Biasing of LTP and grid-clipping

[anson]

Hello,

I've got a question though about biasing and the long tail pair. I've read in Merlin's book (which I managed to get my library to buy) that to get more output swing out of a LTP, it's best to bias in the mid to warm range (I presume that means around - 0.5v to 1v or so), and that even though their may be clipping at the grid of inverting stage of the LTP, it doesn't show up in the power section because that power valve is in cut off at that time.

But it seems to me that while the grid-clipped part of the waveform from the inverting LTP stage may not show up in the corresponding power valve(s), the grid-clipped part of the waveform from the inverting stage will still be sent to the non-inverting stage via the coupled cathodes, where it then becomes amplified and send to the power valve(s) corresponding to the non inverting stage. Except this time the clipped portion will be fully amplified by the output section.

So in the end it seems that "warm biasing" will still lead to a clipped waveform showing up in the non-inverting output stage. Is this right?

Thanks
Anson

[Chuck H]

No, Merlin is correct because the output of the other half of the LTP is a mirror image, but in opposite phase. So the positive swing is still the relevant factor on the inverted side of the PI since the power tubes will be in cuttoff (in AB1) on the negative swing. It's a good bit to wrap your head around but it all adds up.

[anson]

No, Merlin is correct because the output of the other half of the LTP is a mirror image, but in opposite phase. So the positive swing is still the relevant factor on the inverted side of the PI since the power tubes will be in cuttoff (in AB1) on the negative swing. It's a good bit to wrap your head around but it all adds up.

Thanks chuck. But I'm still not clear about why. So I'll have to be led by the hand...

First, the down-going half on the input signal:

So lets use a hypethetical scenerio where the input signal into the LTP is a sine wave, 2v peak to peak, and an LTP that is biased at 0.5v. That means that the negative-going signal will cause the inverting LTP grid to go down to -1.5v, which will manifest as a positive going wave at the inverting LTP anode that will be sent to and amplified by one half of the AB1 push pull pair, while the other half of the push pull pair is in cut off.

Now the other half of the signal:

But then the positive-going signal going into the inverting LTP grid will go up to 0v, and the rest of the waveform will be clipped. This manifests as a negative-going clipped wave at the inverting LTP anode, into the corresponding push-pull half and desn't get perceived by the speaker becasue it in cut off. Am I right so far?

But whenever one half of the push-pull section is in cut off, the other must be amplifying, because that's the whole idea of using the push-pull system. Isn't it? And what else can it amplify at that moment other than the portion of the sine wave that was clipped by the inverting LTP grid? As far as I can tell, that part of the since wave that got chopped off is lost information. Once the grid clips at the inverting LTP grid, you can't get a sine wave out of that anymore. The current signal that then goes to the non-iverting stage via the cathodes will then have that clipped form, which will manifest out of the non-inverting stage as positive going chopped sine wave that will get amplified by the non inverting LTP's corresponding half of the push pull pair, which is not in cut off.

So that's how I still understand it. If incorrect, then there has to be a point somewhere in my explination that is fauly, and hopfully it can be pointed out.

Thanks!!

[kg]

how is this LTP set up?

very long tail to negative supply rail, input grids at ground?

medium length tail to ground, input grids reference floated above ground?

in any event, imo the primary job of a driver is to provide adequate output swing voltage. keeping that in mind, i always optimize plate voltage output swing by making it around 50-66% of Vak, using whatever bias voltage or plate/cathode resistor ratios i need to achieve it.

[anson]

how is this LTP set up?

Well the configuration I'm talking about is pretty much the "industry standard" version, with 1M grid resistors, 100k and 82k ohm anode resistors, 820 ohm bias resistor, 10k ohm tail, 500ohm bias shunt resistor, and 100n grid-to-ground cap for the non inverting stage. B+ is 400v. That's what's on my amp right now. I've changed the bias resistor value throughout the years from R470 to 1.2k, trying to get the most "clean" voltage swing from the LTP. I'm driving 4 KT 88's.

My main curiousity here is what happens when the bias resistor is changed from R820, to R470 ohm, to an even lower value like 100 ohm, resulting in warmer bias. In a normal preamp stage, this would mean that the in-going signal would be clipped by the grid long before the anode clips, but Merlin's book suggests that it doesn't, at least the way I read it anyway.
A

[Chuck H]

It's about the voltage on the grid as it relates to the cathode... As it related to the plate voltage as well... The tube needs to see maximum amplification efficiency on the positive swing (for maximum output as it relates to the original Q). The negative swing is only significant if it clips while the power tubes are conducting. Which it shouldn't if the bias is set properly.

What are you using the "500ohm bias shunt resistor" for??? If this is the shunt for the NFB that's a very different thing from bias and has a very different criteria.

[anson]

What are you using the "500ohm bias shunt resistor" for???

My mistake. What I really meant to say was "500 ohm global NFB shunt resistor"

[kg]

i think chuck's explained it pretty well.

the "bias voltage" that you're concerned about exceeding only comes into play during grid positive/plate negative situations.

during that time the output stage will likely be cut off, and any "extra" swing in that direction is effectively doing nothing.

similarly, depending on the magnitude of negative bias on the output stage then any driver grid negative/plate positive swing that exceeds the bias voltage will be clamped by grid conduction. unless of course you make provisions for handling it...

[jmaf]

I think he's talking about the bias on the grid of the triode with the 82k plate resistor. He's wondering something I was discussing with you a while back, kg. If he biases that grid at -0.5v, the positive swings coming from a high Z source will clip on the positive grid region.

IMO, correct me if I'm wrong, that bias is set by the first resistor after the common cathode junction. Elevating it to 1.5K should give him -1.5 V bias. Donno if I'm totally of the mark here, but that seems to be what he's talking about.

[Merlinb]

So in the end it seems that "warm biasing" will still lead to a clipped waveform showing up in the non-inverting output stage. Is this right?

You are right, but for 'normal' biasing this doesn't matter because the power valves need 10-30V of positive swing to fully drive them, but the non-inverting output of the LTP will be delivering way more than that, so its clipping will always be masked. Max clean swing still occurs with centre biasing though, as ever.

Really I should have said "centre-to-warm biasing" to avoid confusion.

But be warned; if the LTP is very warm biased and overdriven, the output signals will undergo a big shift in duty cycle that can cause one power valve (the one driven by the non-inverting output) to redplate! This is something I need to add to the second edition.

[anson]

I think he's talking about the bias on the grid of the triode with the 82k plate resistor.

Yes

the "bias voltage" that you're concerned about exceeding only comes into play during grid positive/plate negative situations.

during that time the output stage will likely be cut off, and any "extra" swing in that direction is effectively doing nothing.

As I understand it, it'll be cut-off for that particular PP half, but the other half will be conducting the clipped waveform. It won't be noticed in most output configurations, but if the headroom of the output vavles is huge and or/ there is lots of attenuation between PI and power valve grids, then the clip may show.

But be warned; if the LTP is very warm biased and overdriven, the output signals will undergo a big shift in duty cycle that can cause one power valve (the one driven by the non-inverting output) to redplate!

Yes, I was thinking that something like that might happen. Although I couldn't articulate so well.

Max clean swing still occurs with centre biasing though, as ever.

So I guess my first post was really made to make sure that center biasing is the way to get the most clean output swing from an LTP. It seems that it still is.

Thanks for all your posts!

[kg]

As I understand it, it'll be cut-off for that particular PP half, but the other half will be conducting the clipped waveform.

not following you here.

what "other half?" the other half of the splitter?

It won't be noticed in most output configurations, but if the headroom of the output vavles is huge and or/ there is lots of attenuation between PI and power valve grids, then the clip may show.

in that case, you have a driver that has insufficient pk-pk output swing to properly drive the finals.

[mooreamps]

But be warned; if the LTP is very warm biased and overdriven, the output signals will undergo a big shift in duty cycle that can cause one power valve (the one driven by the non-inverting output) to redplate! This is something I need to add to the second edition.

Redplate a triode ? at what, a few milliamperes ? How stupid is that ???? Fine, you go ahead and write it up that way...


You know what PT Barnum once said ?

"There's a sucker born every minute"....

-g

[Robert M. Martinelli]

Redplate a triode ? at what, a few milliamperes ? How stupid is that ???? Fine, you go ahead and write it up that way...


You know what PT Barnum once said ?

"There's a sucker born every minute"....

-g

The max. POWER (not current) an ECC83/12AX7 can handle is 1W, and it becomes way less than that when both the triodes are used at the same time, because of the limited ability of the envelope to dissipate heat.

If you go over the max dissipated POWER of a certain valve/tube, ANY valve/tube can "redplate", if the voltage drop is high enough, even an ECC83/12AX7, even with a few mAmps.

Oh, maybe a little off topic, but I think you had that coming....I would like you to be able to express your opinion without having to be impolite and insulting like you've been in your post. We all know about your beef with Merlin, but, if you believe your point to be a good and valid one, you could just as well try to express it in a more gentle way...but maybe I'm asking too much of you...

If you can't cope with this simple wish of mine, then maybe Steve can do something about it.

Isn't there enough hatred around already? Let's try to keep this "place" as peaceful and nice as possible.

JM2CW

Best regards

Bob

[Amp Kat]

I aproove Bob's message ^^^^^^