00:04 Advanced biasing techniques in integrated circuits 02:25 The Scaled bandage reference and adjustable voltage PVT independent references enforce similar currents and voltages. 08:07 Enforcing equal currents for substantially similar voltages. 10:21 Diode connection impacts voltage regulation. 15:39 Different configurations for VT reference 17:40 Adjustable voltage PVT independent references 21:39 Discussion on creating a constant voltage using adjustable voltage references. 24:13 The need for a stable solution in chip design. 28:23 Using op-amp for voltage equalization 30:44 Analyzing voltage references using equations 35:48 Voltage cancellation at 1.1 volts 37:48 Discussing the issue with the voltage and finding a solution 42:25 Generating a VT reference by dropping voltage across a resistor. 45:01 Explanation of scaling factor and voltage references 50:11 Adjustable voltage PVT independent references Crafted by Merlin AI.
40:52, In the technology I am working with the resistors are also temperature dependent. When I generate current using this method(low voltage bandgap reference) It is dependent on temperature because of resistor. How to generate a current that is independent of supply without using resistors. At 48:26, resistors R1, R2 and R3 are not temperature independent
As far as I know all the resistors in the integrated circuits are temperature dependent. At 48:26, yes, the resistors R1-R3 are not temperature independent, but the ratio is. So by doing a good layout, the resistors ratio are temperature independent. The equation 48:26 is also independent of supply voltage. By choosing the ratios, you can pretty much get any voltage you want.
Hello Professor, About the floating mirror: When we talk about resistance seen from node, we are talking about small signal resistance right? So in small signal model, even if we connected VDD to ground with diodes, I think it is okay, since VDD is ground in small signal model. Sorry, I still don't understand why we can't connect VDD to ground with only diode connected MOS. Ang above is my doubt. Thank you very much.
Dont think about small signal in this case. Small signal is only valid if it makes sense to linearize the non-linear components around an operating point, which is not the case here. We chose the floating mirror because it made the voltages below it equal and independent of the supply voltage. With only diodes above it the voltage would be very much dependent on the supply.
If the currents in the two branches are equal, then if they flow through unequal impedances, the voltages on the sources of nmoses in the floater cannot be the same, which means they have different overdrives, which, in it's turn, implies unequal currents in the branches
Am I right that by making pmoses "stronger" we can make the currents more or less equal, but still, we would end up having unequal voltages on the given nodes?
@@brantran3754 This line of questioning came to my mind too. If the combination of the pmos current mirror and the floating current mirror forces the currents to both branches to be equal and the source voltages to be the same, are we saying then that the impedance of the circuit underneath is forced to be the same even if we make them different ? The obvious answer is no so there must be a flaw in the reasoning. The gate voltages of the floating current mirror transistors are the same because they are the same node. For the source voltages to be the same, the VGS's should be the same. For the VGSs to be the same, the floating current mirror transistors should have the same current AND the same W/L ratio (or it is the same as saying their overdrive voltages should be the same). So the source voltages are the same if the W/L ratio is the same. Otherwise they are not. The condition should be, same current AND same W/L ratio of the floating current mirror transistors to force the source voltages to be the same, with both source voltages one VGS below the gate voltage.
@@brantran3754 increasing the length of the pmos current mirror will make the mirroring more accurate because by increasing the length, we are increasing its output impedance r[o] = 1 / lambda I[D]. The key to forcing the source voltages to be equal is to make the W/L ratio of the floating current mirror equal which makes their overdrive voltage equal and therefore their VGSs
@@brantran3754 I think the proper statement of what the combination of the pmos current mirror and the floating current mirror does is - the currents through both branches are made equal by the pmos current mirror AND the floating current mirror TRANSMITS the source voltage from one side to the other side. The source voltages will be equal only if the W/L ratio of the floating mirror transistors are the same (same overdrive voltages) otherwise the source voltages will be different.
I don't know how to connect op. amps in the circuits you drawn (right connections of positive and negative inputs) in such a way that they are inside a negative (and not positive) feedback loop. How did you chosen the right way in so fast manner? It not seems to be intuitive. Thank you.
Towards the end, we end up with a voltage depending on Vbe... but we didn't want this to happen, 'cause then the temperature dependence isn't proportional.
wished our professor would do such a good job! Splendid, excellent teachning skills Sir thank you
00:04 Advanced biasing techniques in integrated circuits
02:25 The Scaled bandage reference and adjustable voltage PVT independent references enforce similar currents and voltages.
08:07 Enforcing equal currents for substantially similar voltages.
10:21 Diode connection impacts voltage regulation.
15:39 Different configurations for VT reference
17:40 Adjustable voltage PVT independent references
21:39 Discussion on creating a constant voltage using adjustable voltage references.
24:13 The need for a stable solution in chip design.
28:23 Using op-amp for voltage equalization
30:44 Analyzing voltage references using equations
35:48 Voltage cancellation at 1.1 volts
37:48 Discussing the issue with the voltage and finding a solution
42:25 Generating a VT reference by dropping voltage across a resistor.
45:01 Explanation of scaling factor and voltage references
50:11 Adjustable voltage PVT independent references
Crafted by Merlin AI.
40:52, In the technology I am working with the resistors are also temperature dependent. When I generate current using this method(low voltage bandgap reference) It is dependent on temperature because of resistor. How to generate a current that is independent of supply without using resistors. At 48:26, resistors R1, R2 and R3 are not temperature independent
As far as I know all the resistors in the integrated circuits are temperature dependent. At 48:26, yes, the resistors R1-R3 are not temperature independent, but the ratio is. So by doing a good layout, the resistors ratio are temperature independent. The equation 48:26 is also independent of supply voltage. By choosing the ratios, you can pretty much get any voltage you want.
Hello Professor,
About the floating mirror:
When we talk about resistance seen from node, we are talking about small signal resistance right?
So in small signal model, even if we connected VDD to ground with diodes, I think it is okay, since VDD is ground in small signal model.
Sorry, I still don't understand why we can't connect VDD to ground with only diode connected MOS. Ang above is my doubt.
Thank you very much.
Dont think about small signal in this case. Small signal is only valid if it makes sense to linearize the non-linear components around an operating point, which is not the case here. We chose the floating mirror because it made the voltages below it equal and independent of the supply voltage. With only diodes above it the voltage would be very much dependent on the supply.
But what if, in the floating mirror, we have different impedances "under" the floater?
If the currents in the two branches are equal, then if they flow through unequal impedances, the voltages on the sources of nmoses in the floater cannot be the same, which means they have different overdrives, which, in it's turn, implies unequal currents in the branches
Am I right that by making pmoses "stronger" we can make the currents more or less equal, but still, we would end up having unequal voltages on the given nodes?
@@brantran3754 This line of questioning came to my mind too. If the combination of the pmos current mirror and the floating current mirror forces the currents to both branches to be equal and the source voltages to be the same, are we saying then that the impedance of the circuit underneath is forced to be the same even if we make them different ? The obvious answer is no so there must be a flaw in the reasoning. The gate voltages of the floating current mirror transistors are the same because they are the same node. For the source voltages to be the same, the VGS's should be the same. For the VGSs to be the same, the floating current mirror transistors should have the same current AND the same W/L ratio (or it is the same as saying their overdrive voltages should be the same). So the source voltages are the same if the W/L ratio is the same. Otherwise they are not. The condition should be, same current AND same W/L ratio of the floating current mirror transistors to force the source voltages to be the same, with both source voltages one VGS below the gate voltage.
@@brantran3754 increasing the length of the pmos current mirror will make the mirroring more accurate because by increasing the length, we are increasing its output impedance r[o] = 1 / lambda I[D]. The key to forcing the source voltages to be equal is to make the W/L ratio of the floating current mirror equal which makes their overdrive voltage equal and therefore their VGSs
@@brantran3754 I think the proper statement of what the combination of the pmos current mirror and the floating current mirror does is - the currents through both branches are made equal by the pmos current mirror AND the floating current mirror TRANSMITS the source voltage from one side to the other side. The source voltages will be equal only if the W/L ratio of the floating mirror transistors are the same (same overdrive voltages) otherwise the source voltages will be different.
35:03
It should be R2/R3 not R1/R3 in front of logarithm there at Vout expression
I don't know how to connect op. amps in the circuits you drawn (right connections of positive and negative inputs) in such a way that they are inside a negative (and not positive) feedback loop. How did you chosen the right way in so fast manner? It not seems to be intuitive. Thank you.
Towards the end, we end up with a voltage depending on Vbe... but we didn't want this to happen, 'cause then the temperature dependence isn't proportional.
Scale the Vbe term so that its Vbg term results in the voltage you want. Then scale the PTAT term to cancel the CTAT term in Vbe term
What if I need Iref from the first circuit (ptat using self-bias)
Are there any problem sets available for this course?
Is this a bandgap reference?
Yes
PTAT current flows through BJT, of which voltage is CTAT.
Thanks 😄
Bandgap* perhaps.
I saw the cheat and was bothered, till he mentioned it. haha
7:25 a culprit is talking
Who is he?