Read M7/M10 where I wrote M6/M7.i cant see the difference, you can explain it for me?
since you have all the simulations, can you tell me what pulls the gate of M2 low and the gate of M10 high right after your clock goes high?Thanks again.
I am attaching a simulations of input voltage from 0 to VDD with and without M10.
No difference is seen in the output voltage for dc input, and for ac the output voltage more accurate without m10.
Can you tell me for which input voltages the circuit will not work without m10?
During track mode, ’sn’ is high and transistor M7 turns on. as a result, transistor M2 opens and forces Vdd on transistor M5 gate, consequently, the positive plate of the MOS capacitor connects to the gate of M0 (same gate like M10) and the negative plate connects to the input voltage so its gate potential is equal to the source voltage plus the supply voltage (Vdd). Thus, the Vgs of transistor M0 is equal to Vdd, and, therefore, its resistance, is independent of the input voltage.since you have all the simulations, can you tell me what pulls the gate of M2 low and the gate of M10 high right after your clock goes high?
'sn' push M7 gate, I connected only the bulks of M7, M10 to gnd.Read M7/M10 where I wrote M6/M7.
Note that you connected the M7 gate net to the M10 gate and body nets on your implemented circuit.
View attachment 179743
From your simulation, if I see this correctly you don't have Vgs of Vdd on M0. Could there be a problem on node Cap_p shouldn't it go higher Vdd+vin and not only as in your simulation to around 5.6 V.During track mode, ’sn’ is high and transistor M7 turns on. as a result, transistor M2 opens and forces Vdd on transistor M5 gate, consequently, the positive plate of the MOS capacitor connects to the gate of M0 (same gate like M10) and the negative plate connects to the input voltage so its gate potential is equal to the source voltage plus the supply voltage (Vdd). Thus, the Vgs of transistor M0 is equal to Vdd, and, therefore, its resistance, is independent of the input voltage.
During track mode, ’sn’ is high and transistor M7 turns on. as a result, transistor M2 opens and forces Vdd on transistor M5 gate, consequently, the positive plate of the MOS capacitor connects to the gate of M0 (same gate like M10) and the negative plate connects to the input voltage so its gate potential is equal to the source voltage plus the supply voltage (Vdd). Thus, the Vgs of transistor M0 is equal to Vdd, and, therefore, its resistance, is independent of the input voltage.
'sn' push M7 gate, I connected only the bulks of M7, M10 to gnd.
From the theoretical analysis there should indeed be Vin+Vdd at the gate of M0, but in practice there is a voltage drop as a result of charges that open M5, M0 or leak through M3.From your simulation, if I see this correctly you don't have Vgs of Vdd on M0. Could there be a problem on node Cap_p shouldn't it go higher Vdd+vin and not only as in your simulation to around 5.6 V.
I agree that M7 is critical to circuit function. I opened the discussion regarding the role of M10. I think M7 alone can do the trick. (The simulations are also with and without M10)So, you are confirming what I said in my first comment in #2 that right after sn goes high it is the M7 that pulls down the gate of M2, after which M10 takes over.
Now, if you remove M7, what will pull the gate of M2 down initially?
Yes it won't reach exactly Vin+Vdd but only 5.6 is way to low your problem is M3 when the drain voltage goes above Vdd your bulk is below the drain Voltage and the drain bulk diode opens that is also the reason you don't see the effect of M10 because M5 doesn't open fully and node C_n is less than Vin. You need M10 for high input Voltages when the node C_n is vdd-vth and M7 shuts off and the gate of M2 starts floating this prevents M10 as he takes over and pulls the gate of M2 to vin.From the theoretical analysis there should indeed be Vin+Vdd at the gate of M0, but in practice there is a voltage drop as a result of charges that open M5, M0 or leak through M3.
I agree that M7 is critical to circuit function. I opened the discussion regarding the role of M10. I think M7 alone can do the trick. (The simulations are also with and without M10)
I agree about connecting the body of the M3 transistor to the source. Regarding the role of the M10 transistor, this was my theory, but as you can see from the simulation for Vin=Vdd=5V>Vdd-vt is still obtained at the output Vout=5v even without the M10 transistor.Yes it won't reach exactly Vin+Vdd but only 5.6 is way to low your problem is M3 when the drain voltage goes above Vdd your bulk is below the drain Voltage and the drain bulk diode opens that is also the reason you don't see the effect of M10 because M5 doesn't open fully and node C_n is less than Vin. You need M10 for high input Voltages when the node C_n is vdd-vth and M7 shuts off and the gate of M2 starts floating this prevents M10 as he takes over and pulls the gate of M2 to vin.
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?