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circuit with two directy coupled FETs

https://www.researchgate.net/publication/3845973_A_04_mm_33_V_1T1C_4_Mb_nonvolatile_ferroelectric_RAM_withfixed_bit-line_reference_voltage_scheme_and_data_protection_circuit

The circuit is a fixed voltage reference generator with aim to provide a constant VRE.

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    \$\begingroup\$ Since you are asking how this works, I assume it isn't your own design. To help provide context for an answer, and to comply with the site rules for How to reference material written by others, please edit the question to include a link to the source of the reference generator circuit. \$\endgroup\$ Commented yesterday
  • \$\begingroup\$ I have added link of the source here \$\endgroup\$ Commented yesterday
  • \$\begingroup\$ The source link is not accessible without requesting a download from research gate so, I suggest you add background info to your question taken from that file in order to provide context to your post. Of course, you could perform a simulation of the circuit and find the answer yourself. \$\endgroup\$ Commented yesterday
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    \$\begingroup\$ "The circuit is a fixed voltage reference generator" It isn't. "with aim to provide a constant VRE." It doesn't. That circuit doesn't regulate to any specific voltage. Replace the parts, and you'll get a different voltage. Change the temperature, and you'll get a different voltage. Put a load on it, and you'll get a different voltage. A dead giveaway is that it uses academic symbols for MOSFETs instead of the engineering symbols. \$\endgroup\$ Commented yesterday

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In a very rough explanation, T2 act as regulator.
When you power up the circuit T1 go ON, then there's a current that go in the branch R2-T1-R3 (call it \$I1\$), and then VRE is the voltage across R3 (R3*I1) plus the T1's \$V_{DS}\$.

If the Vcc go up, for example, the current \$I1\$ increase ⇒ increase the voltage across R3 ⇒ increase the current drawn by T2 ⇒ decrease the current \$I1\$ ⇒ decrease the voltage across R3.

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  • \$\begingroup\$ thanks for the answer :) I have one follow up on this ... , I got the point that when I1 increases with VCC , it increase voltage across R3, and so the transistor T2 draws more current because the gate voltage increases. Does that mean that the extra I1 will just pass through T2 towards the ground ? \$\endgroup\$ Commented yesterday
  • \$\begingroup\$ Yes. And obviously the same happens if the Vcc decrease.: the current I1 decrease ⇒ decrease the voltage across R3 ⇒ decrease the current drawn by T2 ⇒ increase the current I1 ⇒ increase the voltage across R3. \$\endgroup\$ Commented yesterday
  • \$\begingroup\$ Much like any R divider is a fixed Vref but with Vt offsets controlled by NFB on gates 😎 \$\endgroup\$ Commented 12 hours ago
  • \$\begingroup\$ NFB? What means? \$\endgroup\$ Commented 10 hours ago
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In this system, T1 passes some current, and it's unimportant how much current, but the consequence of that current is a voltage across R3. This voltage is \$V_{GS}\$ for T2, and as you should know, \$V_{GS}\$ is what determines the "on" (or "off") state of a FET.

If T2 switches "on", it tends to connect T1's gate more closely to ground, 0V, directly reducing the \$V_{GS}\$ of T1, switching T1 more "off". This reduces R3 current and T2's \$V_{GS}\$. This then causes T2 to switch more "off", in opposition to the initial perturbation of T2's state.

This relationship forms negative feedback. The two transistors operate together to maintain a stable condition, in which both \$V_{GS}\$ setlle at exactly the value required to maintain that state. If either \$V_{GS}\$ were to change, even by the tiniest amount, this would cause the other \$V_{GS}\$ to change also, in a direction that would oppose the initial perturbation, and restore equilibrium.

In this way, currents in the system are held constant, and consequently \$VRE\$ is also constant. You could no doubt demonstrate that many nodes in this circuit have constant potential, not just \$VRE\$, but \$VRE\$ seems to be adjustable, by varying R1 and/or R2. It is negative feedback that produces constant and stable state, but resistances and FET \$V_{GS(TH)}\$ that determine the actual values of potentials and currents.

Of course, since \$V_{GS(TH)}\$ varies with temperature, potentials and currents are only as "constant" as temperature.

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  • \$\begingroup\$ Thanks for the answer , explaining with respect to negative feedback is appreciated !! \$\endgroup\$ Commented 15 hours ago
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From the provided circuit, it is clear that the static VRE voltage, here indicated by Vo, is desired, as there are no signal sources. I therefore proceed with a static analysis of the circuit. It contains two N-channel enhancement MOSFETs, for which the analytical relationship between drain current and gate-source voltage VGS is known, given the threshold parameters KMOS and VGST.

enter image description here

enter image description here

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  • \$\begingroup\$ -1 this design was not intended for 20 A and states no purpose here, the link in question defines it uniquely , but nice math \$\endgroup\$ Commented 16 hours ago
  • \$\begingroup\$ In the example, IDON is not the design drain current. IDON (and VGSON) is the drain value provided by the manufacturer (VGST threshold voltage as well), which I used to calculate the constant Kmos. The data refers to the Si4830ADY dual N-channel MOSFET; I don't think I've made any confusion. \$\endgroup\$ Commented 16 hours ago
  • \$\begingroup\$ Thanks for providing a mathematical overview :) \$\endgroup\$ Commented 15 hours ago
  • \$\begingroup\$ We know that for a MOSFET enhancemt ID=Kmos(VGS-VGSTh)^2. Kmos is determined by considering the known values ​​of the coordinates of a point on the transcharacteristic (IDON and VGSON) of the device. \$\endgroup\$ Commented 14 hours ago
  • \$\begingroup\$ Yes but we don't know why drawing so many Amps is a good regulator. It has similarities to the BJT CC limiter but less accurate due to Vt tolerance. Whereas in mA level it is useful for biasing ferromagnetic RAM \$\endgroup\$ Commented 13 hours ago
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This novel voltage reference was presented for sensing ferromagnetic RAM with a self-trimmed bias voltage reference Vre.

I ran it on Falstad's sim and found some novel properties.

  1. Vre matches Vgs(th) of T1 plus the very low current and voltage drop across R2.
  2. When R2 matches R3 then any noise in Vdd has a 98% reduction if both FETs are matched, which was the optimal R ratio point.
  3. If T2 has a much lower Vgs(th) then the reduction is more than 99.6% reduced.

This puts both T1 and T2 into the saturation high resistance zone.

Read the Vpp levels.

enter image of her

In another simulation using R2/R1=50, I vary R3 and record Vref using V+=10V. But due to the heat loss at high Vref/V+, efficient low current occurs with low R3/R2 <<1 while R2/R1=50.

enter image description here

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  • \$\begingroup\$ any questions? . \$\endgroup\$ Commented yesterday
  • \$\begingroup\$ Thanks for the answer :) , Noise analysis with respect to resistor and transistor is interesting :) I want to follow up ,like for high noise reduction , you mentioned the Vth of transistor T2 should be lower as per simulation . Isnt that case will cause unstability to provide equillibrium to VRE since transistor are not matched and somehow we are controlling VGS of both transistors to have equillibrium , which is intended in the circuit. \$\endgroup\$ Commented 15 hours ago
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    \$\begingroup\$ Low Vt on T2 was a special case to conduct more and thus lower Vref and noise with more gain. But still stable. Do you have goals specs?? If you know the NPN Nch FET CC design, this is similar except sensing R2 current as well as R3 current \$\endgroup\$ Commented 12 hours ago
  • \$\begingroup\$ You can use matched FETs . I chose small signal FETs with Vt=1.5, β=20m ,but depend target IO specs (much like CD4xxx CMOS \$\endgroup\$ Commented 12 hours ago

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