Help with load line for LTP phase inverter

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SixStringBender
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Help with load line for LTP phase inverter

Post by SixStringBender »

Hello fellows, I want to do the load line design for the LTP phase inverter in a single channel, no tremolo blackface type build. I'm seeking better balance when the amp is pushed. Also wanting less drive to the output tubes.

I read Aiken's LTP phase inverter load line article.

I have 435VDC B+3 (supply to the PI plate resistors). Do I use the B+3 when calculating the load line plot and deduct 80V to arrive at 355VDC? The 80V deduction is what I plan to drop with the tail resistor.

For example, using a 12AT7 with 47K plate resistors.
355V/47K=7.55mA.

So, I believe, The first end point of the load line is the intersection of the 355V plate supply voltage and the zero current line.
And the other intersection of the plate voltage of 0V and the plate current of 7.55mA is the other end point of the load line.

Is this correct? 7.55mA is within the 10mA max spec of the 12AT7.

Changing focus:
The 12AX7, for example, has a max plate voltage of 330VDC. This is the plate voltage on the plate and not the B+ supply, right?

Because I have a blackface with 428VDC B+3 (LTP PI supply) and 245VDC on the PI plates. 82K and 100K plate resistors.
The B+4 (preamp supply) is 408VDC and 310VDC is on the preamp plates in this amp. It has the 1K and 4.7K drop resistors in the power supply.
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Re: Help with load line for LTP phase inverter

Post by Stevem »

First off what output tubes are you running?
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Not screaming like the passengers in his car!😊

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pdf64
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Re: Help with load line for LTP phase inverter

Post by pdf64 »

Yes it's the HT to cathode voltage that is relevant for load lines.
The limiting plate voltage will be plate to cathode voltage; as the plate voltage can swing up toward the HT with signal, the HT should be taken into account.
Note that one or another tube limiting values are often breached by guitar amps; probably best not to breach too many simultaneously.
SixStringBender
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Re: Help with load line for LTP phase inverter

Post by SixStringBender »

The first amp I mentioned I want to redesign the PI for isn't built yet, but it is running in LTSpice. It drives a pair of MOSFETs, that drive the simulated 12W 6V6 output tubes biased at 70%. I won't use the 6V6 in the real amp because they can't handle the B+. I like the JJ 6V6S. I'm just using these because the simulated output transformer we found for LTSpice matches the impedance of the 6V6. I don't know of a simulated JJ 6V6S, or I would use it instead. If I had a simulated 6L6 output tranny I would use it with a pair of simulated 6L6GC's.

The PI is balanced with low volume and MV settings, until the simulated volume and MV is pushed, and not very hard before balance is reduced. This is why I want to redesign the PI with a larger tail. There are two bias pots and both output tubes are biased at idle, volume and MV at 0% with no signal, biased at 70%.
SixStringBender
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Re: Help with load line for LTP phase inverter

Post by SixStringBender »

Thanks Pete! So the larger blackface amps with the 1K and 4.7K drop resistors are over spec on the HV. Should I reduce the voltage a bit? Or is it safe because the plate to cathode voltage is within the max spec? I'm assuming it is okay since blackface amps aren't melting down preamp tubes.
pdf64
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Re: Help with load line for LTP phase inverter

Post by pdf64 »

As your LTP cathode voltage is elevated above 0V then as long as the HT to cathode voltage isn't much in excess of 330V then it should be ok.
But bear in mind the limit on cathode to heater voltage, some tubes are only 100V; hence your may need to elevate the heaters above 0V too.
SixStringBender
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Re: Help with load line for LTP phase inverter

Post by SixStringBender »

Yes, thanks. I have a 28VDC power supply in my schematic. I already have the artificial center tap of the filament supply at it. HAHA It was there to power the drains of two MOSFETs, so I used it.
tubeswell
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Re: Help with load line for LTP phase inverter

Post by tubeswell »

SixStringBender wrote: Tue Jun 20, 2017 5:02 am Hello fellows, I want to do the load line design for the LTP phase inverter in a single channel, no tremolo blackface type build. I'm seeking better balance when the amp is pushed. Also wanting less drive to the output tubes...
The recipe for this is higher tail resistance and matching plate resistances.

Bear in mind that:

1) The tail (including bias resistor) sees the current for both triodes, but each plate resistor only sees the plate current for its own triode.

2) The plate-to-cathode voltage does not need to be high, and in fact the lower it is, the better balance - at the expense of gain.

3) It pays to keep the bias on the cool side when attempting to get 'more perfect balance' in a LTP, but don't go tool cold either. (i.e. choose a bias resistance that doesn't force the inverting triode to go into grid current limiting or plate cutoff too soon - often the cause of mismatch in balance under big signal conditions).
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SixStringBender
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Re: Help with load line for LTP phase inverter

Post by SixStringBender »

Thank you tubeswell. Every bit of knowledge helps and I do appreciate it.

Plus I like your signature. HAHA
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jjman
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Re: Help with load line for LTP phase inverter

Post by jjman »

Since you are not averse to employing SS devices to the PI you could check on using a Constant Current setup on the tail. I don't know if those force more balance but it's worth checking.
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pdf64
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Re: Help with load line for LTP phase inverter

Post by pdf64 »

Also don't overlook Merlin's page on the LTP http://www.valvewizard.co.uk/acltp.html
or the ampbooks calculator https://www.ampbooks.com/mobile/amplifi ... iled-pair/
Yes, a CC source as the tail has very high ac impedance so improves the balance.
See how Marshall implemented it http://schems.com/schematicheaven.net/m ... 2x100w.pdf
SixStringBender
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Re: Help with load line for LTP phase inverter

Post by SixStringBender »

Thanks fellows. I have read Merlin's article before, but until I read it after reading Aiken's I didn't know how to apply some of it. Reading it again was a nice help. Aiken's explination of the load line got me started.

Before I found Aiken's LTP load line article with equal plate resistors and a large tail I was using the amp books LTP calculator to balance the PI. Once in LTSpice that balance went sour when the amp is pushed. I was using the 22K Fender tail.

I found a mistake I almost made when I reread Aiken's article for the humpteenth time. I was going to drop 80V for my amp too, but I caught where he said 25-30% of the total B+ for the PI supply must be dropped for balance. For my amp 25% of the B+3 is 108.75V. So I calculated the PI with a drop of 110V across the tail.

With the resistors I calculated from the load line the PI is now running better balanced in LTSpice.

My B+3 is 435V. 435*.25=108.75V.

So 435V-110V=325V. 325V is a safe voltage for the 12AY7 I plan to use. I plan on using 82K plate resistors because it is the lowest 2W standard value I know of that puts the plate current in a safe area.

325V/82K=3.96mA.

If I'm correct, at 325V the 12AY7 can handle 4.6mA plate current. The 12AY7's max plate dissipation is 1.5W. So 1.5/325V=4.6mA. So 3.96mA (4mA) should be safe.

My friend plotted the load line and went with -3V and 1.8mA.

Long story shorter, I calculated an 833R tail (825R 1% metal film) and a 30.5K tail (30.9K 1% metal film).

Believe it or not I'm a blind dude using a powerful screen reader that gives me control of my PC. I have a buddy helping me plot the load lines. He tells me what looks best. and I make the calculations. So feel free to tell me if I messed up. lol My buddy is also the one running LTSpice, and he said these resistors, with the 12AY7, gave us our best balance yet.

I'll find info to study on the SS constant current source. Thanks.

By the way, I calculated 2W plate resistors because the voltage is 325V and the current is 4mA. So using Ohmsy, 325*.004=1.3W.
tubeswell
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Re: Help with load line for LTP phase inverter

Post by tubeswell »

SixStringBender wrote: Fri Jun 23, 2017 5:38 pm ... he said 25-30% of the total B+ for the PI supply must be dropped for balance. For my amp 25% of the B+3 is 108.75V. So I calculated the PI with a drop of 110V across the tail.
Good start if you are going to drive 6V6s or EL84s - which don't take much input signal to overdrive. (If you're driving 6L6s, its better to limit the tail to 50V or so for more gain.)
SixStringBender wrote: Fri Jun 23, 2017 5:38 pmSo 435V-110V=325V.
Yes. 325V becomes the effective HT voltage reference point for the DC load line on the x-axis (for calculating the LTP load line)
SixStringBender wrote: Fri Jun 23, 2017 5:38 pm325V/82K=3.96mA.
3.96mA becomes the maximum possible current that is achievable if you are dropping all the HT voltage across an 82k plate resistor. i.e. it is the reference point for the DC load line on the y axis - It is not the idle current, and it is not the reference for calculating the power rating needed for the plate resistor. (See further replies below)
SixStringBender wrote: Fri Jun 23, 2017 5:38 pmIf I'm correct, at 325V the 12AY7 can handle 4.6mA plate current.
A 12AY7 triode has a maximum plate dissipation of 1.5W. And 4.6mA is the current required to achieve this at 325V. But the dissipation seen by the plate resistor is the average voltage dropped across the resistor by the average current through the plate resistor - (even under signal conditions). Under your scenario of a B+ of 435 and a tail voltage of 109V, the average plate resistor voltage drop will not be anywhere near 325V, and the average plate current will not be anywhere near 4.6mA. Rather, a more accurate estimate of these averages can be obtained by working out the idle voltage with your effective HT of 325 and whatever bias voltage you adopt.
SixStringBender wrote: Fri Jun 23, 2017 5:38 pmMy friend plotted the load line and went with -3V and 1.8mA.
Yes, for a 12AY7 triode with an HT = 325V, Ra = 82k, Vg1 = -3V, I get a plate idle voltage of about 175V with about 1.8mA of plate current (see the intersection of the blue and green lines on the DC load line plot attached. The blue line is the plate resistor load line. The green line is the cathode resistor load line. Note that for the purpose of plotting this load line, I adopted a bias resistor value of 1640R, which is 2 x 820R. This is because in a typical LTP with a 820R bias resistor, you are seeing the tube current of 2 triodes. So in order to plot the same -3V bias voltage for one-triode's-worth of current in the model, I needed to use twice the bias resistance value)
SixStringBender wrote: Fri Jun 23, 2017 5:38 pmBy the way, I calculated 2W plate resistors because the voltage is 325V and the current is 4mA. So using Ohmsy, 325*.004=1.3W.
Using the plate idle voltage obtained from the DC load line I plotted:

325 - 175 = 150V dropped across the plate resistor

150V x 0.0018A = 0.27W.

A 1/2W resistor will be okay, and a 1W resistor would be more than adequate.

FWIW, the plate dissipation for each triode under these conditions will be (175-109)V x 0.0018A = 0.12W.

The size of the tail resistance (including the 820R bias resistor) would need to be 109/0.0036A = 30.3k. Try 27k tail plus 820R bias to start with.
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Last edited by tubeswell on Sat Jun 24, 2017 11:57 am, edited 2 times in total.
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martin manning
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Re: Help with load line for LTP phase inverter

Post by martin manning »

Looks like the plot above came from Blencowe's load line plotter. Link for those who want to play (scroll to "downloads" at the bottom of the page): http://www.valvewizard.co.uk/links.html
tubeswell
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Re: Help with load line for LTP phase inverter

Post by tubeswell »

Indeed it was. :)
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