Thats why I am getting more and more in troubles!
Maybe I am rich in my head but getting more and more poor, concerning money
So, I must do another things, the next days...
Franz
I measure a person's wealth by their generosity so Franz is indeed rich! 😉
Franz, I find not going to work can be a problem in that I spend far too much time doing DIY hi-fi. It truly is addictive!
Nuuk said:Franz, I find not going to work can be a problem in that I spend far too much time doing DIY hi-fi. It truly is addictive!
If I didn't go to work, I would have more time for this.
But less money.
Oh, I need a partner to finance this DIY vice.
Any takers?😀
Nuuk said:
Franz, I find not going to work can be a problem in that I spend far too much time doing DIY hi-fi. It truly is addictive!
Guys, when will you stop modding the modded LM GC.
I can't resist implementing them in my setup - they are so good!!! 😎
My focus is this DIY hifi - my means is my work 😀
Huh, am I addicted??
Joe Rasmussen and other VBIGC owners:
What is the operation point of your 6DJ8 input tube (or what type do you use)?
How many volts do you measure at the cathode?
What cathode resistor do you use actually?
Thanks
Franz
What is the operation point of your 6DJ8 input tube (or what type do you use)?
How many volts do you measure at the cathode?
What cathode resistor do you use actually?
Thanks
Franz
For an E88CC?
What is the calculation behind and how much Ug- do you expect by this setup?
Please, help me! It must be so simple, but I dont get it!
The tube has its Rp and is in series with Rc. So, a simple voltage divider. Isn't it?
The voltage at Rc should be higher than 50% of Vcc, to get the Ug-.
Wrong?
I think, the E88CC has a Rp to high for this voltage (is more than the double of an 6DJ8).
Franz
To Franz : When I see at curves of ECC 88, I see that probably most linear is by this condition : Ua = 90 V and Ia = 16 mA, Ug = - 1 V. So you must go to higher voltage ( better will be to make voltage doubler ( + / - 70 V instead + / - 35 V ) and higher Ia, which signify, that you must to use lower value of Rc ( from cca 2k2 up to 3k3 ) - Ug must be hold from cca - 0.5 V up to - 1 V.
Franz, I'm not a specialist in this field, but my father was a little puzzled when I talked about using the E88CC at +/-35v.
He knows it very well, and in the 60s this valve was used at around +/-100v.
He knows it very well, and in the 60s this valve was used at around +/-100v.
Hi,
My valve buffer and my Valve preamp both use ECC88 tubes. I have +/- 50V and have the bias set at -0.6V which seemed to give the best sound. I experimented quite a bit and this sounded most natural to my ears. In order to get the bias to this point I had to make the supply rails asymetrical with resistance. My VB has the enhancement of the constant Voltage Mosfet on top. My Valve preamp is a full implementation of the CC-CV-VB, based upon Joes design.
Shoog
My valve buffer and my Valve preamp both use ECC88 tubes. I have +/- 50V and have the bias set at -0.6V which seemed to give the best sound. I experimented quite a bit and this sounded most natural to my ears. In order to get the bias to this point I had to make the supply rails asymetrical with resistance. My VB has the enhancement of the constant Voltage Mosfet on top. My Valve preamp is a full implementation of the CC-CV-VB, based upon Joes design.
Shoog
O.K.
I think I learned it now:
It is best, to choose Rc in this circuit similar to Rp of the tube. So in the case of an E88CC 10K.
When I choose another value for Rc, then I have DC on the input pot (causing some speaker moving, when turning the pot).
Only solution in this case: use input coupling cap and connect the grid by a resistor to ground.
Or balance the Rc to the same value as Rp. I prefer this solution.
But then, you have to vary the Vcc, to obtain the choosen optimal workpoint of the tube.
So, for the E88CC, the optimal value must theoritical be an Vcc about 130V (+-65V), Ia about 6.5mA.
Unfortunately, I have just 63V caps in this part of the PSU. I will try +-50V, E88CC, Rc 10K, what should result in Ia about 5mA.
Franz
P.S.
Ug- is not important, as the Uc will move the same way (and hopefully the same amount). So there will be no grid current, if input voltage is higher than the Ug-.
I think I learned it now:
It is best, to choose Rc in this circuit similar to Rp of the tube. So in the case of an E88CC 10K.
When I choose another value for Rc, then I have DC on the input pot (causing some speaker moving, when turning the pot).
Only solution in this case: use input coupling cap and connect the grid by a resistor to ground.
Or balance the Rc to the same value as Rp. I prefer this solution.
But then, you have to vary the Vcc, to obtain the choosen optimal workpoint of the tube.
So, for the E88CC, the optimal value must theoritical be an Vcc about 130V (+-65V), Ia about 6.5mA.
Unfortunately, I have just 63V caps in this part of the PSU. I will try +-50V, E88CC, Rc 10K, what should result in Ia about 5mA.
Franz
P.S.
Ug- is not important, as the Uc will move the same way (and hopefully the same amount). So there will be no grid current, if input voltage is higher than the Ug-.
Lets resume my findings (Heureka!?*):
- for the configuration input pot => gridstopper => grid it is an advantage to use Rc=Rp, for the E88CC 10K. This way I avoid DC at the pot.
- a good working point is choosen by the supply voltage. For the E88CC between +-45 to +-60V must be optimal, with Rc 10K.
It would be a nice idea, to supply the anode voltage by a regulated adjustable PSU. So, it will be possible, to adjust in the above named range and decide by listening!
Searching in my hobby box, I found a new ECC86 (equiv. 6GM8), wich works with a max anode voltage of 30V and good values between +-6 to +-12V!
Of course, I will try this tube (as there are two 50V caps in the psu) with Rc 3.3K (wich sits already on the board for tests with higher current with the E88CC, but the pot is scratching, because DC on it) and +-9V 😎 (later I will try the above mentioned adjustment with reg psu).
Franz
*Enfin?
- for the configuration input pot => gridstopper => grid it is an advantage to use Rc=Rp, for the E88CC 10K. This way I avoid DC at the pot.
- a good working point is choosen by the supply voltage. For the E88CC between +-45 to +-60V must be optimal, with Rc 10K.
It would be a nice idea, to supply the anode voltage by a regulated adjustable PSU. So, it will be possible, to adjust in the above named range and decide by listening!
Searching in my hobby box, I found a new ECC86 (equiv. 6GM8), wich works with a max anode voltage of 30V and good values between +-6 to +-12V!
Of course, I will try this tube (as there are two 50V caps in the psu) with Rc 3.3K (wich sits already on the board for tests with higher current with the E88CC, but the pot is scratching, because DC on it) and +-9V 😎 (later I will try the above mentioned adjustment with reg psu).
Franz
*Enfin?
Hi,
Rc is not equal Rp, the internal resistance (Ri) of the ECC88 and its cousins isn't 10K.
Ri of the USSR cousin 6N1-P comes close to 10K but that's not the point.
If you're wondering how you could have 0 V at the cathode's output, look at the following formula:
AC resistance = rp + (µ + 1)Rk
For ECC88/6DJ8 etc. this would give us: 3.200 + (33 + 1) * 200 =
10.000
So with a split rail you would find balance at the connection point of the 200R cathode resistor and the 10K voltage divider.
Cheers, 😉
Rc is not equal Rp, the internal resistance (Ri) of the ECC88 and its cousins isn't 10K.
Ri of the USSR cousin 6N1-P comes close to 10K but that's not the point.
If you're wondering how you could have 0 V at the cathode's output, look at the following formula:
AC resistance = rp + (µ + 1)Rk
For ECC88/6DJ8 etc. this would give us: 3.200 + (33 + 1) * 200 =
10.000
So with a split rail you would find balance at the connection point of the 200R cathode resistor and the 10K voltage divider.
Cheers, 😉
What I dont want is some DC or current at the grid, to avoid a input cap and a grid resistor.
I dont completely understand (as it is not in an excel sheet 😉 ).
Could you show me an example for an E88CC with +-35V or more?
Thanks for your explanation, Frank!
Franz
P.S.
The amp is running right now (Marla Glen) with an ECC86 Rc 3.3K and a very small print tranny, 2x6V 135mA, what gives +-11VDC. I think, satisfying, up to now!
Oh, I think I made another mistake: Rp isnt Ri? 😕
I thougt, in english, Ri is called Rp. Wrong?
Franz
I thougt, in english, Ri is called Rp. Wrong?
Franz
Why not? But for just 2x>10mA? Is it not a little bit overdimensioned?
It would be very nice, to adjust the own amp with two voltmeters by ears!
Imagine: when changing valve, you find the new optimal point...
Dreaming
Franz
BTW:
With your regulated amp, Carlos, you could easily replace the OPA627 with an ECC86, add a tranny for filament and go 🙂
Wrong, dear Carlos!
The difference between input and output must be within 37V.The chip itself is floating and independent of the voltage.
You could even use LM1083.
Franz
Hi,
There always has to be a gridleak resistor or current will tend to develop as the grid potential charges positive.
In the case of this circuit the 50K B pot will double as a gridleak resistor but it is not really advisable to use a pot this way as they really don't like to see DC on the wiper.
In practice it probably works but if you hear the pot making funny noises you now know where it comes from.
A solution is the use a switched attenuator or, horror of horrors...a DC blocking cap in conjunction with a gridleak resistor.
Rp is not Ri even though I have replaced Rp with Ri in the formula I posted just to show how the voltage division works out.
Rp (aka Ra) is what Americans call the plate resistor and we would call the anode resistor (Ra).
Ri, the internal resistance of the valve is not a fixed number as it varies with the amount of current run through the valve.
For most basic calculation you can often just use the Ri as given in the manufacturer datasheet or calculate it yourself.
About the only factor that's pretty constant for a tube is the amplification factor mu.
So, in this circuit, a cathode follower, there is no Rp as there is no anode/plate load.
The circuit can't amplify the voltage presented at its grid but will present it to the next stage as coming from a low source impedance with increased current capability to drive the input source impedance.
Re: ECC86; running it at a +/- 35 rail is already rather "high voltage" for this valve. It was designed to run from low voltage car battery supplies.
Cheers, 😉
There always has to be a gridleak resistor or current will tend to develop as the grid potential charges positive.
In the case of this circuit the 50K B pot will double as a gridleak resistor but it is not really advisable to use a pot this way as they really don't like to see DC on the wiper.
In practice it probably works but if you hear the pot making funny noises you now know where it comes from.
A solution is the use a switched attenuator or, horror of horrors...a DC blocking cap in conjunction with a gridleak resistor.
Rp is not Ri even though I have replaced Rp with Ri in the formula I posted just to show how the voltage division works out.
Rp (aka Ra) is what Americans call the plate resistor and we would call the anode resistor (Ra).
Ri, the internal resistance of the valve is not a fixed number as it varies with the amount of current run through the valve.
For most basic calculation you can often just use the Ri as given in the manufacturer datasheet or calculate it yourself.
About the only factor that's pretty constant for a tube is the amplification factor mu.
So, in this circuit, a cathode follower, there is no Rp as there is no anode/plate load.
The circuit can't amplify the voltage presented at its grid but will present it to the next stage as coming from a low source impedance with increased current capability to drive the input source impedance.
Re: ECC86; running it at a +/- 35 rail is already rather "high voltage" for this valve. It was designed to run from low voltage car battery supplies.
Cheers, 😉
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