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Questions about Cascode Bandgap

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pseudockb

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Cascode Bandgap

Hi all, I have some questions regarding the cascode bandgap.

1. Could someone please explain to me how the start-up circuit works? My understanding is that the NMOS acts as a pull-down resistor to pull the voltage at the gate of Mpinj to GND so as to inject a current into the bandgap core. How does the start-up circuit turn off? Or is it operating at all times? How does Mpstart come into play?

2. What is the reason of connecting Q3/Q5 at the bottom of cascode biasing transistors?

3. I can diode connect transistors MP2 so that I can remove the M3 transistors to save a branch of current. Any bad effect of this?

4. Intuitively speaking, the PSRR of this bandgap is a resistive devision between the output impedance of the cascode current mirror and the R2. I have heard that bandgap circuit utilizing an opamp clamping the at the top of Q1 and top of R1 can achieve a better PSRR compared to the cascode bandgap. Why is it so?

Thanks.
 

mpig09

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Re: Cascode Bandgap

Dear :

1.When VDD rise to supply voltage(3.3V or 5V...)slowly, Mnstart will first on then other MOS, when Mnstart on, Mpinj will on, so the bandgap will into desired operation point.

2. What is the reason of connecting Q3/Q5 at the bottom of cascode biasing transistors?
==>for high accuracy.

3.remove the M3 transistors to save a branch of current.
==>that ok, when your design & calculate is right.


4. I don't understand.


Best Regards,
mpig
 

pfd001

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Cascode Bandgap

M3 tansistors can not be removed. It provides bias for the PMOS transistors near the power supply
 

jjsnail

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Cascode Bandgap

M3,M5 provides biass for the cascode M1,M2

but I think we can use current miror with M1,M2 to save the bias. am I right?
 

corel

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Re: Cascode Bandgap

1. I also can't understand how the startup circuit cut-off from the main circuit after it start
4. use opa is voltage negative feedback to make two nodes' voltage equal.
This circuit is use current negative feedback to realize two nodes' voltage equal.
But I can't explain how this current negative feedback work,Can somebody explain it?
 

davidwong

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Re: Cascode Bandgap

Mpinj will be off after startup
Vgs of Mpinj will be close to 0V
as Mnstart is long channel device.
 

she_long

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Re: Cascode Bandgap

Hi, pseudockb
I think in the schematic the resistor R1 should be under the source of Mn1.
If that, for R1, Mn1 Mn2 Mn3 Mp1 Mp1cas and Mn1 consist of a negative feedback.
Mn3 can increase the open loop gain to increase PSRR.
 

waosai

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Re: Cascode Bandgap

I think the following structure can move Mp3, Mn3, Mp5, Mn5, Q3 and Q5.
I don't know whether I am right.

67_1164878652.JPG
 

niezimei

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agree with she_long

change the polarity so as to make the whole circuit in negative feedback control.BTW, this circuit need a capciator which is added between the gate of Mn3 and gnd to make a dominate compensation.
 

liuyonggen_1

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Re: Cascode Bandgap

first, M3 an M5 is used to bias the mp1 and mp2,respectively,which is very necessary for low power supply applications,because the voltage drop across the whole cascode current mirror is only 2 times Vov. that's to say, if the power supply voltage is large enough, you can use other structures to replace M3 and M5.

second, i think the start-up circuit is not good. in my opinion, it's difficult to turn off the current-injection banch. i suggest you use other start-up structures. there are many simple start-up structures.
 

fanrong

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Re: Cascode Bandgap

she_long said:
Hi, pseudockb
I think in the schematic the resistor R1 should be under the source of Mn1.
If that, for R1, Mn1 Mn2 Mn3 Mp1 Mp1cas and Mn1 consist of a negative feedback.
Mn3 can increase the open loop gain to increase PSRR.

If so , the loop gain will large than 1 , which will lead to unstable !?
 

devop

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Re: Cascode Bandgap

davidwong said:
Mpinj will be off after startup
Vgs of Mpinj will be close to 0V
as Mnstart is long channel device.
there will always be I flowing through Mstart?
 

she_long

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Re: Cascode Bandgap

fanrong said:
she_long said:
Hi, pseudockb
I think in the schematic the resistor R1 should be under the source of Mn1.
If that, for R1, Mn1 Mn2 Mn3 Mp1 Mp1cas and Mn1 consist of a negative feedback.
Mn3 can increase the open loop gain to increase PSRR.

If so , the loop gain will large than 1 , which will lead to unstable !?

Yes, to the plus of R1, the loop gain will large than 1, but it's a negative feedback, it's the same in 'waosai's picture. R1 always in the negative feedback loop. That's more clearly in the BGR using an op amp.
 

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