Determining necessary bias voltage from tube data sheets

[Michael Allen]

I have no idea how to interpret tube data sheet graphs, or draw lines to determine current and bias points and all that jazz.

I'm building a submini express using 5899 output tubes. I want to know where I should bias them at initial startup, so I can determine the circuitry for my bias tap. I'm using back to back 12V transformers and I've got a HV+ of 140V with no load. My existing bias tap circuit gives me a max negative voltage of about -12V. I don't think this is low enough so I'll put together a voltage multiplier to get me in the right range.

But I need to know that range first. How do I figure out my bias voltage from the datasheet info?

[Old Tele man]

...it's a miniature semi-remote cutoff pentode, but the first question should be "...how are you planning on using it?" and, then these:

1) which: triode or pentode operation?

2) which: small signal or 'power' amplification?

3) which: single-ended (SE) or push-pull (PP)?

4) which: class-A or class-AB1?

5) which: cathode bias or fixed bias?

6) what plate- (Vp) and screen-voltages (Vs) to be used?

...from the data sheet, it looks like you'll need a "fixed" bias of about -1.0 to -1.5Vdc with Vp = Vs = 100Vdc (EbIb chart on page 6) for pentode operation; or, about -2 to -2.5Vdc if triode operation at Vp = 150Vdc (EbIb chart on page 9).

[Alf]

Hi Teleman , this is interesting.

Maybe you could say a bit more about these datasheets ?

e.g. Transconductance.

Does 4200 umhs mean 4.2 mA/V ? Is it the same ?

And the platecurrent number in mA means how "strong" the tube is when it's new . I mean how much current it draws then ?

For people like me who haven't had a technical education this is quite difficult sometimes.


Thanks , Alf

[Old Tele man]

Howdy, Alf!

Yes, the transconductance value of 4500 µmhos is the same as 0.0045 A/V...however, in europe, it's defined as Siemens (S), so the value could also be called 4500 µS or 0.0045 S. Notice that the 'old' USA units of mhos is simply ohms backwards, named so because transconductance (A/V) is simply the inverse unit of resistance (V/A).

The plate current value shown as Ip = 7.2 mA is just the value at the stated operating condition: Vp = Vs = 100Vdc, Rk = 120 ohms, and Is = 2.0 mA, it is NOT a value indicating tube strength...that's the transconductance value = figure of merit or goodness.

With that tube, the two design-controlling (limiting) conditions to be mindful of are: (1) the LOW plate and screen dissipation wattages of 1.1W and 0.55W, respectively; and, (2) the "absolute" maximim Cathode Current value of 16.5 mAdc. The best way to do these are to: (1) draw-in a 1.1W dissipation curve over the EbIb curves (triode and/or pentode) and (2) also draw a straight line from the 16.5 mA point on Y-axis (Ib) horizontally across the EbIb curves...together, they show you the tubes operating boundaries because we need to keep both idle and maximum power points at/below BOTH lines for Class-A operation.

With push-pull Class-AB operation, however, you can approximately double the 16.5 mA value because with PP operation each tube conducts on alternating halfs of the input signal and rests during the other half, so the "average" current remains at 16.5 mA, like this:

Ik(avg) = (2*16.5 mA)/2 = 16.5 mA

...where the first "2" is because you've doubled the maximum value, while the second "2" value is because its occurs on alternate cycles...so, they effectively cancel.

[Alf]

Thanks man for the reply , hope I can "bother you" some more whenever I don't understand these things.

Regards, Alf

[Old Tele man]

...to add a 'final' answer to the orginal question--Yes, there *IS* an way to estimate idle bias voltage, but it's only a rough approximation (±20%)!

...using the tubes' published transconductance (gm), triode & pentode amplification factor (µ1 and µ2) values, and the intended idle plate (Vpq) & screen (Vsq) voltages, it's possible to backsolve the Child-Langmuir 3/2's Law equation to determine an approximate control grid idle bias voltage (Vg.dc):

Vg.dc ~ 0.9*[(3/2)*(Ipq/gm)] - [Vsq/µ1 + Vpq/µ2]

...so, assuming these values and Class-A pentode operation:

Ipq = 7.3 mA (so, Ppq = 1.1W)
Isq = 1.9 mA (so, Psq = 0.3W)
Vp = 150 Vdc
Vs = 150 Vdc
gm = 0.0045 A/V
µ1 = 30 (guessed value, not given value)
µ2 = 1170 (260K*0.0045A/V)

Vg.dc ~ 0.9*[(3/2)*(0.0073A / 0.0045A/V)] - [150Vsq/30 + 150Vpq/1170]
Vg.dc ~ 0.9*[0.0219A / 0.0090A/V] - [5V + 0.13V]
Vg.dc ~ (2.19V - 5.13V) = -2.94 Vg.dc

...or, you could use the charts on page 10 and page 11, which also show Vg ~ 2.9-3.0Vdc.

[Michael Allen]

Old man tele, thanks so much for the explanations! I never would have come across those equations on my own I think.

To further elaborate, this amp will be class AB push-pull. The 5899 power section will be fixed bias, pentode operation, with plate value of around 130? I'm guessing because I don't know how far my unloaded 140V supply will pull down when it gets loaded. Screen voltage should be about 10V lower I think.

To bring up another question, how do I determine plate to plate load resistance? Other tube datasheets I've looked at list this value, but this datasheet does not. Is there a way to figure this out?

[Old Tele man]

...the "rule-of-thumb" for push-pull pentodes is plate-to-plate impedance (Zpp) is approximately 1/6th to 1/8th of the tubes dynamic plate resistance (rp), so that'd be something between:

Zpp ~ 260K/6 = 43.3K ohms
Zpp ~ 260K/8 = 32.5K ohms

...that's way out of the typical OT region, so you might have to look for a center-tapped differential line-balancing transformer or something similar?!?

...also, with PP operation (the Vg.dc example above was for SE Class-A), the idle plate current will be lower, but the idle plate voltage MUST be kept at/below 165Vdc for continuous operation...hence, the "recommended" use of 150Vdc as a maximum intended voltage in all the specs and curves.

...so, if we arbitrarily assume Ip.max = 16.5 mA (for now) and Vp.q = 150, that'd be an effective plate load resistance (Ra or RL) of 9090 ohms so the Zpp value (which is 4*RL) would be about 36K ohms, which is roughly in-between the above 32.5K-43.3K estimated values. Since Vs should be well filtered (but Vp isn't so critical due to OT cancellation of 60/120Hz ripple), it's common that Vs WILL be lower than Vp due to "smoothing" (either choke or RC filtering). Which do you plan to use?

...now, to estimate idle plate current, lets assume 70% plate dissipation: Ppq = (0.7*1.1W) = 0.77W or about 0.8W, so at Vp.q = 150 Vdc, that'd be:

Ip.q = (Pp.q / Vp.q) = (0.77W/150V) = 0.0051A, or about 5 mA ...and Is.q would be roughly 1.3 mA (~¼th of Ip.q)

...with "fixed" bias, Vg.dc = -3.8 Vdc, or with "cathode" bias Rk ~ 302 ohms, shared by both tubes.

...screen grid voltage CAN be the same as plate voltage (for maximum tube efficiency and power), but yes, Vs certainly can be lower than Vp, but doesn't HAVE to be. The "danger" is when Vs is HIGHER than Vp, which can lead to current runaway (and screen 'melt down') during periods of high power output.

[raiken]

If you look at the plate curves for Ec2 - 150V, a good load line that intersects the "knee" of the curves would put you around (130V/24mA) = 5.4K, so for push-pull class AB1 you'd want 4x that or around 22K p-p.

[Old Tele man]

Originally Posted by raiken

If you look at the plate curves for Ec2 - 150V, a good load line that intersects the "knee" of the curves would put you around (130V/24mA) = 5.4K, so for push-pull class AB1 you'd want 4x that or around 22K p-p.

Randall brings up an excellent point, ie: running the effective RL load line through the "knee" is where optimum output occurs. The rule-of-thumb values aim more for voltage gain over power gain.

...but (correct me if I'm wrong) I assumed Mr. Allen isn't familiar with "reading" the EbIb curves yet, so I just stuck with the basic equations.

[tubeswell]

Originally Posted by Michael Allen

Old man tele, thanks so much for the explanations! I never would have come across those equations on my own I think.

Yup you can say that again. I find these posts most useful (if a little technical)

[raiken]

Originally Posted by Old Tele man

...but (correct me if I'm wrong) I assumed Mr. Allen isn't familiar with "reading" the EbIb curves yet, so I just stuck with the basic equations.

I wasn't intending to slight your equations! Just offering an alternative method of finding a good load impedance that maximizes output power - he's only going to get a watt or so...

[Old Tele man]

Originally Posted by raiken

I wasn't intending to slight your equations! Just offering an alternative method of finding a good load impedance that maximizes output power - he's only going to get a watt or so...

...I didn't feel you were slighting me at all, because, actually, you "jumped ahead" to the answer that I was (eventually) going to present--using the curve to find the numbers...using the "..thru the knee..." method.

...I was the HARE and you were the RABBIT (ha,ha)!

[Michael Allen]

Starting to get this I think. To draw the load line, you start with one point at 0 current, 150V. Where is the second point on the Y axis? At 16.5mA or 33mA? And then you just adjust that point on the horizontal axis until the line intersects the knee of where we want idle current to be?

So as Mr. Aiken suggests, adjust the horizontal voltage to 130V and draw a line to 33mA? Then at about 12mA would be the intersection of the -1.5V grid voltage?

[raiken]

Originally Posted by Michael Allen

Starting to get this I think. To draw the load line, you start with one point at 0 current, 150V. Where is the second point on the Y axis? At 16.5mA or 33mA? And then you just adjust that point on the horizontal axis until the line intersects the knee of where we want idle current to be?

So as Mr. Aiken suggests, adjust the horizontal voltage to 130V and draw a line to 33mA? Then at about 12mA would be the intersection of the -1.5V grid voltage?

I just did a quick visual approximation using your plate voltage and looked where it intersected the knee to get a rough estimate of the required plate resistance (which is multiplied by 4 for class AB1 push-pull operation). If you want a more detailed explanation check out the pentode section of these papers (although they all use different methods, so it may confuse the issue more - you may want to find a good textbook explanation

http://www.ax84.com/media/ax84_m225.pdf

http://www.freewebs.com/valvewizard1/pp.html

http://greygum.net/sbench/sbench102/pent.html

...although they all use different methods, so it may confuse the issue more - you may want to find a good textbook explanation here:

http://www.pmillett.com/tecnical_books_online.htm


RA

[Michael Allen]

I'm trying to put this amp together with what I have, and the OT i want to use is a 10W, with 8k, 4k, 2k, 1k primaries and 8 and 4 ohm secondaries. The only speaker cab I have right now is a single 8 ohm Private Jack. So the highest impedance I could get would be 8 ohm cab into 4 ohm secondary to give 16k reflected load. That's short of the suggested 22k. But 22k is well short of the mathematical result of 36k.

Just what determines the acceptable reflected primary load? If I use this 16k load, then with 130V on the plate, the current would be about 8mA. So I would need to bias colder to, say, -2.5V?

[Old Tele man]

Originally Posted by Michael Allen

Just what determines the acceptable reflected primary load?

...one hyphenated-word answer: turns-ratio (TR), ie: TR = SQRT[Np/Ns]

...remember, a transformer does NOT itself set ANY resistance / impedance value, it merely TRANSFORMS up or down (proportional to TR) the value that's presented to the output or secondary winding, ie: speaker(s) in our case. Connect a 1.6K ohm load to the primary of a 20:1 Turns-Ratio OT and the secondary will 'see' a 4 ohm load. Likewise, connecting a 4-ohm load to the secondary will "reflect" a 1.6K ohm load back to the primary winding! The 4X thing is a function of Class-AB push-pull (PP) operation, hence:

Zpp ~ rp/6-to-rp/8
RL = Zo*(TR)^2 ...Zo = speaker impedance
Zpp = 4*RL ...due to Class-AB push-pull operation.
Zpp = 4*Zo*(TR)^2

...we want the "reflected" resistance / impedance that's presented at the PRIMARY windings (where the tubes are) to "match" (roughly) the 1/6-1/8th ratio of the tubes' working plate resistance value (rp).

...so, with some "creative" selection of the available OT output taps and the total speaker impedance value(s), you might be able to come up with a combination that gets you pretty close to the desired OT primary value.

Zi = Zo*(TR)^2 or, RL = Z.spkr*(TR)^2 and thus: Zpp = 4*Z.spkr*(TR)^2

...for example, a 4-ohm speaker load connected to an OT with TR=20:1, will reflect an effective RL = 1.6K and a Zpp = 5.6K, and would sometimes be designed as: 5.6Kpp:4.

[Michael Allen]

Awesome, thanks so much for all the help guys.

So if I plug my 8ohm speaker into the 8ohm secondary and use the 8k primary, i'll get Zpp = 4*8*1000 = 32K. Should be perfectly suitable.

Thanks so much again!

[raiken]

Originally Posted by Michael Allen

Awesome, thanks so much for all the help guys.

So if I plug my 8ohm speaker into the 8ohm secondary and use the 8k primary, i'll get Zpp = 4*8*1000 = 32K. Should be perfectly suitable.

Thanks so much again!

No, if you plug your 8 ohm speaker into the 8 ohm secondary and use the 8k primary, you'll get 8k.

If you plug your 8 ohm speaker into the 4 ohm secondary and use the 8k primary, you'll get 16k.

RA

[Michael Allen]

Right, but then 8k * 4 since it's AB push pull for a Zpp of 32k?

[raiken]

Originally Posted by Michael Allen

Right, but then 8k * 4 since it's AB push pull for a Zpp of 32k?

No, push-pull output transformers are spec'd for plate-to-plate impedance. You'll get 16K plate-to-plate if you connect an 8 ohm speaker to the 4 ohm tap and use the 8K primary.

If you connect a 16 ohm speaker to the 4 ohm tap and use the 8K primary you'll get 32K.

Note, however, that the low frequency cutoff point will change because the transformer was designed to run into the matched impedances set on the secondary. The formula for the lower -3dB point is f = Z/(2*pi*L), where L is the primary inductance. Since the primary inductance is constant and fixed by the number of turns and the core characteristics, the lower -3dB point will go up by a factor of 4 when you change the reflected impedance from 8k to 32k, so your low end will decrease.

For more info:

http://www.aikenamps.com/OutputTransformers.html

RA