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Biasing a cathode follower?

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#62
Re: Heater/cathode voltage

Positron said:
Hi G,

Also want to make sure the heater cathode/voltage isn't exceeded at turn on.

Just a thought.
Click to expand...

That asks for a delayed B+ and B- turn on 😱

Also 20 mA through the tube will get the anode dissipation close the maximum rating of the tube. That’s fine for a power output stage but for a line driver? If 10 mA sound as good it will extend the tube’s life time.

Cheers 😉
 
#63
Golly you guys want to make a simple thing hard.

First, I would agree that 20 ma is total current overkill in this application. I used it as an example in a calculation because that was the number that the poster was starting with. 10 ma would indeed be more appropriate (I think I mentioned this earlier in connection with 6SN7s), but I'm not the one buying the tubes.

Second, the heater-cathode voltage of nearly all small dual triodes is 200 V peak, which doesn't get exceeded on turn-on. And the cathode resistor certainly limits the current. You don't need slow turn-on or fancy protection. You DO need some way of keeping the turn-on and turn-off away from the input of an already-on power amp, but that's true of nearly every tube circuit. If 200 V at turn on scares you, you can always bias the heaters with a resistive string to, say, 50-60V, but again, this is an unneccessary complication- the peak heater-to-cathode voltage is not exceeded. And biasing the cathode negative is asking for hum trouble.
 
#64
heater/cathode voltage

Hi Sy,

200 volts is the peak, 100 average. With initial turn conditions, before the heater warms up, the heater/cathode voltage could spike to 250 volts or so, as the supply/transformer is unloaded.

As you mentioned, one could raise filament to limit this differential.

Just wanted to be on the safe side Sy.
 
#65
Yes SY it is hard to make things simple 😀

Think you are right concerning the cathode-heater voltage. Most tubes survive the double of the ratings for a short time but it is not guaranteed. The datasheet does not state a pp rating for it.

Cheers 😉
 
#67
I was told by a knowledgable person that a CF sounds best when run hot and that a 5687 sounds it's best when run in that range. I've never heard a 5687 CF so I don't know. I have no problem with biasing the heater above ground if it will protect my tube. I can use a voltage divider between the + rail and the - rail and get 40 volts of lift on the heaters easily. I have a lot of 5687s but they have become more expensive as word about them got around. Supply and demand you know. Also can someone please explain how the grid voltage will be only 5-10 volts negative to the cathode when I am dropping that much voltage across the cathode resistor? Does anyone have a biasing and B+ recommendation? As far as the power supply goes I already have the majority of the parts and if I don't use them they will continue to collect dust. I'm an overkill kind of guy when it comes to design margins and safety ratings.
 
#68
Also can someone please explain how the grid voltage will be only 5-10 volts negative to the cathode when I am dropping that much voltage across the cathode resistor?
Click to expand...

Sure. You understand that the grid is sitting at ground? This is true because it draws essentially no current, so will have the same potential as the bottom of the grid leak resistor (the volume control in this case). OK, so let's look at the cathode. If the tube's cathode is drawing 20 ma and the cathode resistor is 10K and it's returned to -200, then Ohm's Law forces the cathode voltage to be pretty close to ground, too.

Now, here's where the intrinsic negative feedback of the CF comes in. If the cathode and grid are both at zero, the tube will start drawing more current. That doesn't change the grid voltage, but it makes the cathode more positive wrt to ground (and hence wrt the grid). As the current gets higher, the cathode gets more positive, which bucks the current rise- it's equivalent to making the grid more negative. So the circuit biases itself- you don't have to do anything other than choose the cathode resistor. The cathode will sit at whatever voltage above ground it takes to bias it up wrt to the grid (at DC ground). 5V, 10V, it doesn't matter, that's what it will end up being.

The stuff about running hot is pretty suspicious. You need enough current to get the tube into a reasonably linear region and to drive a typical cable capacitance. Either way, 10 ma is more than enough (and that means a 20K cathode resistor). Regarding heater-to-cathode, I don't have experience with the tube you're using, but I've used similar circuits with 12AT7s, 6SN7s, and the like with zero problems. If it worries you, just use a protection diode- no signal passes through it, so you can maintain your vacuum purity. You haven't shown your heater supply, but if there's a center tap on the heater winding, you don't even need the diode, just connect the CT to the cathode through a large resistor, say 470K/0.5W.
 
#69
Soft starting the B+/B- is not too difficult. A pair of negative temperature coefficient thermistors wired to the ends of the power trafo's B+ winding will get the job done. You want to select thermistors that are HOT at the operating current drawn by both channels. That way, the voltage drop during operation, as opposed to start up, in the NTC devices is close to zero.

Both Allied Electronics (www.alliedelec.com) and Mouser (www.mouser.com) carry the Thermometrics brand of NTC parts.

If sourcing 5687s becomes problematic in the future, JJ's current production ECC99 will be a GOOD alternative. The ECC99 is a high gm, low Rp, twin triode similar to the 5687 that has a mu in the low 20s.
 
#70
CT filament

Sy's way is a good way, G as the filament will follow the cathode. (I take it Sy's value of the resistor is ok for that tube.)

During initial conditions (turn on), the cathode will drop to the negative rail voltage, some -250 volts or so while the grid is zero volts unless some sort of precaution is used. I personally don't like that condition.

Take care G.
 
#71
SY said:
No, the grid will sit at ground, the cathode will be at roughly 5-10 volts, the plate will be at 200 volts, and the tube will be operating in a happy place.

I think you're a little overboard on the supply, but that's relatively harmless. Once you have the basic circuit up and running, you can start playing around with current sources, servoes, and all kinds of wonderful creatures.
Click to expand...

OK. My brain is starting to hurt. I'm missing something pretty basic here SY and I'm sure that when I realize what it is I will feel really silly. I have a peak to peak difference in potential between the + and - rails of 400v with the grid sitting in the middle at a 200v difference from either the + or - rail. A 20K resistor with a 10mA current flow through it will drop 200 volts, which would seem to raise the cathode to the same potential as that of the grid, 0 volts (ground). Since the voltage drop in a series circuit is cumulative there would have to be a 200v drop across the tube itself also. My question is where does the 10 volt difference in potential between the cathode and the grid come from? You must understand I'm not accustomed to bipolar supplies nor to CFs.
 
#72
Well, you're assuming that it's exactly half the voltage dropped across the plate-to-cathode. It's not, it's more like 0.49. The drop from plate to grid is indeed half. The calculation of resistor size based on the cathode being halfway between the two rails is just an approximation- we make that approximation because we know that the proper bias voltage will be something very small compared to 200V (like 5 or 10 volts), so we can ignore it for purpose of estimation. Now when we set it up in a real circuit, we find that the drop across the resistor isn't 200V, it's 200V plus the bias voltage (which is still pretty close to 200V). And the cathode will be sitting at 5-10V, which is pretty much at ground compared to 200V- the estimation is a valid one.

That deviation at the cathode from the halfway point is due to the self-biasing nature of the feedback from the cathode resistor. Again, the feedback takes care of it so you don't have to. Whether it's 0.49, 0.48, or 0.495, or whatever, the cathode will draw sufficient current to bias itself properly. The voltages/currents will be slightly different than the estimates, but a 4-5% deviation is absolutely negligible here.
 
#73
One more point- if you do a thought experiment and substitute a current source for the cathode resistor, the voltages still work out the same way. In fact, the cathode resistor is more or less a current source.

The disadvantage of the sort of estimation calculation I did here is that you will get a slightly different current than you expect- instead of (for example) 20 ma, it might be 20.5 or 21. If that's critical to you for esthetic reasons, you can do some fine tweaking of the resistor value (use 10.5K instead of 10K!). It will have NO effect on the audible performance of the circuit, and next to zero effect on the measurements.
 
#74
Thanks SY. I'm going to go ahead and use a 20K resistor and go for a 10mA current. I will make sure to use Thermistors in the Power supply and I will bias up the heater to avoid that problem also. Thanks for your help very much. It will be a month or so before I start work on this project as I am soldering on a new amp as we speak and it will take a few weeks to save enough to purchase the parts I need for the linestage. I will let you know the results of my endeavor as soon as I have something to report. Thanks again.
 
#75
Here is the version of the 5687 Buffer that I plan on building. Thanks SY et al.
 

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