Impedance changes in a tube circuit

GroupDIY Audio Forum

Help Support GroupDIY Audio Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.

Emperor-TK

Well-known member
Joined
Jul 14, 2004
Messages
1,076
Location
NJ, USA
I was wondering in general how the impedance changes in a tube circuit from input to output. I am looking for some general info, but I also have a specific application in mind. I am building an Alembic F2-B preamp (Fender Dual Showman) and would like to add a low-Z transformer balanced output. This was the "fruitcake amp" from this thread:

http://www.groupdiy.com/index.php?topic=5315&start=30

The F2B schematic appears here:

http://www.geofex.com/FX_images/alembpre.gif

My question is first, how do I determine what the output impedance of this circuit is, and second, how do I figure out what ratio of output transformer to use. Looking at Jensen's white papers, I see that a 12:1 transformer is used for 10Kohm to 150 ohm matching. I don't quite understand how the numbers relate. I tried searching the forum, but I think this topic is too general to get good hits. I've also been bugging Gus a lot about this build, so I figured I'd give him a break too.

Thanks,
Chris
 
From my limited understanding of tube circuits: the output impedance of that circuit is Rp||rp, where Rp = 100K and rp = the dynamic plate resistance of the tube under those specific conditions. Without a value for B+, you can't calculate rp, but for a 12AX7, it will be somewhere in the ballpark of 50 to 100 KOhm. That's a big ballpark! The 1M resistor is big enough that you can ignore it for all practical purposes.

After you figure out the Zout of that last triode, divide it by the square of the turns ratio of the transformer you use and BANG! That's your output impedance.

I may be all wrong here, I'm sure NYDave or someone will correct whatever mistakes I may have made.

Good luck!

Peace,
Al.
 
Chris, I'm going to give you the "teach a man to fish..." answer.

First of all, to predict the output impedance of the circuit, you'll need to learn how to use load lines. One of the best tutorials I've seen is in part one of Norman Crowhurst's "Designing Your Own Amplifier" series:
http://www.audioxpress.com/resource/audioclass/

You'll find that the impedance at the 12AX7 plate is rather high. You'll be hard-pressed to find a stepdown transformer of high enough ratio that will pass full-bandwidth. Luckily, since this is a guitar preamp, you don't need full audio bandwidth. Rolloff starting any higher than, say, 5kHz should be acceptable. The low end is more of a problem; our hypothetical stepdown transformer would need a VERY high primary inductance to preserve low frequency response.

The turns ratio of a coupling transformer has a square root relationship to the impedance ratio. For instance, say you want to transform a source impedance of 100K to a source impedance of 600 ohms:

100000 / 600 = 166.7

sqrt 166.7 = 12.91

So you'd need a transformer with about a 13:1 turns ratio.

In the particular case of this preamp, it would be more productive to add another tube to allow a cathode follower output. That would improve performance by isolating the plate of V1 from the load, and would greatly increase your available choices in output transformers.
 
The output impedance with a cathode bypass cap should be the value of the plate load resistor in parallel with the plate resistance.  We know the plate load resistor is 100K, but we don't know the Rp, nor do we know the B+.

So, first I drew a load line by assuming 300 volts B+, and finding the maximum possible current through the 100K plate load resistor in a dead short, 300V/100K=3 mA.  So I drew a line from 300V to 3 mA.

Next, I tried to find the idle point of the tube with a 1500 ohm cathode resistor, and I lucked out by guessing it would be a bias of -1.5V.  I checked this by figuring 1.5V/1500ohms = 1 ma.  At that point, everything (nearly) lines up on the graph, so I'll assume I'm close enough.  For some reason, I can't think of the math to figure this out exactly.  

Then I calculated the plate resistance (Rp). Rp is defined as a change in plate voltage divided by a change in plate current with the grid voltage held constant, so I changed the plate voltage by +10 volts and followed the grid voltage curve for a reading of plate current, which showed a change of .125mA.  So 10V/.125mA = 80000, or 80K.  (Don't forget to move the decimal!  .125mA = .000125A)

So the output impedance with the cathode bypass capacitor is the plate load resistor (100K) in parallel with the Rp (80K), which comes out to about 44K.  If we didn't have the cathode bypass cap, we'd have to add Rk(µ+1) to the Rp value.

Back to the load line to determine µ, which is defined as a change in plate voltage divided by a change in grid voltage with the plate current held constant.  Now we take readings along the horizontal plate current lines, and we can see that changing the plate voltage by 30 volts produces a change in grid voltage of .5 volts.  30volts/.5volts = 60.  Just plain old 60, with no units of measure.  (I chose to change by 30 volts because it was easier to make approximations on the graph than with the 10 volt change used earlier.)

Without the cathode bypass capacitor, Rp= 80K + 1500ohms(60+1), or 171500 ohms.  Put that in parallel with the plate load resistor and I get about 63K.

So with the cap, Zout = 44K.
Without the cap, Zout = 63K.

-E.
 
At least at first read, it sounds like you did it correctly. Your final figure seems realistic. I'll read it over again later and will speak up if I spot any errors.

Notice that I never gave even an approximate figure for the output impedance of the preamp. I just said "For instance, say you want to transform a source impedance of 100K...", just using a convenient number to illustrate an example of how you'd pick out a transformer. I didn't mean to imply that the actual source impedance of the circuit in question would be 100K.
 
Leaving the cathode resistor unbypassed allows degenerative current feedback, which reduces distortion and tendency toward grid overload (signal rectification). This comes at a penalty of lower gain and increased output impedance.
 
[quote author="electronaut"]I gave this a shot, and I'm not sure if my answers are close or totally wrong. They seem lower than NYD's and alk509's numbers.

[...]

So with the cap, Zout = 44K.
Without the cap, Zout = 63K.[/quote]

That actually sounds about right, and it's in accordance with Dave's and my math.

Peace,
Al.
 
I took a look at the Tube Register, and Electronaut's numbers look like they're in the right ballpark. But let's turn the question around.

First, to the original poster: you're looking for a low-impedance balanced output. To get that from a transformer, you need a stepdown transformer, so of course you'll decrease output level. If you're looking for a mike-level output, like you'd get from a DI, then that's cool. (If not...well, I'll get to that in a second.)

So the question becomes, what will you be driving? Typical mike preamps have an input impedance of about 1500-3000 ohms; to be on the safe side, let's take the lower number. If you use a 12:1 stepdown transformer, in effect you're stepping up[ the load impedance by the square of the ratio. 1500 * 144 = 216k, which in parallel with the 1M on the output of the Alembic yields about 178k. This is a little low for a 12AX7 to drive; the original Dual Showman design loaded the preamp section with about 380k, give or take a bit. It'll work, but the output tube will be working a little hard.

On the other hand, if you want a balanced line-level output, you'll need a 1:1 transformer, and at that point all impedances stay about the same. Trying to drive the typical 10k input impedance of a line input with a 12AX7 voltage amplifier will give you low gain and lots of not-very-pleasant distortion. If a balanced line-level output is your goal, I'd suggest installing a cathode follower made with something like a 12FQ7 or 12SN7, and a good-quality 1:1 xformer.

One thing that's confusing is that most of the audio world does not use impedance-matched connections any more -- certainly not in this application. Instead, the most common setup is a bridging connection, where the load is significantly higher than the output impedance of the driving circuit.

Peace,
Paul
 
Wow, thanks for all the great replies. I really love the "teach to fish" approach. That's what I was hoping for, thanks. Now to digest all that info so I can digest the fish.

Load lines, fishing lines...

-Chris
 
Yes, I am a Stereolab fan. Not a huge fan though, like where I have all the obscure singles, etc. I do think that Emperor-Tomato-Ketchup is one of my all time classics though, besides being the best name for a record I have ever heard. :thumb:

-Chris
 
OK, so back to the Alembic circuit.  For some reason I still can't figure out how to determine the idle point of the tube.  If we assume the 300V B+, and we know the plate load resistor, we can draw the load line.  How do we determine the bias from a given resistor value?

I feel like it must be a simple matter of rearranging a formula or something, but for some reason I can't make sense of it at the moment.  Any ideas?

Thanks.
 
> How do we determine the bias from a given {cathode} resistor value? ... I feel like it must be a simple matter of rearranging a formula or something

It isn't. Or at least, it requires data not normally available.

On the chart, assume a current, compute the voltage, and put a dot at that current and grid bias. Do this over a wide range of possible currents. You end up with a curved line (which is why it isn't a trivial calculation).

Now spot your B+ and draw your plate resistor line from there. The intersection of the plate resistor and cathode resistor lines is your operating point.

But for any 12AX7 bias point useful for driving any significant load (even a 470K hi-fi line input) the plate resistance will be so close to 62K that you don't need to sweat-out the real value. It could be as low as 33K except out there on the zero-bias line it can't swing. It can be over 100K, but at bias points with such low current that it can't do useful work. Inside a high-gain amp where it feeds an inch of wire and a grid, it might work around 150K Rp, but not on a jack to the outside world.

So the output impedance is around 60K||100K||1Meg or 36K.

But that is not the load it can drive. As a general rule, a triode should see a load 2X to 5X its plate resistance. Lower makes distortion. Higher does not suck enough work out of the tube. So we want 120K to 300K load. We already blew that: the 100K load. Not much too low, and we may not ask a guitar amp to be super clean. But we don't want much added load. Over 10X the existing impedance, or 360K, would be nice (and is probably the design load). We could fudge that to 3X or 100K, with increased distortion and lower output voltage. However, in many cases a 12AX7 on simple audio (guitar, not guitar/bass/drums) is tolerable at lower impedances, and once you get down to Rp the distortion does not get worse, you just lose output voltage.

The transformer primary must be designed for the source impedance (36K) for bass response and iron distortion, for 100K-300K for low tube distortion in the bass. Taking 80Hz as the bottom of the guitar, this gives primarly impedance of 75H to 600H. You are not going to find a 600H winding. There has to be some compromise with bass distortion. That may not be bad in guitar duty, but has to be considered.

Do you really want to drive 600Ω? As Paul mentions, few inputs today are 600Ω. You can pretend you will only see 10K loads and usually be right.

If so: you need 8:1 with 75H to 22:1 with 300H. 10:1 is easily found, but either it won't have ~100H inductance, it won't pass line-level cleanly, or it will be exotic and expensive.

Taking 10K loads, you can use 2:1 at 75H to 5:1 at 300H.

You don't just want to match impedances; in fact impedance really should be a secondary concern. In post-1935 design, you put in enough tubes to give excess gain, so you don't HAVE to carefully impedance-match. How much level does this have, and how much do you need?

My guess is that this has enough voltage gain from guitar to make around 1V output in use, maybe over 3V peaks. So say 2V no-load, or 1V loaded in 36K. 1V in 36K is 0.027 milliWatts or -15dBm. You can NOT feed +4dBm 600Ω lines with this: not enough power gain from guitar input. Even assuming that loads WILL be 10K, output level is only -3dBu. That's within the gain-range of most "+4dBu" gear. So a good quality (high inductance) 40K:10K transformer will work. Line impedance is 5K plus the considerable leakage inductance of a hi-Z winding, so don't run it long distances. Even 20 feet might be "far".

If your goal is a mike input: any general-use mike input can take -50dB levels, say 2mV. You have 1V. 500:1 ratio seems called for, and would give extremely low output Z (0.15Ω!). We still need that 75H-300H primary, which is a tough winding. So take a common mike input transformer, backward. To isolate the back-to-back transformers, put a 6dB pad between. We need 4mV out, so with a 1:10 (used 10:1) we need 40mV in. And such a transformer likes to see 10K-20K source on the hi-Z side. We have 1V at 36K. We need a pad with >36K in, 10K out, and 1V/40mV loss. 250K and 10K resistors give the 25:1 loss, a 10K output Z, and a nice 250K input Z. The transformer can be plucked from any old-school gym PA system. Output level will be similar to a close-talked SM57, a level most board-ops understand.

If you must drive "line" inputs: try to find a good 40K:10K line transformer. They are rare beasties. And it won't be a great line output.

For an all-tube good line output: as Dave and Paul hint, you want another tube. 6C4 (half a 12AU7) is a safe choice, can work with inexpensive 10K:600 iron, give voltage gain of about 3, good solid +18dBm peaks.

Alternatively: add a 24VCT transformer and a 5532 chip. It kinda wants a TL071 also, because the input bias current of a 5532 makes 200K inputs difficult. Probably one of the hot FET-input chips with low THD and true 600Ω drive is simpler.
 
Hey Tomato, this is something I've messed about with, I had good luck with an ADK output transformer, the ratio was 15k to 150/600 secondary, but the output was dropped down to more or less Mic level.

I'd like to see some pics of your layout when your done. I have to open mine up yet again and try a different layout cause I've got some hum problems. This amp is a Fender Showman gain and EQ stage so I may try
to do this with a point to point type turret board layout. By the way, the way the midrange of the eq tilts the frequencies around makes for a really awesome bass sounds.

So while I'm at it, I'd definately like to add another stage so it has a line level output. I've been looking into push pull output stages because I have this Triad A-78J trafo which is labelled as follows

A-78J transistor Push Pull output transformer
primary 2k noDC with a center tap
secondary 150/150 ohm dual tie for center tap. 600ohm CT
15-15000 CPS at 20dbm

I've been looking at a load of different amp output schemes and I'm frankly confused. Maybe someone could point me to a topology that would
work for this transformer and I can bang my head against that for a while.

Thanks
Sleeper
 
> looking into push pull output stages

A proper push-pull tube output needs 4 tubes. One push, one pull, one phase-split, and one for gain and feedback over the whole mess. Dual-tubes cut the bottle-count but not the complexity.

A 2K CT winding is way too low for any common tube. That's 500Ω each side, suggesting a plate resistance of 100Ω to 500Ω. In the twin-tubes, 12AU7 is 6,000Ω and AT AX are higher. 6BL7 is 2K and a big bottle. The big TV V-sweep tubes have one 800Ω plate and a high-gain volt-amp, but eat heater power.

Look at it another way. 20dBm is 0.1 Watts. 0.1 Watts in 2K is √(P*R)= √(0.1*2000)= 14 volts RMS or 20V peak across the whole winding, 10V peak on one winding. How are you going to get 10V peak efficiently from a 300V supply? You are wasting at least 290/300= 96% of your supply power. It is worse: the 20mA peaks mean a much fatter tube and bigger heater than if you used a higher impedance.

This was made for 15V-24V supplies in transistors that would pass 20mA at small voltage loss, not for tubes.

6DJ8 in a White Cathode Follower (possibly with a volt-amp in front) would actually push it OK (ignore the center-tap). You'd have to run up close to the tube ratings, short life (hundreds of hours). A couple of 6EM7 working at 100V B+ would work fine, wasteful, but cheap and long-life. A 12AU7 or 6C4 with a 20K:600 5mADC transformer would work about as well with a lot less waste.

Hmmmm... 12AU7 with long-tail bias could drive push-pull center-tapped 2K no phase-split with one bottle and an overall voltage gain of just about unity. THD would be largely 3rd and rise quickly above 5% at +20dBm. Output impedance on the "600Ω" tap would be more like 3600Ω. Workable, not happy.

Output transformers for tubes working below 1 Watt output want to be 10K or more. Going any lower drowns you in either THD or heater power. Even up at 100 Watts and more, we rarely go below 4K and 6K-10K are much more common, even out to 10,000 Watts. In tubes, voltage is cheaper than current, so impedances want to be high. 2K is not high for a tube.
 
[quote author="PRR"]On the chart, assume a current, compute the voltage, and put a dot at that current and grid bias. Do this over a wide range of possible currents. You end up with a curved line (which is why it isn't a trivial calculation).

Now spot your B+ and draw your plate resistor line from there. The intersection of the plate resistor and cathode resistor lines is your operating point. [/quote]


The curves aren't curved.  :?

CathodeLines.gif
 
[quote author="PRR"]> If you must drive "line" inputs: try to find a good 40K:10K line transformer. They are rare beasties. And it won't be a great line output.

For an all-tube good line output: as Dave and Paul hint, you want another tube. 6C4 (half a 12AU7) is a safe choice, can work with inexpensive 10K:600 iron, give voltage gain of about 3, good solid +18dBm peaks.[/quote]

I don't know about the 6C4 version, but if you're talking about using a cathode follower, I've had bad luck with CF's and 12AU7's unless the filament is floated to +90V or so. They act weird with the filament down at normal levels. 6SN7's and 6FQ7's seem to have fewer problems.

Alternatively: add a 24VCT transformer and a 5532 chip. It kinda wants a TL071 also, because the input bias current of a 5532 makes 200K inputs difficult. Probably one of the hot FET-input chips with low THD and true 600Ω drive is simpler.

The problem with the FET units is that the output impedance of the Alembic is high enough to start causing common-mode distortion in the opamp. One possibility might be the old LM310 buffer amp. No common-mode distortion, and only 2nA typical input bias current. Run it into a 1:1 transformer

Or you could do an inverting FET amp using the OPA604, with Rin and Rfb = 360k. Stick a 100-ohm resistor in series with the output pin and take the feedback and output on the right side of the resistor, just in case you run into stability problems. Run the 604 into a 1:1 transformer, remembering to reverse the output leads to compensate for the inverting amp.

Peace,
Paul
 
Here's my general-purpose White cathode-follower output stage I've been using in most of my recent tube projects. It has some real "grunt" and can drive 600 ohms directly to about +24 or +25dBM before clipping. With higher impedance loads, it can do well above +30dBU.

12BH7WCF.gif


I present it just as a point of interest. It probably wouldn't be practical for your circuit without a complete redesign of the power supply, since it draws about 8 times the plate current of your existing preamp circuit. You probably don't need this much drive capability anyway.
 
Hi Dave,
I'll have to look at the datasheets on that tube. I don't know about the emperor, but I have so much spare plate current that I was a bit worried...

I'm thinking this type of a gainstage might be perfect, I may not need quite so much grunt coz im thinking the signal 241- transformers that CJ scoped out recently and I just scored might do the trick.
I may be able to get a little bit of transformer gain or else have a slightly higher transformer input impedence depending on which whether i use it normal or reversed.

The white cathode can be made with any dual triodes yes?
I mean any dual triodes that meet basic criteria for the given power supply etc.

P.s. thanks PRR, going to have to save this trafo for use in a transistor push pull... hmmm funny thats what it says on the can...
 
[quote author="Sleeper"]The white cathode can be made with any dual triodes yes?[/quote]

Yes, but watch that Vhk rating!

Peace,
Al.
 

Latest posts

Back
Top