EQP1A with cascode amp circuit

[ruffrecords]

@ 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

My poor man's EQPA is designed to present a 10K impedance  at its input which is why it uses a 10K:10K  transformer. The original EQP1Ais designed tp present a 600 ohm impedance at its input which is why it uses a 600:600 input transformer. So, if you are using the original EQ circuit ypi should use a 600:600 transformer.

Cheers

Ian

 

[PRR]

> why 27K can't go well?

Set up a Voltage Divider with 600 Ohm useful load and 27,000 Ohm of series resistance. How much of the input (supply) voltage gets to the load? About 0.02 or 1/46th of it.

Or come to my house. The power line from the street is too-long. Lamps sag when toaster or well-pump start. Say I have 1 Ohm of "useless" resistance in the long line, and 10 Ohms of useful toaster resistance. 120V at the street is only 109V at my toaster and lamps. With 27,000:600 the wasted voltage is much-much more.

Sometimes it "can't be helped". At my house, I could upgrade or transform my too-long power line, but the cost would be many thousand bucks, which is more than I care to spend to reduce lamp-dimming. (So far the PC VCR etc are OK with 108V dips, or it would be more annoying.) In Vacuum Tubes, tubes do NOT conduct near as well as metals, and need "some" load impedance to work well.

IMHO trying to find your Voltage Gain and your Power Output in one lump is not a best path, especially when the load impedance is smaller than the internal impedance of most vacuum tubes.

> a way to save some $

The output transformer is surely a large part of the total cost. I would buy bigger bottles to avoid bigger iron. Or even use a tube for voltage gain then use chips to do the hard work of driving a load (my rabbi does not prohibit that), though mixed 300V+30V systems can be awkward.

 

[Deepdark]

Ok. With this design, I applied Zout = 1/gm to find approx. the output impedance and I got around 560ohm. That's why I Wonder what is wrong with my design, or maybe my math. I tried to lower RI to 10k but I always finished with a really high current. I don't know how Ian managed to achieve this. I think we use the same tube

 

[ruffrecords]

A mu follower will not drive a 600 ohm load directly at anything other than a very small signal level without distortion. Remember, the calculated output impedance is no indicator of drive capability. My design will drive a 10K load to high levels (over +20dBu) which is why I recommend it is only used with a bridging load.  A 10K load reflected to the primary of a good quality 600:600 transformer will be close to 10K of most of the audio range.

For better drive capability use the same tube in an SRPP configuration. This is a rue push pull circuit and will drive a 2400 ohm load to very high levels. With a 2400:600 transformer fitted it will drive +20dBu into 600 ohms. The downsides are the stage has lower gain and higher distortion that the mu follower and you lose another 6dB of gain in the transformer. That's the price you pay if you want to drive a 600 ohm load.

Like all engineering, it's a compromise.

Cheers

Ian

 

[Deepdark]

To be honnest, i just want to use it with the original filter stage. I don't mind using a 10k:10k transformer at the input if it's what it take to had the required performance. But has you said, the original filter has been made for a 600ohm load, probably why We have an interstage to elevate 600 to what is needed. In my case, if i want to use the original filter section and the mu follower, is it a possible consideration? Can i put a 10k:10k at the input instead of the 600:600, or put a 10k load resistor across the secondary of the input transformer to drive it? What will happening if i run the original input filter section with a 10k:10k transformer?

By the way, i Check at the spec of my apogee ensemble and the input impedance is 10k so I think I'm ok to rout it and get my 10k load. Probably lot of equipements have too.

 

[Deepdark]

Ok.

I redid my math with the good advice from you, Ian. I Bias both triode the same, at around -5v, just to see what happen. So my HT is 300v. I looked at the -5v curve and tried to pick a Ia which is not too low. I ended up with a value of 2.95ma, VA of 130V and a resistor RK1 of 1K5 (approx.). Then, I substract HT-VA1 = 300-130= 170V. If I bias the upper triode the same, I pick the same point at -5v, 2.95ma at 130V (anode voltage). then I draw my load line from 170V, cross the bias point and ended up at around 12.2 ma. So, 170V/12.2ma = +/- 13K.  Which is not that far from yours. It's a totaly different way of doing it than the method in Valve wizard website. Are my maths ok?

 

[ruffrecords]

There is a simpler way to work out the middle resistor. You say you chose Va = 130V and Ia = 2.95mA. With a 300V supply that  leaves 300 - 260 = 40V across the resistor. Current through it must be 2.95mA so resistor value is:

40/2.95 K  = 13.56K

Cheers

Ian

 

[Deepdark]

Damn hahaha. You're right it's simpler that way XD So I guess my maths are ok, though. Do I bias it right to ensure good performance?

 

[Deepdark]

Ok here is a revised version taking in count what you said to me Ian and PRR, about the 27K load and the input load, etc.

HT = 310v
Bias = -5v
Ia = 3ma
Va = 135v

Tel me if it looks right. The last thing I need is the value of the output cap (the one at the OPT primary). Is 4u7 pf a good guess?

 

[ruffrecords]

Ok here is a revised version taking in count what you said to me Ian and PRR, about the 27K load and the input load, etc.

HT = 310v
Bias = -5v
Ia = 3ma
Va = 135v

Tel me if it looks right. The last thing I need is the value of the output cap (the one at the OPT primary). Is 4u7 pf a good guess?

That should work. The best thing is to build it and see. Real tubes never perform quite the same as the simulation. I still think the idle current is a little low. I usually start the design by deciding the idle current as this is what determines the output drive capability. Then work out the drop across the middle resistor and subtract this from the HT voltgr. Divide this by two and you have the plate voltage. Find where that plate voltage intersect the chosen current to determine the bias and hence tha value of the cathode resistor.

A 4.7uF output capacitor should be fine.

Cheers

Ian

 

[Deepdark]

Awesome. Thank you. Is it ok to put the 10k resistor across the secondary of the input transformer? My guess is that it Will act as a load.

 

[Deepdark]

following your method (it works great by the way) I can determine a lot of point really fast I tried my design with 5ma at 310ht. it gave me a bias of around  -3.85v of bias. If I lower to 4.5ma, I got a bias at around -4v. I think this one looks good.

 

[Deepdark]

Is the input load impedance you are referencing to (10K) is in relation with our load resistor of 10K?

I know lot of device has a 10K load across their output (ie, G9). But if I use it in insert of my apogee ensemble, the output from my ensemble goes to the input of the pultec and then the output of the pultec goes to the input of the apogee. The apogee impedance at the output is 100ohm and the line input is 10K (accoding to the website). So I should have trouble using it that way, no? If I strap a 10k load resistor across the sec. of the input transf. does it simulate a 10k load? I saw this on the SRPP version of gyraf.

 

[ruffrecords]

Is the input load impedance you are referencing to (10K) is in relation with our load resistor of 10K?

I know lot of device has a 10K load across their output (ie, G9). But if I use it in insert of my apogee ensemble, the output from my ensemble goes to the input of the pultec and then the output of the pultec goes to the input of the apogee. The apogee impedance at the output is 100ohm and the line input is 10K (accoding to the website). So I should have trouble using it that way, no? If I strap a 10k load resistor across the sec. of the input transf. does it simulate a 10k load? I saw this on the SRPP version of gyraf.

In the old days, most devices had a 600 ohm output impedance and a 600 ohm input impedance. One problem with this scheme is that you  lose 6dB of level every time you connected two items of equipment together. Another problem is you can only drive one input with an output. If you wanted to drive two 600 ohm inputs in parallel they would look like 300 ohms. Later, the idea of 10K bridging inputs came about. These inputs have a 10K input impedance so they represent a negligible load to a 600 ohm source and you can connect several inputs in parallel to one output without problems. Later still, especially when transistor circuits became common, the actual output impedance of circuits become much lower than 600 ohms - typically 100 ohms or less.

So, today, a 600 ohm output tends to mean that an output can drive a load as low as 600 ohms but in most cases it will be connected to a 10K load.

The original EQP1A was designed in the days of 600 ohm out and 600 ohm in so it needs a 600:600 input transformer with a 600 ohm load to work as designed. The filter components values are selected with this assumption. So if you use the original EQ design you must use a 600:600 input transformer loaded with 600 ohms at the input.

As you are driving modern equipment which has 10K bridging inputs, then the output of your EQP1A does not need to be able to drive a 600 ohm load. My gain make circuit will not drive a 600 ohm load but it will drive a  10K bridging load.

 

[Deepdark]

Is the input load impedance you are referencing to (10K) is in relation with our load resistor of 10K?

I know lot of device has a 10K load across their output (ie, G9). But if I use it in insert of my apogee ensemble, the output from my ensemble goes to the input of the pultec and then the output of the pultec goes to the input of the apogee. The apogee impedance at the output is 100ohm and the line input is 10K (accoding to the website). So I should have trouble using it that way, no? If I strap a 10k load resistor across the sec. of the input transf. does it simulate a 10k load? I saw this on the SRPP version of gyraf.

In the old days, most devices had a 600 ohm output impedance and a 600 ohm input impedance. One problem with this scheme is that you  lose 6dB of level every time you connected two items of equipment together. Another problem is you can only drive one input with an output. If you wanted to drive two 600 ohm inputs in parallel they would look like 300 ohms. Later, the idea of 10K bridging inputs came about. These inputs have a 10K input impedance so they represent a negligible load to a 600 ohm source and you can connect several inputs in parallel to one output without problems. Later still, especially when transistor circuits became common, the actual output impedance of circuits become much lower than 600 ohms - typically 100 ohms or less.

So, today, a 600 ohm output tends to mean that an output can drive a load as low as 600 ohms but in most cases it will be connected to a 10K load.

The original EQP1A was designed in the days of 600 ohm out and 600 ohm in so it needs a 600:600 input transformer with a 600 ohm load to work as designed. The filter components values are selected with this assumption. So if you use the original EQ design you must use a 600:600 input transformer loaded with 600 ohms at the input.

As you are driving modern equipment which has 10K bridging inputs, then the output of your EQP1A does not need to be able to drive a 600 ohm load. My gain make circuit will not drive a 600 ohm load but it will drive a  10K bridging load.

ok. Since I'm a visual person, the tricks of impedance is a little hard to understand (I really feel newby right now hahaha).  What you mean is: The original filter section was made for a 600ohm load (because it was in/out 600ohm at the time). That's what explain the bridging load resistor of 600ohm at the input and output. So later came the 10K bridging which represent a negligeable load for a 600ohm source. Am I allright?

So in our case, I need to feed the eqp into a 10K load, ie, the input of my apogee which has 10K represent the 10K load I need. Is that it? In this configuration, I'll get the max gain available, with a good 600:600 output transformer, without any 600ohm load resistor at it's output. So when you said the make up gain will not drive a 600 load but a 10K bridging load, you are refering to the input I fed the output of the eqp in. Is that it? So what is important here, is to connect the eqp in a 10K input, not connecting a 10k device into the eqp. I think I imagine the scenario in reverse. I thought I had to fed a 10K device into the eqp to get the load, but I need to fed the eqp INTO a 10K input. That's my understanding. So, looking at the input of the eqp, I need a 600:600ohm because I use the original filter section which was designed to be fed by a 600ohm output device. In my case, there is 2 scenarios possible.

1. I record some tracks with a, ie, Gyraf G9 and fed the output of the G9 INTO the input of the eqp, then the eqp output to the apogee input (10K). the output of the G9 being 600ohm, will I loose some gain or I'll be ok? I have my 10K load so I guess I should be ok.

2. I mix a track and put the eqp in insert of my apogee. The output of the apogee is 100ohm, fed INTO the eqp (which is 600) and go back to the apogee input rated 10K. I should be ok. Feeding 100ohm into 600ohm does not drive in some loosing gain, isn't it?

The last thing I would like to enlight is the resisor we strap at the secondary of the input transformer. The original eqp had a 620r. The Gpultec has a 10K (with a 600:600 input transformer). Do I need such a resistor. is it there to simulate a load to the transformer?

Finally, 600:600 input and output, feeding into a 10K device and I should get the max gain available. is that it? Thanks for your patience Ian. I feel guilty to monopolize your time with such newby questions.

 

[ruffrecords]

ok. Since I'm a visual person, the tricks of impedance is a little hard to understand (I really feel newby right now hahaha).  What you mean is: The original filter section was made for a 600ohm load (because it was in/out 600ohm at the time). That's what explain the bridging load resistor of 600ohm at the input and output. So later came the 10K bridging which represent a negligeable load for a 600ohm source. Am I allright?

Correct!

So in our case, I need to feed the eqp into a 10K load, ie, the input of my apogee which has 10K represent the 10K load I need. Is that it?

Correct again!!

In this configuration, I'll get the max gain available, with a good 600:600 output transformer, without any 600ohm load resistor at it's output. So when you said the make up gain will not drive a 600 load but a 10K bridging load, you are refering to the input I fed the output of the eqp in. Is that it?

Correct!!

So what is important here, is to connect the eqp in a 10K input, not connecting a 10k device into the eqp.

Correct!!

I think I imagine the scenario in reverse. I thought I had to fed a 10K device into the eqp to get the load, but I need to fed the eqp INTO a 10K input. That's my understanding. So, looking at the input of the eqp, I need a 600:600ohm because I use the original filter section which was designed to be fed by a 600ohm output device. In my case, there is 2 scenarios possible.

1. I record some tracks with a, ie, Gyraf G9 and fed the output of the G9 INTO the input of the eqp, then the eqp output to the apogee input (10K). the output of the G9 being 600ohm, will I loose some gain or I'll be ok? I have my 10K load so I guess I should be ok.

It will be OK. The Gyraf G9 will drive 600 ohms with little or no loss

2. I mix a track and put the eqp in insert of my apogee. The output of the apogee is 100ohm, fed INTO the eqp (which is 600) and go back to the apogee input rated 10K. I should be ok. Feeding 100ohm into 600ohm does not drive in some loosing gain, isn't it?

Correct!!

The last thing I would like to enlight is the resisor we strap at the secondary of the input transformer. The original eqp had a 620r. The Gpultec has a 10K (with a 600:600 input transformer). Do I need such a resistor. is it there to simulate a load to the transformer?

The 620 resistor is probably not essential. It was in the old days when the load had to be 600. I suggest you use the 10K specifies in the Gyraf design.

Finally, 600:600 input and output, feeding into a 10K device and I should get the max gain available. is that it?

That is exactly right. The two 600:600 transformer both have unity gain so there is no loss. All the gain from the make up amp will therefore be available.

Thanks for your patience Ian. I feel guilty to monopolize your time with such newby questions.

My pleasure.

Cheers

ian

 

[Deepdark]

;D awesome

Looks like I've learn something haha.

Now, I continued the calculation of some other factor. First thing first, is 4.5ma a good idle current in this case? Anyway, here is my scenario with that idle current:

HT: 310Vv
Ia: 4.5ma
Bias: -4v
Load: 10K
Anode: 132.5v/anode
H2%: 1.5% (second harmonic distortion)
- swing: 42.5v
+ swing: 45v
Gain: I managed to look at the formula: A = Vout/Vin. According to my graph, I got a Vin of around 17.5v (where I took my values) and a Vin of 2. So, for 2v grid increase, I have 17.5v out. Is that it? So I got around 9.5v gain. This value is for 1 anode, isn't  it? So for the 2 triode, I got around 19. Am I allright? So it gave about 25.5db of gain.

Is there other point I missed, or other factor I should take in account?

For what I see, I should have a good performing design.

 

[ruffrecords]

4.5mA idles current is fine.

For working out the gain you need to think about the actin of the mu follower. The voltage from the first anode goes to the second grid. The top triode is a cathode follower so its cathode follows the grid. Since the middle resistor is connected between the first anode and the top cathode, it has signal voltage at both ends so the current through it is almost constant and equal to the idle current. Therefore, the load line you need to use to work out the gain is a horizontal line at the idle current. If you do this you should fine the gain is equal to mu, which is why it is called a mu follower.

Cheers

Ian

 

[Deepdark]

allright. So I got a gain of around 20.

Next. If I want to adjust the amp at unity gain, I guess a pot would be best. A log or linear? And how do I manage to find the needed value?

 

[ruffrecords]

allright. So I got a gain of around 20.

Next. If I want to adjust the amp at unity gain, I guess a pot would be best. A log or linear? And how do I manage to find the needed value?

470K linear would do it. The Pultec has an insertion loss of about 20dB and the amp has about 26dB so with a 470K LIN at half way you should be close to overall unity gain.

Cheers

Ian