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This source claims that putting a metal plate in between the capacitor plates greatly reduces the capacitance. How is this possible?

Two equal capacitances in series decreases the capacitance by half, but the distance is also decreased by half, so the overall capacitance must not change right?

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In the case of the valve/tube, the purpose of the screen grid is to reduce the effect of the capacitance between the grid and anode.

The screen grid is not just a floating piece of metal, it's connected to a low impedance supply (don't remember offhand whether it's low or high voltage).

Without it, when the anode changes voltage, the anode to grid capacitance induces a current into the grid which fights the control signal. This is the so called Miller capacitance effect. This loads the grid with an effective capacitance that's enhanced by the gain of the valve.

With the screen grid, the anode to screen capacitance induces that current into the screen grid. Being supplied with a low impedance, that current flows into the supply with little change in voltage. The screen to grid capacitor sees little change in voltage, and so the grid has much, much less current induced it, possibly being only enhanced by a very small gain, often less than 2.

If the screen grid is left unconnected, then sure, anode to grid capacitance remains essentially unaltered, as does the Miller effect when the valve is used as an amplifier.

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    \$\begingroup\$ High voltage; it splits the anode current. In a pentode, there is a suppressor grid between screen grid and anode. This is connected to 0V (or the cathode, sometimes internally); it repels "secondary emission" charge (electrons bounced off the anode by incoming ones) preventing them reaching the screen grid. Good explanation : perhaps worth pointing out that the tetrode and pentode are what transistor people would call cascode circuits, with the screen grid being the base voltage of the upper transistor \$\endgroup\$ – Brian Drummond Mar 4 '20 at 13:45
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    \$\begingroup\$ Great answer. Although I would add that the screen grid, when biased positive, acts to suck electrons away from the control grid and reduce the space charge effects, greatly increasing the gain of the tube. \$\endgroup\$ – electrogas Mar 4 '20 at 22:20
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The capacity remains the same, as seen by the calculation below.

$$ C1+C2=\dfrac{\dfrac{ϵA}{d1} \cdot \dfrac{ϵA}{d2}}{\dfrac{ϵA}{d1}+ \dfrac{ϵA}{d2}}=\dfrac{ϵA}{d1+d2}=C $$

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    \$\begingroup\$ Let \$d_1,d_2\$ be the distances from the middle metal plate to the outer capacitor plates and say \$d=d_1+d_2\$. Then \$C_1\| C_2 = \dfrac{\dfrac{\epsilon A}{d_1}\dfrac{\epsilon A}{d_2}}{\dfrac{\epsilon A}{d_1}+\dfrac{\epsilon A}{d_2}} = \dfrac{\epsilon A}{d_1+d_2}=C\$ \$\endgroup\$ – across Mar 4 '20 at 7:27
  • \$\begingroup\$ Yes, you are right, will correct my answer. \$\endgroup\$ – xeeka Mar 4 '20 at 7:44
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You are forgetting one factor about elements in series, the voltage is also split between the capacitors while the charge stays the same, imagine you got capacitor A and capacitor B

$$ V_{total}= V_a+V_b=\frac{Q_T}{C_A}+\frac{Q_T}{C_B} $$

if we rearrange $$ \frac{V_{total}}{Qt}=\frac{1}{C_T}=\frac{1}{C_A}+\frac{1}{C_B} $$

which ends up looking like resistors in parallel operation wise. If they are identical you get half.

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  • \$\begingroup\$ Voltage on each capacitor is half, but distance between plates in each capacitor also reduced by half, so overall capacitance doesn't change right? \$\endgroup\$ – across Mar 4 '20 at 6:56
  • \$\begingroup\$ voltage and distance is half, but charge stays the same \$\endgroup\$ – Juan Mar 4 '20 at 6:57
  • \$\begingroup\$ If we let the capacitance without metal plate is \$C_o\$. From your equation above, after introducing the metal plate \$C_T = \dfrac{1}{C_A}+\dfrac{1}{C_B}\$. Doesn't this equal to \$C_o\$? \$\endgroup\$ – across Mar 4 '20 at 7:01
  • \$\begingroup\$ I agree that \$C_T\$ will be less than both \$C_A\$ and \$C_B\$. But my question is about the capacitance without the metal plate \$C_o\$.. \$\endgroup\$ – across Mar 4 '20 at 7:05
  • \$\begingroup\$ I think the most convincing way to see what the source is telling you about the capacitance without the plate is to calculate energy in both cases and see how it changes. \$\endgroup\$ – Juan Mar 4 '20 at 7:07
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The screen grid does not do much screening in the true sense of the word. It does the same job as the 1st accelerator grid in a CRO tube or picture tube. It pulls the electrons from the cathode & fires them onto the anode. This allows the anode to be placed much further away (a large round tube instead of a small rectangular one), as it only needs to collect the electrons, not attract them. This greatly reduces the capacitance between it & the control grid.

It also has the advantage, for output valves, of getting better saturation. The anode can get down to a lower voltage, as the 'so called', screen grid attracts the electrons & throws them onto it.

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