EQP1A with cascode amp circuit

[Deepdark]

Thanks Ian and CJ. Ian, i just saw that your poor mans and my design are really close. Can you explain a little about the impedance required to drive the unit at 26db. If it's possible to give me a little explanation. Thanks for helping me

Sorry, I don't understand the question - not sure what you mean by 'impedance to drive the unit at 26dB'.  Can you explain a bit more what you want?

Cheers

Ian

"If you only plan to drive 10K bridging load then you could use a good quality 600:600 transformer and have the full 26dB gain available."

I want the full 26db, but don't quiet understand how to achieve this. Do you mean using an input transformer of 600:10K and an output transformer of 600:600?

 

[ruffrecords]

Thanks Ian and CJ. Ian, i just saw that your poor mans and my design are really close. Can you explain a little about the impedance required to drive the unit at 26db. If it's possible to give me a little explanation. Thanks for helping me

Sorry, I don't understand the question - not sure what you mean by 'impedance to drive the unit at 26dB'.  Can you explain a bit more what you want?

Cheers

Ian

"If you only plan to drive 10K bridging load then you could use a good quality 600:600 transformer and have the full 26dB gain available."

I want the full 26db, but don't quiet understand how to achieve this. Do you mean using an input transformer of 600:10K and an output transformer of 600:600?

OK, now I understand. The mu follower gives you the 26dB. The 600:600 transformer at the output does not lose any of it as long as you only load with with no less than 10K. There's no need for a transformer at the input. Basically the circuit I posted will do it.

Cheers

Ian

 

[Deepdark]

Ok. So how can i assure to load at minimum 10k? If i use it, for exemple, after a Compressor, is there a kind of load resistor i can put somewhere to be sure to get the right load?

Why do i don't need an input transformer?  Does the gpultec could have Been made without the input transformer? I though the input tranny was useful to bloc dc, rfi, match impedance, etc

 

[ruffrecords]

Ok. So how can i assure to load at minimum 10k? If i use it, for example, after a Compressor, is there a kind of load resistor i can put somewhere to be sure to get the right load?

The load impedance I was referring to is the one you feed the EQ into - in other words the input  impedance of the equipment connected to the EQ output. Most equipment these days has an input impedance of 10K or more. Some older gear has 600 ohm inputs - like the EQP1A.

Why do i don't need an input transformer?  Does the gpultec could have Been made without the input transformer? I though the input tranny was useful to bloc dc, rfi, match impedance, etc

Sorry, I thought you meant a transformer just before the tube. You do need an input transformer to provide a balanced input. This needs to be a 600:600 type if you are using the original EQP1A  EQ section.

Cheers

Ian

 

[Deepdark]

Thanks a lot. Yes i use the original filter section. So i need an input and output at 600:600. May i know how to evaluate the input transformer impedance? 

Also, i was aski g myself how to calculate the upper grid by-pass cap (RG2). Is there a general roule of thumb ie some maths I can apply?

Is a ht of 300v too high for this tube? And does my resistors values look ok, according to my maths?

Thanks for the help

 

[ruffrecords]

Thanks a lot. Yes i use the original filter section. So i need an input and output at 600:600. May i know how to evaluate the input transformer impedance? 

It is usually specified by manufacturers. The original EQP1A was designed to present a 600 ohm load at its input.

Also, i was aski g myself how to calculate the upper grid by-pass cap (RG2). Is there a general roule of thumb ie some maths I can apply?

The top triode is a cathode follower.  Its input impedance is therefore approximately the grid resistor multiplied by the mu of the tube so the input impedance is likely to be in excess of 10M ohms. The capacitor couples the output of the bottom triode to the grid of the top triode so it needs to be large enough to pass the lowest frequency of interest, say 20Hz. For a 3dB drop at 20Hz, a 1M input needs a coupling cap of at least 1nF. 100nF will be plenty.

Is a ht of 300v too high for this tube? And does my resistors values look ok, according to my maths?

Thanks for the help

300V is fine for this circuit. In a mu follower the middle resistor is always a compromise. It needs to be large enough so that, when multiplied by the mu of the top triode, it produces low levels of distortion. Larger values, however, drop more voltage so the output swing is reduced. You also want a reasonably high current to be able to drive a 10K load so again the middle resistor should not be too large else it will drop a lot of voltage. As a rule, if the effective anode load resistance is 10 times the tube's anode resistance the distortion will be low. Since the mu of the 6SN7 is 20 and you will probably operate it at a current that means its plate resistance is about 10K, the a middle resistance of 10K will be plenty. Any larger than that gives a tiny improvement in distortion but a large reduction is output swing.

 

[Deepdark]

Thanks a lot Ian. Actually, I run RI at 27K. The bias is about -3v at 3.95ma. I set the lower anode voltage at around 85V. I calculated the value of mu at around 17.5/18, not that far from 20. Does it looks right?

Also, I would like to calculate my power supply needs. How do I find my total current load? Do I apply the HT / load resistor (around 28K) = around 10.8ma??

 

[ruffrecords]

Thanks a lot Ian. Actually, I run RI at 27K. The bias is about -3v at 3.95ma. I set the lower anode voltage at around 85V. I calculated the value of mu at around 17.5/18, not that far from 20. Does it looks right?

You need to watch grid current. With -3V bias your peak to peak input is about 4V before you run the risk of grid current and the severe distortion it causes. This gives you a maximum of 80V pp output or 28V rms. I tend to bias both tubes at -5V and that gives me a good 40V rms at the output before grid current. To be honest I have ever quite understood the reason for running the two tubes at different bias points.

Also, I would like to calculate my power supply needs. How do I find my total current load? Do I apply the HT / load resistor (around 28K) = around 10.8ma??

It is a class A circuit. Average supply current equals idle current.

Cheers

Ian

 

[Deepdark]

So my idle current is the sum of the lower and upper plate current? When i bias the lower plate, i though it was common to set the plate voltage at around 1/3 HT. So you est it at 1/2 ht for both?

 

[Deepdark]

Also, I'm not sure how to get the ic current. By the graph? I looked at it and I'm not sure how to read it...Thanks Ian for helping me.

 

[ruffrecords]

So my idle current is the sum of the lower and upper plate current? When i bias the lower plate, i though it was common to set the plate voltage at around 1/3 HT. So you est it at 1/2 ht for both?

The two triodes and you 27K are in series therefore they must all carry the same current (the idle current). Each one may have a different voltage across it but they all carry an identical current. It is just like three different resistors across a supply. The 27K is fixed and the tube resistance depends on where you bias it.

So the first thing you do is pick the idle current. The work out the voltage drop across the resistor V = I * R and subtract this from your power supply voltage. The remaining voltage you have to divide somehow between the two triodes. To maximise output voltage swing I choose to make them equal.

The Valve Wizard has a different approach and also explains the use of tube graphs:

http://www.valvewizard.co.uk/mufollower.html

Cheers

Ian

 

[Deepdark]

Thanks Ian. Actually, i designed mine by the valve wizard method. Thats why i adjust the lower plate 1/3ht. ta bias point,-3v, i have around 3,95ma. Is it my idle curent? To achieve this, i est my lower triode bias at 85v, draw a vertical Line up. -2v curve. It gave me 3,95ma. Then, i sibstract 85v from my ht, it gave me 215v/2 = 107,5v. I then draw a vertical Line once i cross the 3,95ma point. So, the bias is -3v. To find the load resistor, i draw a Line from 215v, i cross the bias point of -3v and gave me around 7,6ma. My load is 27k.

Is my idle is 3,95ma? Or 7,6ma. That what i try to understand.

 

[Deepdark]

Ok. I read again the valve wizard tuto. 3,95ma is the current that run through the 2 triode because they are in series. Is it what We call when We speak about quiescent valve?

 

[ruffrecords]

Ok. I read again the valve wizard tuto. 3,95ma is the current that run through the 2 triode because they are in series. Is it what We call when We speak about quiescent valve?

Yes, the dc current flowing in the valve with no signal is called the idle current or the quiescent current.

Cheers

Ian

 

[Deepdark]

Thanks for your time Ian. I really appreciate it. How's your health, by the way?

 

[ruffrecords]

Thanks for your time Ian. I really appreciate it. How's your health, by the way?

Up and down. The plastic surgery after the removal of the skin cancer has left me as handsome as ever which is good news. Now I have gone and gotten gout in my left toe!!

Cheers

Ian

 

[Deepdark]

Wish you all the best Ian.

 

[PRR]

> another tentavive with a Mu-Follower topology

At a glance, I see a 27,000 Ohm resistor in series with the tube, and a 600 Ohm output load. This can't go well.

Ian's poor man make up stage is open to similar comments(though as he says, "the middle resistor is always a compromise"). The 10K is less a problem than 27K. I do wonder if it is remotely push-pull, or if it really works as a gain-stage and a cathode follower (with some economy of current).

> looking into a Mu-Follower device

We (both this thread and the Innernet at large) have terminology problems.

We have already confused cascOde and cascAde. They are quite different basic ideas.

I do not know what a "Mu follower" is. Mu is a large negative number. A Mu-based amplifier does not _follow_, it gains-up and inverts.

Mu-amp is often used for a very high impedance gain-stage, such as:
http://www.muzique.com/amz/mini.htm

With any non-gigantic load impedance, the Mu-amp's voltage gain is no better than a single tube.

What you seem to propose is a "SRPP". This is a tricky design, moreso than you would think from all the jabber about it. It can be good for medium impedances. It "can" drive low impedances to low levels with stunningly low THD (or high THD if the design misses the mark). It is overkill for high Z loading (you could make a better amp using two triodes differently).

 

[Deepdark]

This is what inspire me. http://www.valvewizard.co.uk/mufollower.html

Ian suggest the same page some line above. I don't know if it's a wrong terminology, maybe. According to the maths, where I put my bias, etc. it looks like I got a 556ohm impedance at the output of the stage. I should try a 10k and rebias my design, though. But why 27K can't go well? And what is the trouble with this design, and, ie, the poor man's one.???

 

[Deepdark]

@ Ian:

I just saw on the Poor man's thread that your input transformer is a 10K:10K. I would like to know if I'm better off with it or the 600:600? Actually, I planned using a CMLI-4X150D