DIYstompboxes.com

DIY Stompboxes => Building your own stompbox => Topic started by: brett on February 22, 2007, 06:57:58 AM

Title: What generates the grid bias on low voltage tubes with no cathode resistor?
Post by: brett on February 22, 2007, 06:57:58 AM
Hi
There are quite a few schematics around showing low voltage tube preamps.  e.g. 12U7 at 12V or 12AX7s at 30V. 
These schematics often show a variation on the usual cathode bias system.  The cathode goes direct to ground (ie no bias resistor, no cap).  The grid is tied to ground by the usual 470k to 1M resistor.

What is creating the bias voltage in these circuits?  Does the cathode have internal resistance or something?
thanks for any advice

Here's an example from the sopht.ca amp site:
(http://www.sopht.ca/schematics/sopht_12al8.jpg)
Title: Re: What generates the grid bias on low voltage tubes with no cathode resistor?
Post by: db on February 22, 2007, 08:16:53 AM
They are biased on actually, because a grid to cathode voltage of 0 is really a maximum bias voltage, in fact it's the equivalent of IDSS for JFETs.  The cathode resistor serves to reduce this to a required quiescent plate current.  The other main difference between this and the usual HT circtuits is that the plate voltage is much lower so they are operating in a very small corner of the normal output characteristic.  I haven't played with these circuits but I suspect that if a cathode resistor is used, you don't get enough conduction and have to use quite a high value of plate/drain resistance to bias the plate at roughly half supply.
Title: Re: What generates the grid bias on low voltage tubes with no cathode resistor?
Post by: R.G. on February 22, 2007, 08:36:58 AM
Grid leakage. Some electrons strike the grid and stick. This makes for a small reverse bias on the grid. It takes special techniques to measure this because of the very high impedance of the grid.
Title: Re: What generates the grid bias on low voltage tubes with no cathode resistor?
Post by: JonFrum on February 22, 2007, 11:45:37 AM
You'll find grid leak bias in some early guitar tube amps. Look at some early Gibson amps like the GA-15 and the Les Paul Jr. They usually bump up the grid-to-ground resistor up to 5-10M. The higher resistance keeps the charge from going to ground. The sound is either described as "vintage blues" or "ratty". I believe they went to cathode bias to get more headroom.
Title: Re: What generates the grid bias on low voltage tubes with no cathode resistor?
Post by: JonFrum on February 22, 2007, 11:46:28 AM
Quote from: JonFrum on February 22, 2007, 11:45:37 AM
You'll find grid leak bias in some early guitar tube amps. Look at some early Gibson amps like the GA-15 and the Les Paul Jr. They usually bump up the grid-to-ground resistor up to 5-10M. The higher resistance keeps the charge from going to ground. The sound is either described as "vintage blues" or "ratty". I believe they went to cathode bias to get more headroom.

Edit: I think Ted Weber sells a kit with a grid leak input stage.
Title: Re: What generates the grid bias on low voltage tubes with no cathode resistor?
Post by: db on February 22, 2007, 03:57:32 PM
Have I got it completely wrong or something?

Look at the datasheet for any 12AX7/ECC82/83 etc. for example:
http://www.drtube.com/datasheets/ecc82jj2003.pdf (http://www.drtube.com/datasheets/ecc82jj2003.pdf)

You will see that maximum plate current flows when VG is more positive i.e. closer to zero.  The lack of a cathode resistor sets VG (voltage at gate wrt cathode) to 0V!  Any cathode resistor will result in -ve VG and hence less current.  A VA of about 4V will result in a current somewhat less than 1mA which is probably ideal.  In fact, the use of 22K anode resistors in the circuit suggest an actual plate current of about 200uA.

So, this is the main reason why plate current flows, not grid leakage.

Please correct me if I am wrong (and I apologise if I am) but it looks pretty clear to me.
Title: Re: What generates the grid bias on low voltage tubes with no cathode resistor?
Post by: R.G. on February 22, 2007, 04:36:04 PM
I don't think you're completely wrong; probably a difference in the way we read the question.

QuoteYou will see that maximum plate current flows when VG is more positive i.e. closer to zero.  The lack of a cathode resistor sets VG (voltage at gate wrt cathode) to 0V!  Any cathode resistor will result in -ve VG and hence less current.  A VA of about 4V will result in a current somewhat less than 1mA which is probably ideal.  In fact, the use of 22K anode resistors in the circuit suggest an actual plate current of about 200uA.

So, this is the main reason why plate current flows, not grid leakage.

That's all correct. But that's not what he asked. He asked "what is creating the bias voltage in these circuits?"

You are correct that a tube conducts maximum with Vgk = 0. What we are saying is that even with no cathode resistor, Vgk is slightly negative because the grid intercepts some electrons which have a finite time that they have to leak away. This makes for a slight reverse bias on the grid, and lets the tube be biased slightly into the active region away from the fully-on side. Otherwise, it would not be able to amplify at all. Tubes have to be biased off, not on.

The clearest things may be impeded by language, which is a terribly clumsy tool.
Title: Re: What generates the grid bias on low voltage tubes with no cathode resistor?
Post by: JonFrum on February 22, 2007, 04:38:09 PM
You seem to be confused about tube biasing. The object of biasing - generally - is to put the grid at the mid point of its possible voltage range. That way, a signal applied to the grid will be free to vary above and below the bias point, giving you the most headroom possible.

Look at the plate curve graph in the data sheet you linked. For a 12AU7 tube, assume the plate is sitting at 100V. Draw an imaginary line from 100V, 0mA, to 0V, 5mA. The curves labeled 0V, -2V, -4V, etc, are grid voltages. In this case, your signal can swing between 0V and -8V before maxing out. So you would want to bias the grid at about -4V to allow for as much symmetrical swing between 0V and -8V.
With a typical cathode bias circuit, you would choose a cathode resistor that would drop 4V, and connect the grid to ground with a 1M resistor. Thus, your grid would be -4V with respect to the cathode, and you're golden.

In the case of grid leak bias, the cathode sits at 0V. Usually, a 5M or 10M resistor is used to connect the grid to ground. Under those conditions, he grid attracts electrons, which cannot escape to ground easily. As a result, a negative charge builds up at the grid. Again, your grid is now negative with respect to ground, and the signal has some room to move.

There are online sites that explain tube bias methods in greater detail.


R. G. beat me to it, but I'll let this through.
Title: Re: What generates the grid bias on low voltage tubes with no cathode resistor?
Post by: puretube on February 22, 2007, 04:54:49 PM
space-charge (cloud)...
Title: Re: What generates the grid bias on low voltage tubes with no cathode resistor?
Post by: db on February 22, 2007, 08:23:17 PM
Thanks for putting me straight!  Should've done my homework properly before jumping in feet first. 
Title: Re: What generates the grid bias on low voltage tubes with no cathode resistor?
Post by: brett on February 23, 2007, 08:22:46 AM
Thanks
That's brilliant.  I hadn't thought at all about the size of the pull-down resistor and its effect on grid accumulation.

So far, my experiments with different tubes at low voltage have been diappointing because of low gain and low transconductance.  I'm hoping that this biasing system (and a few other tricks) will help reduce those problems.
cheers 
Title: Re: What generates the grid bias on low voltage tubes with no cathode resistor?
Post by: JonFrum on February 23, 2007, 12:23:24 PM
Have you seen this?


http://amps.zugster.net/projects/subminiature (http://amps.zugster.net/projects/subminiature)
Title: Bias
Post by: Papa_lazerous on February 23, 2007, 01:02:54 PM
Wow I've just thought i'd jump in and say thanks. I've been trying to get my head round this for a while and now its all logical to me.