Tube Biasing Questions
« on: October 08, 2013, 04:11:37 PM »
I want to discuss tube biasing schemes/options. I know there is self bias and fixed bias. Self bias is also known as "cathode" bias, correct?

There are many different ways to bias a tube. Oliver Archut lists them out in one thread on the internet. Here are a few applicable ways...

-External source to cathode, like U47 current bias from the filament.

-External negative voltage to grid, like U67 and C12.

-Mix of both- external source to cathode in conjunction with voltage divider, like the M49/M49B.

-Self biasing via cathode resistor, like M49c and the like.

I am looking at some schematics and trying to make sense of it all. It looks like in the U47 the cathode gets voltage from the B+. 36V goes to the heater, and it seems as though 1.1V is tapped off and fed to the cathode and the rest is fed through the R3 29ohm resistor to ground. Or maybe I'm reading this incorrectly? Is it a full 36V at f1? Or is it 0V? The design of the EF800 tube in the AMI circuit seems to have a 6V heater and 0V at the other end.

Does a cathode have resistance?

I can see that in the M49b the heater voltage gets tapped and sent to the cathode, although i don't understand why the schematic says "1.6V" in reference to the grid. I'm also not sure what the voltage divider is doing, as Oliver expresses this part is important/significant. Maybe more voltage is being applied to the grid than is tapped off for the cathode?

I can see in the M49c that there is no connection between the heater and cathode. The cathode is tied to a capacitor and resistor to ground. Is this right?

Now, the Tab-funkenwerk U47 seems nothing like any of these... 5V goes to the heater and is tied to a voltage divider, which is connected to the cathode. This sounds like the M49b, except that it is the heater end that is tied to the voltage divider, not the heater end tied directly to the cathode. Maybe it's not all that different? It also differs from the U47, because it is the heater end that is ultimately tied to the cathode, as opposed to the opposite end/pin like in the U47, that seemingly has the 0V pin tied to the cathode. Am I making sense?

SCHEMATICS (in case you don't have them memorized!)
AMI: http://www.tab-funkenwerk.com/sitebuildercontent/sitebuilderpictures/U47ArchutAltTubewithRemote.jpg
M49b: https://dl.dropboxusercontent.com/u/43869772/D-M49BC/M49b_Schematic.pdf
M49c: http://www.geocities.jp/dave_group_japan/m49c.gif
U47: http://gyraf.dk/gy_pd/g7/u47.gif

Since the cathode of the EF800 in the AMI is receiving a voltage (I think), this would be fixed bias, right?

Secondly, all of these designs are plate followers. Can anyone describe how a cathode follower works? The cathode seems to be closer to the heater, and I guess a plate follower just makes more sense "in my head" in terms of working as an amplifier.

Thanks!


Matador

Re: Tube Biasing Questions
« Reply #1 on: October 08, 2013, 05:19:20 PM »
It's best to start off with where the bias points come from, then you can figure out how to deliver those biases. ;)

Take a look at a regular 12AY7 plate curve chart from an old tube manual:



This represents only transconductance curves at "convenient to human eyes" grid bias points (0v, -1V, etc).  The real tube operates at all of the spaces in-between as well.

So I picked the regular C12 bias point (cause that was the chart I had in front of me!):  the curve predicts that with a grid to cathode bias of -1V (or grid sitting at -1V with respect to wherever the cathode is sitting), and approximately 50V on the plate (again, with respect to the cathode potential), that 0.6mA will flow from plate to cathode.  The tube doesn't care where these potentials come from, all it cares is the relationship of the grid to cathode and plate to cathode.  The curve above implicitly has a measurement point assuming that the cathode is sitting at "ground", but we are free to define ground anywhere we want to!

So here is where biasing comes in to play:  let's say yea verily that the cathode shall be declared to be ground/0V.  If we ground the grid as well, and keep the plate at 50V, then plate current shoots up to about 2mA (just draw a line straight up from 50V, and find where it intersects the 0V grid line, to follow it over to the left and it hits about 2mA).  This is a legal bias, right?  It is, and actually the tube will work here.  The bad part is that the grid cannot go much higher than ground and still remain a "linear" amplifier.  So there's much less signal swing positive then there is negative.  The bias isn't in a good place at 0V=grid=cathode for amplification purposes.

So if we ground the cathode, we need to bias the grid negative by a volt or two.  Hence we have fixed bias if the grid bias is not a function of plate current.  The C12 design doesn't follow this rule.  It isn't fixed bias.  But one can certainly create a supply that stays fixed regardless of what the tube is doing:  that would be fixed bias (regardless of how the circuit delivers the bias).

So let's say we want to have a grounded grid and we aren't interested in generating a negative voltage.  We need to make the grid about 1V less than the cathode, so why don't we raise the cathode?  This works.  If the cathode is sitting at +1V, then the grounded grid is -1V with respect to the cathode, and then the plate would be sitting at 50+1V = 51V.  The tube chart is made happy again, and nothing about the tube operation changes (all other things being equal).

So the other biasing options are schemes to make this 1 or 2V bias appear on the cathode.  There are many ways to to this.  A resistor works.  If we want the tube to sit idling at this point, we need 1V at about 0.6mA plate current.  Ohms Law says this is a 1.66k resistor.  This is self bias.  If the tube is powering up, it starts off with the grid at 0V, no plate current which means cathode is at 0V, and the plate voltage starts rising.  As the plate voltage rises, it follows the 0V grid curve upwards, and plate current starts flowing.  This causes the cathode to start rising, which makes the tube want to lower it's plate current.  This continues until equilibrium is reached and the plate is at it's final resting spot.

It's "self bias" because the tube itself turns itself on just enough to shut itself back off again.  It's negative feedback at the cathode.  Any wiggling of the grid upwards causes plate current to rise, which causes bias voltage to rise, which causes the plate current to want to fall again.

Again, the tube doesn't care where this bias comes from.  On the U47, a 40mA heater current is forced through a 29ohm resistor:  again, Ohm's Law says this will cause a 1.1V drop.  Declare one side of the resistor ground, and there you have the 1.1V cathode bias.

However this case is different:  both the heater and plate current flow through this resistor.  If the plate current increases by 1mA, then the total current goes from 40mA to 41mA.  This causes only a 1mA * 29ohm = 29mV increase in the cathode voltage:  barely a change at all!  This biasing scheme is starting to look pretty independent of plate current, so if you asked the tube, it looks like a fixed bias scheme.

There are other ways too:  two silicon diodes in series look like about 1.4V, provided they have enough forward bias to turn on.  After that, they approximate a fixed bias, provided the plate current doesn't drop to a point where the diodes shut off.  This can lead to really bad distortion if you don't ensure the tube is always in conduction under all planned operating points.  But it's a good fixed biasing scheme (and cheap!) otherwise.

As for cathode follower, just re-read the section above about the cathode resistor and the negative feedback, and you'll see why a cathode follower works.  The cathode voltage "follows" the direction of the grid voltage.  The rest of the details will have to be left to another post. ;)

gary o

Re: Tube Biasing Questions
« Reply #2 on: October 08, 2013, 06:10:43 PM »
This is very interesting cant grasp it all but some is sinking into my brain thanks guys.....

ruffrecords

Re: Tube Biasing Questions
« Reply #3 on: October 09, 2013, 07:10:23 AM »
There are some sneaky techniques going on in some of these bias schemes. Looking at the U47, the bias is determined by the current flowing though R3 which is a combination of the cathode current and the heater current fed in via R2. The biggest current by far comes from the heater supply. This means R2 is quite small which means there is very little negative feedback in the circuit. The 'normal' way a self bias a tube like this is just using the cathode current but in this case it would mean using a much higher value cathode resistor which would introduce a lot of negative feedback and reduce the gain. Normally you would overcome this by bypassing the cathode with a large value capacitor. The method used in the U47A avoids the large capacitor but minimises the negative feedback.

The M49b is quite very similar but the current due to the heater voltage is much smaller and the grid voltage is tapped off the divider rather than the cathode voltage being tapped off. To be honest, the values for R6 and R7 don't make sense to me.


The M49c is conventional cathode self bias.

The U47 is slightly different but is basically a resistor chain consisting of R4, the tube heater and R3. R3 sets the bias voltage.

Just as an aside, none of these is an anode follower in the  normally accepted meaning of the term. The U47 is a conventional common cathode circuit with a transformer anode load. The M49s do have some negative feedback to the capsule but not in the manner that is normally called an anode follower.

Cheers

Ian
www.customtubeconsoles.com
https://mark3vtm.blogspot.co.uk/
www.eztubemixer.blogspot.co.uk


'The only people not making mistakes are the people doing nothing'

Re: Tube Biasing Questions
« Reply #4 on: October 09, 2013, 09:34:10 AM »
Matador, thanks for the reply! I'm going to have to read it several times I think to let it sink in, but my initial question right off the bat is where are you getting 50V from? The only thing int he circuit that I can think of that's close to that is the capsule polarization.

Thank you ruffrecords, as well!

gary o

Re: Tube Biasing Questions
« Reply #5 on: October 09, 2013, 09:41:11 AM »
Very intereing reading I would like to learn more aboat how much neg feedback these classic cicuits use or dont use & how us DIYers can adjust to our taste......I would like to adjust a DIY M49C circuit for less neg feedback , how can I do that..... change bias scheme to be more like the U47.....?

Are some cathode bias schemes noisier than others ..... I guess filiment connected ones could be ?

great info fellas , thanks

Re: Tube Biasing Questions
« Reply #6 on: October 09, 2013, 12:37:08 PM »
The best way is when the cathod is grounded. Its most noiseless way. For example: U67.

Matador

Re: Tube Biasing Questions
« Reply #7 on: October 09, 2013, 03:08:16 PM »
Matador, thanks for the reply! I'm going to have to read it several times I think to let it sink in, but my initial question right off the bat is where are you getting 50V from? The only thing int he circuit that I can think of that's close to that is the capsule polarization.

Thank you ruffrecords, as well!

It's the plate voltage...it drops across the plate resistance (the load line is the line in my chart whose slope is proportional to 1/plate resistance), which comes from B+ (above 100V).

dmp

Re: Tube Biasing Questions
« Reply #8 on: October 30, 2013, 05:17:08 PM »
Quote
I can see that in the M49b the heater voltage gets tapped and sent to the cathode, although i don't understand why the schematic says "1.6V" in reference to the grid. I'm also not sure what the voltage divider is doing, as Oliver expresses this part is important/significant. Maybe more voltage is being applied to the grid than is tapped off for the cathode?

If you look at R6 and R7 with Ohm's law, the values start to make sense. First off, the heater voltage is firm at 4v in this design, so the cathode voltage is always constant, since it is directly connected to the heater supply. The heater is supplying about 100 mA, which goes through the heater of the tube, (so the heater is equivalent to about a 40 ohm resistor... and a very small amount of current goes through R6 & R7). Since the plate current is less than 1mA, let's ignore it, since it is << than the heater current. Now we have 4v across R6 and R7, which give us a current of 0.008 mA using Ohm's law. This gives a voltage drop of 1.6v across R6, again using Ohm's law. That is the bias voltage (the difference between the grid and cathode).
Now, think about a change in plate current when the tube has a signal applied at the grid. With the heater voltage firmly set at 4v, the cathode doesn't change, the bias doesn't change, and the extra plate current will just go through the 40 ohm equivalent resistance of the heater. The plate current effectively helps heat the tube, and the heater supply doesn't have to supply quite as much current.
So this is a fixed bias scheme, even more so than a U47, since the U47 cathode voltage can wiggle a little bit for changes in plate current.  The difficulty is having a very quiet, stable heater voltage. All filament connected biasing schemes are very sensitive to power supply noise in the heater.

dmp

Re: Tube Biasing Questions
« Reply #9 on: October 30, 2013, 05:26:28 PM »
Another thing to add: the m49b mic circuit incorporates the psu heater voltage circuit, and the ideal assumption that the heater voltage is firm is a detail to think about. What is the component in the PSU connected to the cathode/heater node? A vary big capacitor. So the arguments about a cathode bypass capacitor similarly apply to the m49b design. It just moves the cap into the PSU.


Re: Tube Biasing Questions
« Reply #10 on: December 09, 2013, 08:43:09 PM »
It's best to start off with where the bias points come from, then you can figure out how to deliver those biases. ;)

Take a look at a regular 12AY7 plate curve chart from an old tube manual:

This represents only transconductance curves at "convenient to human eyes" grid bias points (0v, -1V, etc).  The real tube operates at all of the spaces in-between as well.

So I picked the regular C12 bias point (cause that was the chart I had in front of me!):  the curve predicts that with a grid to cathode bias of -1V (or grid sitting at -1V with respect to wherever the cathode is sitting), and approximately 50V on the plate (again, with respect to the cathode potential), that 0.6mA will flow from plate to cathode.  The tube doesn't care where these potentials come from, all it cares is the relationship of the grid to cathode and plate to cathode.  The curve above implicitly has a measurement point assuming that the cathode is sitting at "ground", but we are free to define ground anywhere we want to!

So here is where biasing comes in to play:  let's say yea verily that the cathode shall be declared to be ground/0V.  If we ground the grid as well, and keep the plate at 50V, then plate current shoots up to about 2mA (just draw a line straight up from 50V, and find where it intersects the 0V grid line, to follow it over to the left and it hits about 2mA).  This is a legal bias, right?  It is, and actually the tube will work here.  The bad part is that the grid cannot go much higher than ground and still remain a "linear" amplifier.  So there's much less signal swing positive then there is negative.  The bias isn't in a good place at 0V=grid=cathode for amplification purposes.

So if we ground the cathode, we need to bias the grid negative by a volt or two.  Hence we have fixed bias if the grid bias is not a function of plate current.  The C12 design doesn't follow this rule.  It isn't fixed bias.  But one can certainly create a supply that stays fixed regardless of what the tube is doing:  that would be fixed bias (regardless of how the circuit delivers the bias).

So let's say we want to have a grounded grid and we aren't interested in generating a negative voltage.  We need to make the grid about 1V less than the cathode, so why don't we raise the cathode?  This works.  If the cathode is sitting at +1V, then the grounded grid is -1V with respect to the cathode, and then the plate would be sitting at 50+1V = 51V.  The tube chart is made happy again, and nothing about the tube operation changes (all other things being equal).

So the other biasing options are schemes to make this 1 or 2V bias appear on the cathode.  There are many ways to to this.  A resistor works.  If we want the tube to sit idling at this point, we need 1V at about 0.6mA plate current.  Ohms Law says this is a 1.66k resistor.  This is self bias.  If the tube is powering up, it starts off with the grid at 0V, no plate current which means cathode is at 0V, and the plate voltage starts rising.  As the plate voltage rises, it follows the 0V grid curve upwards, and plate current starts flowing.  This causes the cathode to start rising, which makes the tube want to lower it's plate current.  This continues until equilibrium is reached and the plate is at it's final resting spot.

It's "self bias" because the tube itself turns itself on just enough to shut itself back off again.  It's negative feedback at the cathode.  Any wiggling of the grid upwards causes plate current to rise, which causes bias voltage to rise, which causes the plate current to want to fall again.

Again, the tube doesn't care where this bias comes from.  On the U47, a 40mA heater current is forced through a 29ohm resistor:  again, Ohm's Law says this will cause a 1.1V drop.  Declare one side of the resistor ground, and there you have the 1.1V cathode bias.

However this case is different:  both the heater and plate current flow through this resistor.  If the plate current increases by 1mA, then the total current goes from 40mA to 41mA.  This causes only a 1mA * 29ohm = 29mV increase in the cathode voltage:  barely a change at all!  This biasing scheme is starting to look pretty independent of plate current, so if you asked the tube, it looks like a fixed bias scheme.

There are other ways too:  two silicon diodes in series look like about 1.4V, provided they have enough forward bias to turn on.  After that, they approximate a fixed bias, provided the plate current doesn't drop to a point where the diodes shut off.  This can lead to really bad distortion if you don't ensure the tube is always in conduction under all planned operating points.  But it's a good fixed biasing scheme (and cheap!) otherwise.

As for cathode follower, just re-read the section above about the cathode resistor and the negative feedback, and you'll see why a cathode follower works.  The cathode voltage "follows" the direction of the grid voltage.  The rest of the details will have to be left to another post. ;)

Matador, I feel as though I am finally just barely beginning to understand some of this. However, I have some more questions now. I understand that if you look at plate voltage, and the bias (whether it is a positive cathode in relation to a grounded(?) grid, or a negative voltage applied to the grid, you can read the current of the plate. But what does plate current tell us? I think the answer to this fundamental question will help my understanding a lot.

You mention if we ground the cathode and the grid and drop 50V on the plate, we get 2mA of plate current. However "The bad part is that the grid cannot go much higher than ground and still remain a "linear" amplifier.  So there's much less signal swing positive then there is negative." Is this something I'm supposed to just know? Or is this somehow inherent to the graph and I'm not seeing it? Does this mean we're tryign to keep plate current very low and close to 0mA or we wont have a linear amp?

What is the red line in your image/graph?

Thanks a lot for your patience offering up this knowledge!

Matador

Re: Tube Biasing Questions
« Reply #11 on: December 09, 2013, 11:33:35 PM »
Matador, I feel as though I am finally just barely beginning to understand some of this. However, I have some more questions now. I understand that if you look at plate voltage, and the bias (whether it is a positive cathode in relation to a grounded(?) grid, or a negative voltage applied to the grid, you can read the current of the plate. But what does plate current tell us? I think the answer to this fundamental question will help my understanding a lot.

It tells us nothing other than the current under those bias conditions.  It's a single operating point in a sea of possibilities.

You mention if we ground the cathode and the grid and drop 50V on the plate, we get 2mA of plate current. However "The bad part is that the grid cannot go much higher than ground and still remain a "linear" amplifier.  So there's much less signal swing positive then there is negative." Is this something I'm supposed to just know? Or is this somehow inherent to the graph and I'm not seeing it? Does this mean we're tryign to keep plate current very low and close to 0mA or we wont have a linear amp?

What is the red line in your image/graph?

Thanks a lot for your patience offering up this knowledge!

So the tube charts stop at EG = 0.  It doesn't mean that the tube knows where the chart is drawn.  The curves above continue for positive grid voltages, but they start to "squish together".  Thus the device looks much more non-linear in this region than is generally useful (hence they aren't drawn).

"Linear" in this sense means that a delta input signal causes a fixed-proportional output signal, multiplied by the gain.  You'll notice that as the grid bias gets more and more negative, the curves start to bend.  Hence they only approximate a linear relationship at that point.  Operating in these regions causes distortions (maybe desired!).

The red line is the load-line.  The output signal will follow this line.  If you look at the two axis in this graph, the X axis is voltage and the Y axis is current:  Ohm's Law says V/I = R, hence the graph represents the dynamic impedance of the tube at all bias points.  Hence the slope of the load line is actually just the plate resistance (100K in that chart).

Re: Tube Biasing Questions
« Reply #12 on: December 11, 2013, 12:09:09 AM »
The best way is when the cathod is grounded. Its most noiseless way. For example: U67.

Agree 100%. It's a matter of taste, but after extensive tinkering and listening that is my favorite method. Bass response seems "right."

Re: Tube Biasing Questions
« Reply #13 on: January 04, 2014, 01:27:46 AM »
Still learning quite a bit. How much voltage does a condenser capsule typically generate? In other words, how much voltage is going into the tube grid typically?

Re: Tube Biasing Questions
« Reply #14 on: January 08, 2014, 11:45:42 AM »
A couple more thoughts/questions...

If you design a U47 with a separate heater, the biasing of the tube is inherently different. With the U47, as Matador mentions, the plate current can increase, which effects the voltage on the cathode. Granted, this change is very minute, and so U47 is fairly fixed bias. But this is still not as fixed as a U47 with a separate heater that has the cathode voltage tapped off the heater. Am I misunderstanding?

I somewhat understanding the tube data charts now. However, I still don't understand the significance of the Y access. Is there anything inherently important about the plate current at any given bias point? Obviously reading the chart you can see that if you increase plate voltage, you get a higher plate current. Or that if you keep plate current the same, but you make the grid more negative, you get les plate current. But what does it mean? Why is it important? Why does it matter if the plate current is 0.5mA or 20mA?

Thanks!

Matador

Re: Tube Biasing Questions
« Reply #15 on: January 08, 2014, 12:20:47 PM »
It's important insomuch as this is the mechanism used to get useful amplification of a signal voltage.  An output current may not be useful, however Ohm's Law says we can convert a current to a voltage with a resistance.

The curves are important because they establish the transconductance relationship of a tube.  Transconductance is the idea that a voltage is used to generate a current:  or stated more formally conductance is the reciprocal of resistance, which is voltage / current.  gm is the first derivative of this relationship flipped upside down, or delta change in current / delta change in voltage.  This allows us to reason about how changes in grid voltage cause changes in plate current.  In effect, the curve tube shows us the transconductance at every operating point (since gm actually varies at different points, and is not globally constant).

So in effect, as a designer you are free to choose the operating points of the tube based on your circuit and your requirements.

In terms of capsule voltage, it depends on the capsule.  For a LDC capsule it's surprisingly large...perhaps reaching a volt during loud transients.

Re: Tube Biasing Questions
« Reply #16 on: January 08, 2014, 09:13:35 PM »
Very interesting, Matador! And just like that I think I'm in over my head again with the first part of your response. I'm going to have to do some research on transconductance.

One volt for loud transients is very interesting. Maybe I'm oversimplifying, but at 1V intervals, most audio tubes data sheets I've looked at depict a pretty linear amplification at that low level. I assume you start at the idle point on the tube data sheet and then move positive .5V and negative .5V  following the grid lines from the idle point. If you get the same amplification of voltage (by checking the x axis) on the positive and negative side, the amp is pretty linear, right? If the amplification is more on the positive side and less on the negative side, you'd get distortion?

I was inspired to take some measurements inside of one of my DiY mics. At the tube plate, I'm measuring .75mA and 46V. So using V = IR, this tells me the plate resistance of the tube in this mic is 61.6kohm? Is this just the plate resistance at idle?

The cathode is getting 1.33V.

I tried a different tube (same heater) in the same mic, and I got .75mA and 51V at the plate (also 1.33V cathode). This gives about 68kohm plate resistance, assuming my understanding is correct. So both tubes draw the same current, but the voltage at the plate is different because they have different resistances? Does this mean anything in terms of the actual sound of the tube, or is it all dependent on how that plate resistance reacts with the transformer?

Thanks!

Re: Tube Biasing Questions
« Reply #17 on: January 13, 2014, 06:54:47 PM »
If my calculation for plate Resistance Rp is correct in my previous post, can I calculate tube Zout by calculating the parallel resistance of Rp and the plate resistors? Or does Zout change throught he tubes operation? Would I only be measuring Zout at idle? Or does Rp stay constant while current and voltage are what fluctuate witht he incoming audio signal?

In the U47 schematic, do R5 and R6 (100k and 30k) just drop the B+ voltage to the plate? Do they also effect the Zout of the tube? Does only the 100k effect the Zout, or since they are in series, do you have to think of it as a single 130k resistor?

Thanks

Re: Tube Biasing Questions
« Reply #18 on: January 13, 2014, 08:46:03 PM »
Sorry to post again, but I made a new discovery. If my calculation of Rp is correct (take the plate voltage and divide by the plate current), then even a .3V difference in the grid can result in a 30kohm difference in Rp. That seems pretty insane. I think I must be doing something wrong.

An unintentional 0.3V difference in the grid from what the circuit calls for could easily occur just through resistor tolerance differences. 30K difference in Rp seems like a lot. That's as much as a 23kohm difference in Zout of the tube.

moamps

Re: Tube Biasing Questions
« Reply #19 on: January 14, 2014, 05:15:35 PM »
Rp=dVa/dIa (Vg=const.), not just Va/Ia for one operating point.