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Could you please help me with this question? (Because I'm new here, it seems that I couldn't post an image.)

The circuit diagram is the first picture from this lecture

Voltage Vg is the node voltage between Rg1 and Rg2.

My question is, if the DC voltage Vg is much higher than the amplitude of the small variable signal Vsig, then the charge caused by Vg on either plate of the capacitor should be very large (I think), can Vsig still pass through the capacitor and have a component on Vg? If the left plate of the capacitor is full of negative electrons because of voltage \$V_{dd} \dfrac{R_{G2}}{R_{G1}+R_{G2}}\$, how can small signal Vsig still have influence on the voltage Vg? Thank you!

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  • \$\begingroup\$ If you have an image, upload it to a server, and tell us the URL. We'll add it to your post. \$\endgroup\$
    – stevenvh
    Jun 5 '12 at 8:42
  • \$\begingroup\$ I tried to answer the question behind, but this question is full with misunderstandings about how capacitors and voltage dividers work. \$\endgroup\$
    – clabacchio
    Jun 6 '12 at 9:40
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I'm assuming steady state.

If \$V_{IN}\$ has a low impedance, like from a power supply, it won't change because of the capacitor's load. If the other side is constant as well, then you'll have \$V_g - V_{IN}\$ across the capacitor, that is AC with a DC offset.

edit (re your edit)
So \$V_g\$ is a MOSFET's gate voltage? You haven't said anything about a FET in your original question. Please be more clear.

If the small variable signal \$V_{IN}\$ has a low impedance, like when it comes from an opamp, then my original answer still stands. It doesn't depend on the amplitude.
If the signal has a significant resistance, that will form a low pass filter with the capacitor, with a cutoff frequency

\$f_C = \dfrac{1}{2 \pi \cdot RC} \$

If \$C\$ is the gate's capacitance and \$R\$ is about 10k\$\Omega\$ the cutoff frequency is probably larger than 100kHz, so it may or may not influence your signal. A low frequency signal won't be much attenuated, and you'll still have most of \$V_g - V_{IN}\$ across the capacitor.

edit 2 (after your link to the PDF)
It's still not clear what your schematic is! I presume it's this:

enter image description here

Ok, the \$C\$ is not a gate capacitance. But in this case \$V_g\$ is not a constant voltage! It will have a constant DC component, but part of the AC signal will be added to it. My calculation is still valid, only

\$ R = R_{sig} + \dfrac{R_{G1} \cdot R_{G2}}{R_{G1} + R_{G2}}\$.

And

\$ V_C = \dfrac{1}{1 + j \omega RC} \cdot V_{sig} - \dfrac{R_{G2}}{R_{G1} + R_{G2}} \cdot V_{DD} \$

The voltage has an AC component from \$v_{sig}\$, and a DC component from \$\dfrac{R_{G2}}{R_{G1} + R_{G2}} \cdot V_{DD}\$. The DC level at the left side of the capacitor is 0V, at the right side it is \$\dfrac{R_{G2}}{R_{G1} + R_{G2}} \cdot V_{DD}\$, so there's a DC difference of \$\dfrac{R_{G2}}{R_{G1} + R_{G2}} \cdot V_{DD}\$ across \$C\$. The DC component is negative because I took the MOSFET's gate as reference.
So, yes, apart from the attenuated signal you'll also see the \$\dfrac{R_{G2}}{R_{G1} + R_{G2}} \cdot V_{DD}\$ across the capacitor.

\$ V_G = \dfrac{R_{G2}}{R_{G1} + R_{G2}} \cdot V_{DD} + \dfrac{j \omega R_{G1} R_{G2} C}{R_{G1} + R_{G2} + j\omega(R_{G1} R_{sig} + R_{G2} R_{sig} + R_{G1} R_{G2} ) C} \cdot v_{sig} \$

edit (re your edit dd. 2012-06-06)

"If the left plate of the capacitor is full of negative electrons because of voltage Vdd x RG2/(RG1+RG2), how can small signal Vsig still have influence on the voltage Vg?"

Ah, it looks like we're finally getting at your actual question. The left plate will never be "full", you can always add charge to a capacitor.

\$ Q = C \cdot V \$

So adding charge (\$Q\$) to an already charged capacitor will increase its voltage. So even on a 10000\$\mu\$F capacitor at 100V (holding a 1C charge) a superimposed 10mV\$_{P}\$ AC signal will add/subtract charge. At its maximum the capacitor will hold a 1.0001C charge.

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The circuit you describe is this:

enter image description here

You can redraw it ignoring the transistor, as it will give some second order effects that we want to ignore by now.

enter image description here

The signal Vsig is AC coupled, which means that its DC level will be filtered by the capacitor. What passes is centered on 0V, but another DC voltage is given by the voltage divider, bringing it to a good range for the MOS.

What you have is that Vg is just a node voltage, and it doesn't make sense to treat it as a voltage source. But you have Vdd which is a DC voltage source. Its value of course will be filtered by C1, and won't be seen at Vx.

You can say that Vx is dependent on Vg, because if you compute it in the small signals model, they will be dependent on each other. But the only independent source is v_sig, and all the other voltages (again, for small signals) will depend only on it, because Vdd is constant and it's cancelled when considering variations.

Here is a simulation:

enter image description here

enter image description here

You can see the high pass filter behavior that cuts out the DC component. The values of R and C are arbitrary, and can be adapted to the requirements.

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  • \$\begingroup\$ Between which two points is that measured? (Kortuk's question might answer that) \$\endgroup\$ Jun 6 '12 at 12:09
  • \$\begingroup\$ @FedericoRusso: it's Vg, in fact it's not a gain but just a voltage \$\endgroup\$
    – clabacchio
    Jun 6 '12 at 12:14
  • \$\begingroup\$ Let's take the circuit diagram that clabacchio gave for example. I would like to know what the node voltage Vg will be. What I can't understand is why the small signal source V_sig will have a small variable voltage component added to Vdd*Rg2/(Rg1+Rg2)? If the capacitor has a large C, and before the source V_sig is connected, the capacitor has been charged with a large amount of charge on its plates, nearly to its maximum, then small V_sig still has the ability to add/sbutract any charge on the plate? Alough stvenvh has already given the answer.If I press a spring,with just a little strength \$\endgroup\$
    – 濬明 梁
    Jun 7 '12 at 2:11
  • \$\begingroup\$ @clabacchio:, the spring which has already been pressed a lot still can be pressed a little? \$\endgroup\$
    – 濬明 梁
    Jun 7 '12 at 5:42
  • \$\begingroup\$ @濬明梁 the capacitor is not a spring, it has not a theorical maximum. Steven has already an answer for that. But clearly this circuit is too complicated for you, I suggest trying with something simpler. \$\endgroup\$
    – clabacchio
    Jun 7 '12 at 5:57

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