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Old 21st October 2015, 06:49 AM   #1
RhythMick is offline RhythMick  United Kingdom
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Default Hi from South Yorkshire

Hi all. I've been lurking on this forum for a while and am now wanting to get a better grip on some key concepts, to help me to understand and then build my own amp.

I have a question about the charge present at the "out" end of a coupling capacitor. Quite a few of the designs I'm looking at use the anode voltage of V1 as feed to the grid of V2, via a coupling cap. Voltage swing amplified by V1 and fed into V2 grid, I get that.

The cap blocks DC, but lets the AC through. With no signal present, what's the potential on the grid of V2 ? It's the point it was taken from, but I don't get why - the cap isn't letting DC through. I can only think that the charge on the negative side of the cap rises (or falls) to match that on the positive side, hence no DC current flows, but the charge moves with the signal and can be used to set the grid level.

But then a lot of caps are used to bypass to ground - so the + side has a charge and the - side is at ground. If the - side were constantly trying to rise to match the + side, but being drained to ground, then surely that would be DC current flowing ?

I'm confused - and about such a basic component. Someone point me in the right direction to understand this please ?
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Old 21st October 2015, 07:11 AM   #2
Mooly is offline Mooly  United Kingdom
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Hi from South Yorkshire
Welcome to diyAudio

Caps... so you have an anode at say 200 volts and connect to that a coupling cap to the grid of the next valve. That grid will have a high value resistor to ground (usually ground but not always) and it is that voltage point that the resistor returns to which defines the static DC voltage of the grid.

The cap only passes the changing AC signal which then modulates that previously static grid voltage, the value of the cap setting a limit on how low a frequency gets through.

When you connect a large cap across a power supply rail an initial charge current flows, that current decreasing exponentially until the cap is said to be 'charged' and the current has reduced to zero.
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Old 21st October 2015, 12:04 PM   #3
RhythMick is offline RhythMick  United Kingdom
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Quote:
Originally Posted by Mooly View Post
Welcome to diyAudio

Caps... so you have an anode at say 200 volts and connect to that a coupling cap to the grid of the next valve. That grid will have a high value resistor to ground (usually ground but not always) and it is that voltage point that the resistor returns to which defines the static DC voltage of the grid.

The cap only passes the changing AC signal which then modulates that previously static grid voltage, the value of the cap setting a limit on how low a frequency gets through.

When you connect a large cap across a power supply rail an initial charge current flows, that current decreasing exponentially until the cap is said to be 'charged' and the current has reduced to zero.
Ok thanks,but - a resistor will only show a voltage drop over it if there's current flowing right?

The circuit I'm looking at is here, top right...
Tube CAD Journal: Tube balanced phono stage, July 1999

There's no grid resistor to ground. There's a 1M resistor between the grids of the balanced circuit. In red the author indicates the voltage at the grid of the second valve to be 100V,which matches that at the anode of the previous valve. Hence my question.
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Old 21st October 2015, 12:38 PM   #4
Mooly is offline Mooly  United Kingdom
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Hi from South Yorkshire
Yes for the resistor and current flowing through it. Ohms law V= I*R

In your circuit both grids are tied via the 1M. That 1M is then tied to the lower left valve anode via a 75k. Because the grids draw virtually zero current, both grids of the output valves assume the 100 volts potential of the that first anode. There is essentially no DC current flow and so all those points are at the same potential despite being a chain of 'anode' into 75k into 1M. The DC voltage at all three points is 100 volts, the 100 volts being set by the conduction of the first valve.
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Old 21st October 2015, 02:12 PM   #5
RhythMick is offline RhythMick  United Kingdom
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Originally Posted by Mooly View Post
Yes for the resistor and current flowing through it. Ohms law V= I*R

In your circuit both grids are tied via the 1M. That 1M is then tied to the lower left valve anode via a 75k. Because the grids draw virtually zero current, both grids of the output valves assume the 100 volts potential of the that first anode. There is essentially no DC current flow and so all those points are at the same potential despite being a chain of 'anode' into 75k into 1M. The DC voltage at all three points is 100 volts, the 100 volts being set by the conduction of the first valve.

Great thanks. So even though there is no DC flow through the coupling cap the grid end of it is at the same potential. Got it. I'm over thinking it I know - but I don't want to just build it with out understanding it.

So in that circuit the grid of v2 is at 100v potential. Hence choosing the cathode bias of v2 to set the voltage slightly higher.

Thank you.
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Old 21st October 2015, 02:21 PM   #6
Mooly is offline Mooly  United Kingdom
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Hi from South Yorkshire
Imagine the circuit with all the caps removed. All the DC voltages would remain the same. On an idealised circuit (at least idealised as drawn) the voltages of the two symmetric halves of the circuit are shown to be identical whereas is practice they will vary quite a lot. The lower left valve anode at 100 volts (or whatever it is in the real world, 90 volts, 110 volts etc ) will always appear on the two grids. All other valve voltages can and will be different to some degree or other.
 

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