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About pipeline adc--Sub-adc

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leokaven

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Hi! friend

I'm begin study pipeline adc architecture
I met a problem anybody can help me?please!
Pipeline adc have 1.5bit and 2bit sub-adc

1.5 bit sub-adc need Vref between +1/4Vref and -1/4Vref
2 bit sub-adc need Vref between +1/2Vref, 0 and -1/2Vref
How can i produce these voltage reference?

Sorry!my English is poor
Thanks for your help
 

Usually pipeline ADC is completely differential circuit and reference voltage is differential also.
 

    L

    leokaven

    Points: 2
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Thanks for gevy ^^
But how to completely?
1.5 bit sub-adc need Vref between +1/4Vref and -1/4Vref
2 bit sub-adc need Vref between +1/2Vref, 0 and -1/2Vref

I'm read IEEE paper, A 10b 20Msamples/s analog to digital converver, P354
"each core stage contained an ADSC and DAC that shared a common resistor string, the resistor determine the ADSC decision"

I don't know how to completely 1.5bit and 2bit circuit
How many resistor i need in 1.5bit and 2bit sub-adc?

Please help me,friends
 

leokaven said:
Thanks for gevy ^^
But how to completely?
1.5 bit sub-adc need Vref between +1/4Vref and -1/4Vref
2 bit sub-adc need Vref between +1/2Vref, 0 and -1/2Vref

I'm read IEEE paper, A 10b 20Msamples/s analog to digital converver, P354
"each core stage contained an ADSC and DAC that shared a common resistor string, the resistor determine the ADSC decision"

I don't know how to completely 1.5bit and 2bit circuit
How many resistor i need in 1.5bit and 2bit sub-adc?

Please help me,friends
Click to expand...

you should use a refernce , some resistances to divided the voltage, 3k--2k--3k or so ....
 

    L

    leokaven

    Points: 2
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HI!chency
Thank for you reply my question!

This is my 1.5bit and 2bit schematic
use fully differential signal ±Vin and ±Vref
if ±Vref=1, I need to produce those are voltage reference
Can i use this schematic?
Are these schematic correct or incorrect?

Thanks for your help!
 

Vref=1
V1=0.9625 V2=0.7125
V3=1.0875 V4=0.8375
these voltages can be produced simply by the resistor ladders
use a simple substractor base on capacitors
V1-V2=0.25=1/4 Vref
V4-V3=-0.25=-1/4Vref
the resistor ladders are also used to produce common voltage
this is how i design the requested ref voltage
 

    L

    leokaven

    Points: 2
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leokaven said:
HI!chency
Thank for you reply my question!

This is my 1.5bit and 2bit schematic
use fully differential signal ±Vin and ±Vref
if ±Vref=1, I need to produce those are voltage reference
Can i use this schematic?
Are these schematic correct or incorrect?

Thanks for your help!
Click to expand...
hi,
the 2bit section is correct, but the 1.5bit is not,
u can check it in the attachment.
 

    L

    leokaven

    Points: 2
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Btrend said:
hi,
the 2bit section is correct, but the 1.5bit is not,
u can check it in the attachment.
Click to expand...

Btrend is right, but I don't known ,why did you connect gnd! in your circuit? as a diff input flash adc, i think it's no use.
 

    L

    leokaven

    Points: 2
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HI!Btrend, sigmadelta and chency
Thank for you reply my question!

I have two way to produce voltage reference in overall circuit
One way, Maybe i can shared a common resistor string determine resistor value produce these voltage reference
for 1.5bit and 2bit sub-adc (fig 1)
My sub-adc(flash adc) is use fully comparator,should i connect gnd in common resistor string ?
Is this schematic correct ?


Another way, connect 2bit sub-adc voltage reference and use in 1.5bit produce it(fig 2)

Thanks for your help!
Sorry!my English is poor
 

1. the common gnd is the common mode voltage of ur differential input. usually in single rail (0 ~ vcc) application, it is > 0. so in this case u should not connect it to GND. and why u worry about the common gnd ? it is always equal to (+Vref+(-Vref))/2= 0 ....... (ac GND)
2. fig. 1 is correct.
3. but fig.2 is wrong in the generation of +-Vref/2, unless u use some buffer between the right part & left part. or u choose different values of ur resistors
 

    L

    leokaven

    Points: 2
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haha~~
I get it!
Thanks for your help!
 

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