EQP1A with cascode amp circuit

ok. But how do you determine 470K. In other words, what are the maths?

Deepdark said:

ok. But how do you determine 470K. In other words, what are the maths?

Click to expand...

The output impedance of the poor man's EQP1A EQ is close to 56K. You do not want the gain make up circuit to load this else it will change the EQ response so you choose a load about 10 times bigger. 470K is big enough.

If you are using the original EQP1A EQ circuit, its output impedance is about 10K so you should aim for your load to be 100K or so. In this case a 100K pot will do.

Cheers

Ian

Ok. So it take count of the overall filter section x 10 time the impedance. I'll make some research to fin out how to calculate the entire impedance of a circuit.

Deepdark said:

Ok. So it take count of the overall filter section x 10 time the impedance. I'll make some research to fin out how to calculate the entire impedance of a circuit.

Click to expand...

The actual output impedance varies as you alter the controls - that's what passive EQs do. However,  in the original there is a 10K resistor across the output so the output impedance can never be greater than 10K so it is safe to use that for working out the loading.

Cheers

Ian

allright. Thanks for the clarification. Is it better to strap a resistor to ground on pin 3 of the pot or I could strap pin 3 direct to ground?

Deepdark said:

allright. Thanks for the clarification. Is it better to strap a resistor to ground on pin 3 of the pot or I could strap pin 3 direct to ground?

Click to expand...

Just put it direct to ground.

Cheers

Ian

Thanks Ian. So now, I'm looking at the power supply. I would like to design one with a tube rectifier. First thing first, i planned using a 6x4, but since The idle current is really low (4,5ma) i would like to know if there is a better choice. Any tube with 5 or 6,3v heater. Also, how to determine the voltage drop. There a value in the data sheet given for a specific current. Since i dont think the drop is linear, which formula can be applied?

By the way, this is what the design i plan using. http://diyaudioprojects.com/Technical/Tube-Power-Supplies/

If you have better source for designing tube psu, let me know.

According to the process describe on the link, finding the voltage going out the rectifier is a long process. Is there a simpler way to evaluate the voltage going out the rectifier or We always have to evaluate the voltage drop, the internal resistence and all the performance data? Also, the filtering process is more of a trying and error process. Is there a way evaluate the capacitors value? I mean, i plan making the filtering section in cascade. Maybe 3 section of resistor/caps.

Deepdark said:

By the way, this is what the design i plan using. http://diyaudioprojects.com/Technical/Tube-Power-Supplies/

If you have better source for designing tube psu, let me know.

According to the process describe on the link, finding the voltage going out the rectifier is a long process. Is there a simpler way to evaluate the voltage going out the rectifier or We always have to evaluate the voltage drop, the internal resistence and all the performance data? Also, the filtering process is more of a trying and error process. Is there a way evaluate the capacitors value? I mean, i plan making the filtering section in cascade. Maybe 3 section of resistor/caps.

Click to expand...

That is a rather complex design process but probably necessary when designing a relatively high current power supply. Since you only need a few mA we can simplify the process a lot. First which tube. The lowest current rectifier tube I know of is the EZ80:

http://www.r-type.org/pdfs/ez80-1.pdf

Looking at the data sheet, at 5mA, it is going to drop less than 5 volts. All you need to ensure is that the series resistance is not less than 300 ohms and that the first capacitor is no more than 50uF. We are taking very little current so the dc voltage on the reservoir cap will be close to the peak secondary voltage of the transformer. I would recommend a 240-0-240 or a 250-0-250 transformer. This will give you a raw dc supply of about 350V.  With a 5mA load and a 50Hz supply, the ripple will be about 1V peak to peak

You can then add three RC networks in series to reduce the ripple to acceptable limits. As  a rule I set the resistor value to drop about 10V. So we drop a total of 30V leaving us with a supply around 320V. It will be a bit less than this due to the tube and 300 ohm series resistor drops but it will be close enough to what we need.

So, for 5mA load and 10V drop, each resistor needs to be 2K. Now we just need to choose the cap value. Assume we choose 47uF again.  At the ripple frequency of 100Hz, a 47uF cap has a reactance of about 34 ohms so the ripple is attenuated by about 34/2K times or 35dB. Three of these in a row will give a total ripple reduction of about 105dB. So our initial 1V peak to peak ripple is reduced to less than 6uV which is more than adequate. You could get away with 22uF caps which would increase the final ripple to about 50uV which is still fine.

Cheers

Ian

Damn, that's close to what I've calculate yesterday, but instead of taking 4h of reading/calculating, you resume it in a simple paragraphe hahaha. I got around 355v outside the tube, i calculate i needed 2k - 2k5 resistor and around 20uf. Thanks a lot. This confirm my values. Am i better to put a bleed resistor after the rc's? 200k?

Deepdark said:

Damn, that's close to what I've calculate yesterday, but instead of taking 4h of reading/calculating, you resume it in a simple paragraphe hahaha. I got around 355v outside the tube, i calculate i needed 2k - 2k5 resistor and around 20uf. Thanks a lot. This confirm my values. Am i better to put a bleed resistor after the rc's? 200k?

Click to expand...

That's only because I have been doing it for quite a while!

A bleed resistor is a good Idea. I tend to use a 220K 2W. However, if you are using a mu follower then the top cathode is a long way above ground. This can exceed the tubes rated heater/cathode voltage spec. To fix this it is normal to elevate the heaters which means that instead of connecting them to the HT 0V you connected them to a higher voltage. You do this with a decoupled pot divider across the HT to which you connect one side of the heaters. it is normal to elevate the heaters by about 75 to 80 V. I do this with a pot divider across the HT which consists of a 22K 2W resistor at the bottom and two 33K 2W resistors in series at the top. This provides an elevation point that is exactly 25% of the HT voltage. The two resistors in series make 88K and this itself provides a handy HT bleed resistor.

Cheers

ian

Thanks Ian. I wasn't sure if I needed an elevated heaters supply. Here is a first sketch with the results I got (and yours being close) and for the heaters elevation, tel me if I'm ok. I don'T really know how to evaluate the resistors value of the pot divider. If I need around 80v of elevation, I need to drop around 310-80 = 230v ??? so around 50K ?? and for the second one, around 17K ?? is it ok? 

Deepdark said:

Thanks Ian. I wasn't sure if I needed an elevated heaters supply. Here is a first sketch with the results I got (and yours being close) and for the heaters elevation, tel me if I'm ok. I don'T really know how to evaluate the resistors value of the pot divider. If I need around 80v of elevation, I need to drop around 310-80 = 230v ??? so around 50K ?? and for the second one, around 17K ?? is it ok?

Click to expand...

You need to watch the dissipation in the heater elevation resistors and also their total current draw. 50K with 230V across it dissipates over 1 watt and draws over 4mA current (OK you can afford it). As I mentioned earlier I use a couple of 33K 2W resistors i series for the top resistor, making 66K and I use a 22K at the bottom. That divides the supply by exactly 4 and uses common resistor values but by all means use 50K and 17K they will be fine. The 17K needs decoupling as you showed in your schematic. I normally use a 47uF rated at 100V or more to be on the safe side.

You probably do not need the last 2K in your PSU between the 319 and 310 points unless you have another cap across the HT at the tube itself. You might also need some series reistance between the cathode of the rectifier and the first cap in order to limit the current. Check the tube data sheet for details .

Cheers

Ian

Hi

Thanks for the reply. Looking at the spec for an EZ80, they recall a first cap filtering of 50uf (47 should be fine). and they recall for an 250-250 operation 2x125ohm resistor (Rt). Should I put them in series before the 50uf?

Tell me if I'm right about the heaters. The upper HT is 310v (approx) and the lower is 0v.  if I strap 88K accros HT, then I have 310/88000 = 3.5ma. Am I right? So after the first resistor (33K) I'll have 194v, the after the second I'll have 78.5v (near 80v) and then , through the last resistor, I end up at 0v. Thsi is my understanding. So we evaluate the resistor need by ensuring to not exeding the power transformer capacities? Is that it? In our case, it's not that important looking at the really small current values but in a larger systeme, I guess it's essential to select them according to the transformers spec. Do I have to add the 3.5ma to my idle 4.5ma, to select the resistors values?

That's right. I  most cases the normal load is a lot bigger than the heater elevation  load so you can ignore the latter. In this case they are very similar so you should include it in the idle current total.

Rt is the sum of the secondary winding resistance, the reflected primary resistance and and added series resistance. The transformer resistances obviously depend on the chosen transformer but you wo't go far wrong for a low current supply like this if you assume they are zero so the added series resistor needs to be about 125 ohms (120 will be fine). You only need one between the cathode and the 47uF because only one half of th tube is conducting at any one time and it needs 125 ohms.

Cheers

Ian

Awesome. Thank you. I'll redraw it and repost it tomorrow. I'll probably fine with 2k resistors at 8ma.

Deepdark said:

Awesome. Thank you. I'll redraw it and repost it tomorrow. I'll probably fine with 2k resistors at 8ma.

Click to expand...

I think the time has come to build it!

Cheers

Ian

yup Thanks for your help

Is the XSM600:600 a good candidate for input/output?

I just found, too, a Hammond which could be nice. 140QEX. Is it a good candidate for input or output (or both) ?

Deepdark said:

Is the XSM600:600 a good candidate for input/output?

Click to expand...

Not really. Edcor are very cheap but the winding inductance tends to be rather low. This is fine when being driven by a very low impedance but not when driven by a mu follower. The published inductance is 3 Henries. At 20Hz this represents an impedance of 377ohms so you need to be driving it from a source of lower impedance than this. On the other hand, the Carnhill VTB2281 has an inductance of 12.7 Henries. This means its impedance at 20Hz is 1596 ohms - much better.  This would do for the output. Alternatively you could use something like the Cinemag CMOB-2 or the Jensen JT-123.

At the input you really need to use a screened transformer to ensure it does not pick up local interference for example from the mains transformer. something like the Sowter 3603 would be ideal.

Cheers

Ian