EEVblog Electronics Community Forum
Electronics => Projects, Designs, and Technical Stuff => Topic started by: fabiodl on March 10, 2018, 09:21:05 am
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Hello,
I found this suggested schematic for controlling the 6E2 tube (magic eye)
http://www.kyohritsu.jp/eclib/DIGIT/etc/6e2.pdf
(http://www.kyohritsu.jp/eclib/DIGIT/etc/6e2.pdf)
Listening to this youtube video https://www.youtube.com/watch?v=ncIPR7wjcYI (https://www.youtube.com/watch?v=ncIPR7wjcYI) he finds out that the grid should have voltages in the order of 15 volts.
However, I do not understand how such high voltages can be generated from a ~1.1v of an audio input (which, in this schematic, is at the far right)
All I understand from the schematic is the role of the following parts:
The rightmost potentiometer allows to control the gain
The rightmost capacitor cuts allows only the AC to pass through
The leftmost diode makes only negative voltages pass through
I do not understand the following:
1)How can they achieve the 15v of the video? this seems impossible.
2)what is the role of the rightmost diode? It seems to clamp the positive voltages, but aren't those already blocked by the leftmost one?
3)what is the 100K resistor for? Is it to bias the AC voltage on the left of the capacitor around ground?
4)what do the 470k resistor and 0.47uf cap do? are they there to slow down the input dynamics or do they have another role?
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1. Quite possible, see triangle on the schematics in a video? There are two pins up/down, mark as +-15V. It's not shown, but it's connected to power supply in order to chip (OPA IC) to function. So, OPA could output any voltage between -15 and +15 Volts (or -12 +12 V in worst case scenario).
2. You mean schematic from http://eleshop.jp/ (http://eleshop.jp/) doc. Two diodes form a voltage multiplier, doubler in this case.
More: https://www.electronics-tutorials.ws/blog/voltage-multiplier-circuit.html (https://www.electronics-tutorials.ws/blog/voltage-multiplier-circuit.html)
3. Not really needed.
4. Caps are part of a voltage doubler, 47uF & 0.47uF, though I 'd put equal sizes. 470k is a load, forms decay timing in conjunction with 0.47 cap, RC = 47 millisec.
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The extra resistor is probably just a discharge resistor or something.
The 470K is the tube bias, like a necessary bleeder for the grid since it's high impedance. It's value isn't for timing, it's to set the "operating point" or "bias point" of the tube.
https://tubedepot.com/pages/bias-point
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Thank you MasterT and Cyberdragon
1. Quite possible, see triangle on the schematics in a video? There are two pins up/down, mark as +-15V. It's not shown, but it's connected to power supply in order to chip (OPA IC) to function. So, OPA could output any voltage between -15 and +15 Volts (or -12 +12 V in worst case scenario).
2. You mean schematic from http://eleshop.jp/ (http://eleshop.jp/) doc. Two diodes form a voltage multiplier, doubler in this case.
More: https://www.electronics-tutorials.ws/blog/voltage-multiplier-circuit.html (https://www.electronics-tutorials.ws/blog/voltage-multiplier-circuit.html)
3. Not really needed.
4. Caps are part of a voltage doubler, 47uF & 0.47uF, though I 'd put equal sizes. 470k is a load, forms decay timing in conjunction with 0.47 cap, RC = 47 millisec.
For all the questions, I was referring to the eleshop schematic. That is why I cannot understand where they get -15v.
It's indeed a voltage doubler! Thank you for pointing out.
In this case, isn't the polarized cap backwards? (The eleshop schematic aims at doubling a negative voltage, so the diodes are backwards with respect to the schematic you linked. This would imply a reversed polarized cap as well, isn't it?)
The extra resistor is probably just a discharge resistor or something.
The 470K is the tube bias, like a necessary bleeder for the grid since it's high impedance. It's value isn't for timing, it's to set the "operating point" or "bias point" of the tube.
https://tubedepot.com/pages/bias-point (https://tubedepot.com/pages/bias-point)
Do you know what the target (pin 6) and deflection electrode (pin 7) are used for?
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Right, cap 47uF has wrong polarity, good catch. Regarding voltage -15V , I don't understand Japanese, but this label mark at the audio input must say something like "connect to the speaker". Means there voltage isn't line level 1V RMS, likely 7.5V from power amplifier output. To use it with headphones jack, you need an amplifier, OPA IC or at least single BJT transistor with biasing circuitry.
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Pin 6 is the cathode ray (target) power and pin 7 is the actual electron beam or shadow control forming the actual indicator section. Technically, these are all you need. The triode inside is included for conveniance of having an integrated driver. But you can drive the CRT section with whatever you want. Tubes like the 6AF6 don't even have internal triodes.
https://en.m.wikipedia.org/wiki/Magic_eye_tube (https://en.m.wikipedia.org/wiki/Magic_eye_tube)
As with any other CRT (or even VFDs), they also are affected by magnetism. You can use coils to create effects, just don't warp or magnetize any internal components.
(http://physlab.org/wp-content/uploads/2016/03/Magic_eye3.jpg)
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Thank you guys. Sorry for my complete ignorance about audio equipment and tubes.
Right, cap 47uF has wrong polarity, good catch. Regarding voltage -15V , I don't understand Japanese, but this label mark at the audio input must say something like "connect to the speaker". Means there voltage isn't line level 1V RMS, likely 7.5V from power amplifier output. To use it with headphones jack, you need an amplifier, OPA IC or at least single BJT transistor with biasing circuitry.
Indeed it says "input from the power amplifier", which I wrongly assumed to be about 1.1v.
If it is around 7.5V it completely makes sense.
I have this power amplifier http://a.co/8VwTp8Y (http://a.co/8VwTp8Y) and I measured the voltage across a 4ohm speaker. The AC voltage reported by the multimeter is in the order of 0.4v. Probably there's some correction factor due to the fact that the multimeter assumes a sinusoidal wave while the signal was some random audio, but still I am an order of magnitude away from 7.5v. Where are "power amplifiers" used?
Pin 6 is the cathode ray (target) power and pin 7 is the actual electron beam or shadow control forming the actual indicator section. Technically, these are all you need. The triode inside is included for conveniance of having an integrated driver. But you can drive the CRT section with whatever you want. Tubes like the 6AF6 don't even have internal triodes.
(http://physlab.org/wp-content/uploads/2016/03/Magic_eye3.jpg)
I know that the electrons leave the cathode (pin 3 here) and they reach the plate (pin 9). In this, they are controlled by the grid (pin 1).
Is this what you refer to as the triode? Is the 1M resistor there to limit the current?
Pin 6 is connected to high (positive) voltage and pin 7 too, although through a resitor (why? is it a negative feedback of some sort?).
I tried to change the voltage on the grid, and it works as described in the video: a more negative voltage on pin 1 makes the lightened bars become longer. Why? I would expect a more negative voltage to stop more electrons, why does it correspond to longer bars?
Sorry for this probably really dumb questions
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Pins 1, 3, and 9 form a normal triode, the plate (anode) of which is directly connected to the deflection electrode of the small cathode-ray tube.
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It is stopping the electrons. The target is inverted, the electrons are repelling each other pushing the bars open. When you turn the triode off, the bars are free to close.
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Thank you. I also found the principle of operation of a similar device here
https://radioremembered.org/tuneye.htm
I think I now understood how it works: when the grid does not block the electron flow, the plate, and the connected deflector, are at a low potential w.r.t. the target, and thus a large area is switched off. When the electron flow to the plate is limited by a negative grid voltage, the deflector and the target are at about the same voltage, so all the target gets illuminated, isn't it?