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I have a problem which consist in 1 bit RAM made of 3 MOSFETs. One of the questions is to calculate the maximum voltage that the memory element can receive. I have obtained the result by inspection (it is 4 Volts) but I'm unable to reach the same by applying the Thevenin Theorem.

My understanding is: I have a circuit made of several resistors and one capacitor (which is the memory element). If the circuit is reduced to a \$ (V_{TH}, R_{TH}) \$ Thevening Equivalent, and given that the capacitor behaves like an open circuit for long periods of time (it's fully charged), I can assume that the maximum voltage the capacitor might have is \$ V_{TH} \$. The problem is that the \$ V_{TH} \$ I find has nothing to do with the expected result (4 Volts).

I want to know what I'm doing wrong and how to solve this problem by applying the Thevenin theorem.

The exercise, along with all my schemas and equations, are below:

enter image description here

In a previous question I have calculated the parasitic resistance and it is \$ R_{PA} = 185.347405560882 \$ TeraOhms.

The question I actually need to answer is:

Now, suppose the drain of Q1 is high, and the store line is held at the same voltage as the drain of Q1 . What is the maximum voltage, in Volts, that the gate of Q2 can be charged to? Note, this value must be larger than VOH = 3.5 Volts to satisfy the static discipline.

Now, my 'Tehevening Equivalent' attempt to find the voltage in Q2 is below:

Replacing capacitors by the corresponding RON resistors

Circuits to calculate VTH and RTH

$$ parallel(R_{1}, R_{2}) = \frac{1.00000000000000}{\frac{1.00000000000000}{R_{1}} + \frac{1.00000000000000}{R_{2}}} \\ R_{\mathit{TH}} = {\rm parallel}\left(R_{\mathit{PA}}, R_{\mathit{ON}} + {\rm parallel}\left(R_{\mathit{OFF}}, R_{\mathit{PU}}\right)\right) \\ I_{\mathit{TH}} = \frac{V_{S}}{R_{\mathit{PU}} + {\rm parallel}\left(R_{\mathit{OFF}}, R_{\mathit{ON}} + R_{\mathit{PA}}\right)} \\ e = -I_{\mathit{TH}} R_{\mathit{PU}} + V_{S} \\ \mathit{ITH}_{2} = \frac{e}{R_{\mathit{ON}} + R_{\mathit{PA}}} \\ V_{\mathit{TH}} = -\mathit{ITH}_{2} R_{\mathit{ON}} + e $$

Or in SageMath/Python:

V_S = 5.
R_ON = 2100
R_OFF = 110e6
R_PU = 10e3
R_PA = 185.347405560882e12

parallel(R1, R2) = 1./(1./R1 + 1./R2)

R_TH = parallel(R_PA, R_ON + parallel(R_OFF, R_PU))

I_TH = V_S / (R_PU + parallel(R_OFF, R_ON + R_PA))
e = V_S - I_TH * R_PU
I_TH2 = e / (R_ON + R_PA)
V_TH = e - I_TH2*R_ON

The final result (V_TH) I get is 4.99954549553765 Volts, and it should be 4 Volts.

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  • \$\begingroup\$ In the schematic from the question, there are two \$R_{PU}\$ resistors. Your schematics have only one, why? \$\endgroup\$ – Bimpelrekkie Jan 14 at 13:41
  • \$\begingroup\$ Because the second R_PU is after the mosfet's gate, so is not connected to my equivalent circuit (correct me if I'm wrong) \$\endgroup\$ – Martel Jan 14 at 13:51
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After researching, I think the answer is that when Q3 reaches some voltage, Store disconnects until Q3 discharges a bit, in which case Store connects again, therefore keeping Q3 at a 'controlled' voltage. So the Thevenin Equivalent is not valid for this circuit since Store connects/disconnects depending on the circumstances.

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  • \$\begingroup\$ Assuming 'infinite' Roff below Vth, Is it not simply Vs - Vth? \$\endgroup\$ – Bruce Abbott Jan 28 at 0:20
  • \$\begingroup\$ If you apply Thevenin, yes. Well, there is a small voltage divider due to Roff, but hardly modifies the original voltage. The thing is that V_TH won't reach 5 volts since Store won't allow it (V_gs will be less than V_t) \$\endgroup\$ – Martel Jan 28 at 8:58
  • \$\begingroup\$ I understand what you mean by 'disconnects and reconnects' now. You mean Drain-Source resistance goes up as Gate voltage goes below the threshold voltage, and so 'regulates' the Drain-Source voltage drop to ~1V. This is correct, but your wording is confusing. \$\endgroup\$ – Bruce Abbott Jan 28 at 9:11

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