Why does either side of the waveform determine charge?

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ChrisHansen2Legit2Quit

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I just watched a presentation on how a vacuum triode tube amplifies.

I understand it.

One thing though. The overall charge on the grid (either + or -), is determined by either side of the waveform your inputing.

I dont understand. I just see a waveform as electrons flowing in one direction, then the other. Im not making the connection here. How is the charge dependent upon electron flow in a certain direction.

Thks
 

Under normal circumstances the control grid is always negative with respect to the cathode, the signal you are amplifying adds or subtracts from the negative but never enough to make it go positive. If the grid did go positive with respect to the cathode it would attract electrons from it in the same way the anode would and current would flow, effectively making it a diode.

Brian.
 
Thanks.

That video also mentioned how signals can be sent through several vacuum tubes to get an even stronger signal.

That's great, but why not just feed more power to a single vacuum tube?

If you increase the number of vacuum tubes, then each tube needs a power supply, correct? Now you have 5 wall warts for 5 vacuum tubes. You can run multiple vacuum tubes to a single power supply? See, this is what I dont understand. Power is given by VI. You only get X amount of power from a power supply, so now youre taking X power and distributing it to multiple tubes. Just run X power to 1 tube!
 

ChrisHansen2Legit2Quit said:
That video also mentioned how signals can be sent through several vacuum tubes to get an even stronger signal.

That's great, but why not just feed more power to a single vacuum tube?
Click to expand...
The problem is that tubes have limited voltage gain. Lets say we have a tube amplifier with a gain of 50. Then if we give an input of 1 mV, the output will only be 50mV. Now if we give that 50mV to the input of a second amplifier, the output of the second amplifier will be 50*50mV=2.5V. Total gain with both amplifiers = 50*50=2500. Thus we can use multiple tubes to get very high voltage gain.

ChrisHansen2Legit2Quit said:
You can run multiple vacuum tubes to a single power supply?
Click to expand...
Yes.
 

Godfreyl is right, basically there is a limit to how much you can control the flow of electrons. If you increased the flow (current) to get more 'signal' out, the input would have proportionately less effect on it. The only solution is to amplify in stages, each boosting the signal then passing it on. The other advantage of doing that is you can optimize the first stage for small signals and optimize the later stages for bigger signals.

You don't need a different supply for each stage, normally a single supply would feed them all. Given that vacuum tubes normally use high voltages I'm not sure whether a wall wart supply would be available or advisable to use though.

Brian.
 

betwixt said:
Given that vacuum tubes normally use high voltages...
Click to expand...
Agreed, but a few types were specially designed to run at low voltage. Also, I was surprised a while ago to see this amplifier circuit posted on the forum. It uses ordinary 12AU7 (ECC82) double triodes with a single supply of only 12V!

I mention this because it's possible the video the OP saw actually does use wall-wart supplies for tube circuits.
 

They will work at even lower voltages than that, I'm not a physicist but I would guess that as a vacuum tube diode can rectify microvolt signals without bias, a triode should work almost down to zero anode volts. Half a century ago I experimented with EF74 miniature wire-ended pentodes in regenerative radio receivers running on a 9V battery with great success. I could even pick up US local radio stations across the Atlantic on them. Sadly I don't have any 'bottles' to play with any longer and even less spare time to do experiments.

Brian.
 

The vacuum tube demonstration showed an anode value of 300V and a cathode value of 0V.

If I take a 9V battery and hook the anode up to a plate, will electrons flow through the plate? In other words, will the charge accumulate to 9V?

Now, what if I hook a plate up to the cathode of the battery. Electrons wont flow through such plate, so the plate will accumulate 0 charge?


Intuition tells me each plate gets 4.5V, which gives you the total potential difference of 9v
 
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I'm not sure what you mean. Be careful using the word 'plate' as it can have different meanings, in some countries it is used as the name for the anode.
Can you show a diagram please.

Brian.
 

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