DCR. Or 2K-6K per tube. Times approx 2mA is 12V or less reduction in plate voltage, corrected condition is like 390V, so not worth the effort to correct.

Now set your protractor to the assumed 15K AC-audio impedance (30K per tube). For easy math, lay your ruler from 300V to (300V/30K)= 10mA. Run this angle through the DC idle point 400V 2mA.

> 14ma peak across the transformer 400 / 14,500 = 28ma; 28 / 2 = 14ma per triode  Seems like an odd choice of bias point, but I must be doing something wrong.

Well, not "400V", because we hit the zero-grid line at 50V to 150V depending on load.

And 14mA/28mA one way, 2mA/4mA the other way since we can't flow less than zero current.

As a "Power Amp", this op-point is a real stinker.

The "undistorted" swing is 2mA/4mA peak. The low-THD swing may be 1.5mA/3mA peak. Taking 3mA for both triodes times 15K load, 45V peak, 135mW peak 67mW RMS.

45V peak is really poor utilization of 400V available.

At larger signals, triode current can rise but not fall. Eyeballing 15K load with 5K parallel plates we can swing 15K/20K or 3/4 of 400V or 300V on one side of the signal.

So 45V peak one side, 300V peak the other side. 6:1 asymmetry.

This is an EFFECT. The concept of "THD" is very scrambled after you go through a spring.

> FWIW the primary resistance on Hoffman's reverb transformer is 14.5K.... Hammond shows 22.8K:8 ohm ratio though.  Not a BIG difference, but possibly significant.

Possibly not. The tank input impedance rises about 10:1 from 400Hz to 8KHz. Any "5 ohm" spec is for a specific frequency (perhaps 1KHz), and may be 2 ohms near 400 and 20 ohms above 5KHz. So the "14.5K" is really 7K to 60K.

Taking 7K load and 4mA standing current, the max swing is more like 200V:28V. Taking 60K load and 4mA standing current, the max swing is more like 340V:120V.

[Fresh_Start]

PRR - thanks so much for walking (or crawling) me through this.  I don't "get" how to apply the 300 volt / 10ma point to drawing a line which goes through the idle point of 400V/4ma and 0V at the control grid around 50-150V.  However, the key point is that this is an EFFECT.

Experiments with smaller cathode resistors and/or lower voltage on the reverb transformer primary might be interesting, but this is the reverb circuit everyone loves in vintage Fenders.

Cheers,

Chip

[Systemloc]

Here's how I do all my load lines. Just find the GE datasheet. http://tubedata.milbert.com/sheets11.html I use GE because it has nice graphs, all the data is the same in any brand sheet. Zoom in on the desired graph, then hit "Print Screen." In Windows, that does a screen capture. Then open up MS Paint and hit paste. From there, crop the graph, and draw your load line! Takes 2 minutes.



This is the loadline for the 12AT7 with a 23K:8 transformer. You must make two points on the graph, then just draw a line between them. As PRR said above: First, mark the idle DC condition. I don't bother accounting for DCR of the coil, so 400V at 2 mA. Then, an imaginary point at 0 Volts. This represents a dead short across the transformer in the AC condition. 400V / 46K = 8.6 mA. Note, that's the AC condition; remember that we also have 2 mA of DC as well. You MUST add the DC in, so 8.6+2= 10.6 mA at 0 V. Now draw the line.

I also added a horizontal blue line at grid = 0V to show maximum current swing. The tube can swing current from very close to 0 mA to about 8.5 mA. Notice, this is the MAXIMUM signal output that the tube can generate. In circuit, it is driven with a lower input signal, thus it puts out a lower output signal.

Additionally, the circuit has two triodes in parallel, so the current through the transformer will be double what the load line shows.

Now set your protractor to the assumed 15K AC-audio impedance (30K per tube). For easy math, lay your ruler from 300V to (300V/30K)= 10mA. Run this angle through the DC idle point 400V 2mA.

You need to remember to add in the DC bias current here. 10+2= 12mA at 300V

FWIW the primary resistance on Hoffman's reverb transformer is 14.5K (4 ohm secondary) according to this chart. Hammond shows 22.8K:8 ohm ratio though.  Not a BIG difference, but possibly significant.

Remember, that Hoffman 14.5K : 4 with an 8 ohm load is 29K : 8, which is pretty close to Hammond or Fender's spec. The danger here is that the transformer must be made to handle an 8 ohm load. If the impedance of the primary is too low (too few windings), there would be significant magnetization current. The short answer is that transformer may not perform well. Go with one rated for an 8 ohm load.

[Systemloc]

It's not the power dissipation its the voltage. Because it's transformer loaded, the anode voltage can swing above 400V, theoretically to 800v peak! 12AT7s just aren't built to withstand that, and may arc.

Good point. The 12AT7 appears to be rated for a 550V peak anode voltage. http://tubedata.milbert.com/sheets/155/1/12AT7.pdf

I also found that newer silverface amps used a much lower bias resistor, 680 ohm in an SF Deluxe, raising the power dissipation to very close to maximum! It appears common to replace the bias resistor in a silverface with the 2.2K found in blackfaces.

Edit:
Holy crap, Merlin Blencowe?  Thank you so much! Your webpage is absolutely outstanding. I've learned so much from it! I've got to get your book as soon as I scrape up the cash!

[Fresh_Start]

Systemloc - thanks for bringing this topic up.  Interesting point about the Silverface cathode resistor producing 6 volts of idle bias voltage.  IF I understand this correctly now, a 400 Ea with -6Ec produces a cleaner looking load line; however, it also might allow anode voltage to go WAY above 500 volts. 

FWIW the transformer doesn't care whether you look at it as a 14.5K : 4 ohm or a 29K : 8 ohm unit.  It's just a matter of the ratio of turns between the primary and the secondary windings.  My apologies if you already understand that point.

Respectfully,

Chip

[PRR]

> You MUST add the DC in, so 8.6+2= 10.6 mA at 0 V. Now draw the line.

No.

Forget the tube. We are assuming the transformer reflects an audio impedance which, for analytic simplicity/sanity we assume is constant and resistive. And I happen to like 30K (per tube).

On a V/A chart, a resistance to ground plots as a / slant. A resistance hanging down from a B+ plots as a \ slant.

We may plot 1 Ampere 30,000 volts, or 1 Volt 1/30,000 Amps, and get the same slant. For simplicity, we see that with a 30K load then 300V is a nice round number (lazy to divide) and also in-sight of our likely B+.

Plot a line from 300V to 10mA. This is a 30K slant.

OK, we have a tube and a specific V/A chart (they don't all use the same scales). Plot your 300V 10mA line, then cut a triangle from cardboard.

NOW plot the DC loadline. In most useful cases this is so nearly vertical that we may pretend it IS vertical. From 400V at zero mA to 388V at 2mA is not enough slant to care about.

Slide the triangle to intercept the proposed bias, 400V 2mA.

This IS the AC loadline.



Yes, 30K 400V 2mA gives a really asymmetric loadline. It swings up to 450V and down to 120V. +50V -280V

Here is a factory plotted loadline for another (larger!) triode giving good results. The swing is more symmetrical.

[PRR]

There is some question whether we have the 8V drive signal to swing to zero-grid 120V point.

The AA1164 has gain stage, tone-stack, gain-stage, then feeds 12AT7 grids. Assume gain of 50 * 0.1 * 50 = 250 from jack to 12AT7 grid. We need only 32mV off the guitar to smack 8V on the 12AT7, and this is a fairly weak guitar level.

OTOH it takes 4 to 8 Volts at power section driver grid to slam the 6V6es to maximum output.

Basically the reverb driver is driven very close to zero-grid as the 6V6es clip.

Clearly the 12AT7 is driven to cut-off when the 6V6es approach one Watt output. The reverb drive signal is half-wave at most useful playing levels.

[PRR]

> the transformer must be made to handle an 8 ohm load. If the impedance of the primary is too low (too few windings), there would be significant magnetization current.

In general, transformers don't have impedance.

There is a current which will overheat them. This is never limiting on wide-range audio transformers.

In SE transformers there is a DC current which causes inductance to drop more than designed. This is a fairly soft limit. And AFAICT what Fender used (and is now cloned) is an old-old type '42 radio OT intended to run with 250V B+ at 15mA-20mA. None of the 12AT7 reverb drivers come close to this limit.

There is a voltage/frequency trade-off. For a given frequency, some voltage will cause saturation of iron and abrupt bad-shape waves. If you double the frequency you can double the audio voltage before you get bad-shape waves. IMHO the old radio OT should swing 200V peak at 150Hz cleanly. Reverb drive is generally cut-off at 500Hz, so we could swing far more than 200V. The most we are likely to get is 280V.

If we are talking SPRING REVERB, the "load resistance" is NOT any specific resistance. I guess Accutronics' site has been lost, but the data is out there. The tank coil is a COIL with some DC resistance and an inductance with rising impedance at higher frequencies.