Auxiliary power for the ASIC cards
Ok, so we've got the 1.2V for the asics, now the rest.
I looked at the reference design to see how much current at which voltages would be required.
We need 3,3V for the IO's of the asics and the clock distribution.
And a quiet 1,2V for the PLL's of the asics.
Quiet 1,2V regulator
Lets start with the Quiet 1,2V regulator. The PLL in this chips is not speced, other than the voltage tolerance.
PLL's can be picky eaters, the normal 1,2V, which can be found in this thread, was designed with a larger ripple to improve transient response. The PLL might not like that.
I don't know how much current they'll use so I looked at the reference design where they used a 500mA part for 10 chips. Thus I chose a 600mA pat for my 12 chip board.
Just to be safe. Perhaps it won't need that at all, maybe they just used it because they could reuse it on their controller board.
Yeah, I intend on reusing it aswell on my controller board.
Part chosen: Adjustable LDO AP7165-SPG-13 from Diodes inc. 0,69
Not much to design really with these parts, just a few things to keep in mind.
Transient response, cheap and old parts can be pretty terrible at this, so I pay attention to it. The datasheet checks out, it's not that bad.
Temperature, I will power it with 3,3V so with a 2,1V drop at 600mA you will disapate 1,26W.
I use a SO-8 package with an exposed pad and a pcb with a forced air heatsink cooling. I think it will work out, as the supplier specs it at 40C/W on a 50*50mm board.
PSRR power supply rejection ratio. We are post regulating an SMPS so that is kinda important.
Suppliers almost always show a way to optimistic figure/graph. Not all by the way, more expensive parts, often intended for RF purposes are way better.
Almost all of Analog devices, AMS or expensive ones from Texas instruments and Micrel are pretty good as far as I know.
Anyway lets think about this for a bit.
Most mosfet based LDOs, like this one have, as (most)all mosfets do, a parasitic capacitance from drain to source (drain to bulk and bulk to source in series).
This will allow it to pass high frequency noise.
Also, the regulator feedback loop has a limited bandwidth, it also allows high frequency stuff to pass.
Now the external capacitors will take care of higher frequencies, thus improving PSSR, but due to their construction will start to resemble inductors at higher frequencies.
Thats why I'll put a ferrite bead in series, to get rid of higher frequencies.
For the input and output capacitors I chose a part I used before on this board. The 6,3V 22uF capacitors have about 10uF of capacity at 3,3V and at 1,2V it's 20uF
Parts chosen: 2x 0805 X5R 6,3V 22uF Murata GRM21BR60J226ME39L 0,38 (0,76)
As for the feedback I chose 31,6K for R2 and 15,8K for R1.
Vout = Vref*(1+(R1/R2))
0,8*(1+(15800/31600)) = 1,2V
Combined cost: 0,69 + 0,76 = 1,45
3,3V supply
I went for a cheap, low power part. I didn't know how much all of this would use, so i picked a 2A part. An smps, because dropping from 12V to 3,3V is kinda wastefull.
Also the manufacturer offers relevant design tools without registering first on their website, a big plus.
It powers the previouly mentioned Quiet 1,2V LDO.
And yeah, I can probably reuse this part for the controller board. For it's 3,3V and the 5V for the USB.
I also aimed to reuse components for different purposes to limit the amount of unique parts.
I went for the AOZ1050PI by Alpha&Omega semi 0,69
Just some comments:
-It's pretty obvious that for their lower end stuff, manufacturers either copy regulators from others, licence a design or codevelop it. It gets rather blatent when the only diffrence is the pinout or the internal mosfets Rds(on). Who am I to judge, they just want to minimize cost on low margin stuff.
-The part has pretty high Rds(on) mosfets in it, not necessarily a bad thing. That means the switching losses will be lower due to a reduced gate capacitance.
-It is a current mode feedback design which has it's own set of pros and cons, but for this kind of low power part I think it's allright. Remember, that 1,2V 8A regulator from before was voltage mode.
-Because I don't know how much current will be drawn, I should have taken a part with a pulse skipping or other power saving mode. But that could introduce low frequency noise, and I would like this line to be quiet... So I sacrificed light load efficiency for light load noise performance, I hope it's worth it.
http://s14.postimg.org/hoje404gh/example_design_smps.jpg
I'm going to stick to the reference design as much as possible. It saves time on a part that is not so critical.
Output voltage
To set the voltage I'll take R1 as 31,6K and R2 as 10k.
Vout = 0,8*(1+(31,6/10)) = 3,33V
Softstart
Damn, it's gonna be emberrassing if I get this wrong. I'll just give it a shot anyway.
In the datasheet the softstart is discribed as a capacitor being charged by a 5uA current source till it reaches 0,8V.
Unlike charging with a voltage source, the voltage will increase in a linear fashion when charged with a current source.
C = (I(t)*t)/V(t)
I(t) is a constant, it won't change over time.
t is the softstart time in seconds. I'll go for 1msec.
V(t) I want it to be 0,8V at 1msec.
(5uA*1msec)/0,8V = 6,25nF (I'll go with 6,8nF as that is a more common value)
Input capacitor selection
I don't need a lot, there is quite a bit of capacitence on the 12V line on the board. So I'll take two of the previously used 22uF ceramic capacitors.
At that bias voltage I'll be lucky if I have 10uF left with the two of them in parallel.
Input ripple calculation, perhaps useless in this case, but I'll do it anyway.
Vripple_pp = (Iout/(freq*Cin))*(1-(Vout/Vin))*(Vout/Vin)
(2/(500000*(10*10^-6)))*(1-(3,3/12))*(3,3/12) = 80mV ripple
RMS current calculation
Icin_rms = Iout*sqrt((1-(Vout/Vin))*(Vout/Vin))
2*sqrt((1-(3,3/12))*(3,3/12)) = 0,89A
That shouldn't cause more than 1 degree C of temperature rise.
Part chosen: 2x 1210 X7R 25V 22uF TaiyoJuden TMK325B7226MM-TR 0,85 (1,70)
Inductor selection
The datasheet recommends a 4,7uH part, so lets check it out.
IL_rip_pp = (Vout/(freq*L))*(1-(Vout/Vin))
(3,3/(500000*(4,7*10^-6)))*(1-(3,3/12)) = 1,02A
That is about 50% ripple current, a bit to high in my opinion.
Lets try with 6,8uH.
(3,3/(500000*(6,8*10^-6)))*(1-(3,3/12)) = 0,7A
About 35%, acceptable.
Peak inductor current.
ILpeak = Iout+(IL_rip_pp/2)
2+(0,7/2) = 2,35A (the inductor will have to withstand that current without saturating)
Part chosen: Bourns SRU1048-6R8Y 6,8uH 13,6mOhm 4,1A 0,75
Output capacitor selection
I'll try to reuse the same 6,3V 22uF capacitors used on the 1,2V asic power supply.
At 3,3V three of these will have in parallel about 30uF of capacitence and at worst 2mOhm of ESR.
Output ripple calculation.
Vout_rip_pp = IL_rip_pp*(ESR+(1/(8*freq*Cout)))
0,7*(0,002+(1/(8*500000*(30*10^-6)))) = 7,2mV (acceptable)
Just to be sure lets check the RMS current.
Icout_rms = IL_rip_pp/sqrt(12)
0,7/sqrt(12) = 0,2A
As expected, nothing to worry about
Parts chosen: 3x 0805 X5R 6,3V 22uF Murata GRM21BR60J226ME39L 0,38 (1,14)
Feedback loop compensation
As said before this is a (peak)current mode regulator, it measures the current through the top mosfet and looks at the output voltage. Normal current mode will measure over a very small series resistor located after the LC filter. Or a more advanced scheme which measures over the inductor.
Voltage mode only looks at the output voltage. And needs to carry a lot of high frequency signal magnitude back to the feedback node. To react fast to transients.
That means quite a portion of the switching ripple is brought back aswell, which can create bothersome dutycycle jitter or at worst, instability.
One big benefit of (peak)currentmode is that the loop compensation can be simplified, in the voltage mode regulator we had to compensate for the two poles of the output LC filter.
Now the system is reduced to only a pole and a zero. But it's not all good, the pole is output load dependant. Meaning this pole will be all over the place when output current changes.
fp1= 1/(2*pi*Cout*Rload)
Cout = 30uF
A output of 2A equates a Rload of 1,65Ohm --> fp1= 3215Hz
1A -> 3,3Ohm --> fp1= 1607Hz
0,1A -> 33Ohm --> fp1= 161Hz
That sux but we'll deal with it.
The zero lies way up high due to the low ESR.
fz1= 1/(2*pi*Cout*ESR)
fz1= 1/(2*pi*(30*10^-6)*0,002) = 2,65MHz
And that will double if the ESR turns out to be half, that frequency will double.
Using same value components as used before in this design. The 6,8nF capacitor and the 10K resistor used in the output voltage setting.
Lets see if it will result in something acceptable.
It could be my lack of skill, or the guys making the datasheet didn't know either how to properly optimise the feedback loop.
The formullas they use to get the component values keep contradicting themselves, aswell as showing how different the requirements are for different output loads.
A futile exercise, unfortunatly.
So I grabbed the manufacturers spreadsheet to calculate the loop response.
It does not contain this specific part, but the AOZ1022 has an almost identical feedback loop.
So I used that to simulate it and it worked fine with a 6,8nF cap and a 10k resistor even over a wide range of loads.
@2A load
http://s9.postimg.org/c7v2upoxb/3_3_V_Loop_response_2_A.png
@1A load
http://s18.postimg.org/ml59fvwmx/3_3_V_Loop_response_1_A.png
@0,1A load
http://s13.postimg.org/q9z88fw07/3_3_V_Loop_response_0_1_A.png
So, yeah, I'll stick with those values.
Efficiency, for the most part above 90%, only at under 0,5A would you start to see lower efficiencies. I guesstimate.
Combined cost of the major parts: 1,14+0,75+1,70+0,69 = 4,28
Ok, so we've got the 1.2V for the asics, now the rest.
I looked at the reference design to see how much current at which voltages would be required.
We need 3,3V for the IO's of the asics and the clock distribution.
And a quiet 1,2V for the PLL's of the asics.
Quiet 1,2V regulator
Lets start with the Quiet 1,2V regulator. The PLL in this chips is not speced, other than the voltage tolerance.
PLL's can be picky eaters, the normal 1,2V, which can be found in this thread, was designed with a larger ripple to improve transient response. The PLL might not like that.
I don't know how much current they'll use so I looked at the reference design where they used a 500mA part for 10 chips. Thus I chose a 600mA pat for my 12 chip board.
Just to be safe. Perhaps it won't need that at all, maybe they just used it because they could reuse it on their controller board.
Yeah, I intend on reusing it aswell on my controller board.
Part chosen: Adjustable LDO AP7165-SPG-13 from Diodes inc. 0,69
Not much to design really with these parts, just a few things to keep in mind.
Transient response, cheap and old parts can be pretty terrible at this, so I pay attention to it. The datasheet checks out, it's not that bad.
Temperature, I will power it with 3,3V so with a 2,1V drop at 600mA you will disapate 1,26W.
I use a SO-8 package with an exposed pad and a pcb with a forced air heatsink cooling. I think it will work out, as the supplier specs it at 40C/W on a 50*50mm board.
PSRR power supply rejection ratio. We are post regulating an SMPS so that is kinda important.
Suppliers almost always show a way to optimistic figure/graph. Not all by the way, more expensive parts, often intended for RF purposes are way better.
Almost all of Analog devices, AMS or expensive ones from Texas instruments and Micrel are pretty good as far as I know.
Anyway lets think about this for a bit.
Most mosfet based LDOs, like this one have, as (most)all mosfets do, a parasitic capacitance from drain to source (drain to bulk and bulk to source in series).
This will allow it to pass high frequency noise.
Also, the regulator feedback loop has a limited bandwidth, it also allows high frequency stuff to pass.
Now the external capacitors will take care of higher frequencies, thus improving PSSR, but due to their construction will start to resemble inductors at higher frequencies.
Thats why I'll put a ferrite bead in series, to get rid of higher frequencies.
For the input and output capacitors I chose a part I used before on this board. The 6,3V 22uF capacitors have about 10uF of capacity at 3,3V and at 1,2V it's 20uF
Parts chosen: 2x 0805 X5R 6,3V 22uF Murata GRM21BR60J226ME39L 0,38 (0,76)
As for the feedback I chose 31,6K for R2 and 15,8K for R1.
Vout = Vref*(1+(R1/R2))
0,8*(1+(15800/31600)) = 1,2V
Combined cost: 0,69 + 0,76 = 1,45
3,3V supply
I went for a cheap, low power part. I didn't know how much all of this would use, so i picked a 2A part. An smps, because dropping from 12V to 3,3V is kinda wastefull.
Also the manufacturer offers relevant design tools without registering first on their website, a big plus.
It powers the previouly mentioned Quiet 1,2V LDO.
And yeah, I can probably reuse this part for the controller board. For it's 3,3V and the 5V for the USB.
I also aimed to reuse components for different purposes to limit the amount of unique parts.
I went for the AOZ1050PI by Alpha&Omega semi 0,69
Just some comments:
-It's pretty obvious that for their lower end stuff, manufacturers either copy regulators from others, licence a design or codevelop it. It gets rather blatent when the only diffrence is the pinout or the internal mosfets Rds(on). Who am I to judge, they just want to minimize cost on low margin stuff.
-The part has pretty high Rds(on) mosfets in it, not necessarily a bad thing. That means the switching losses will be lower due to a reduced gate capacitance.
-It is a current mode feedback design which has it's own set of pros and cons, but for this kind of low power part I think it's allright. Remember, that 1,2V 8A regulator from before was voltage mode.
-Because I don't know how much current will be drawn, I should have taken a part with a pulse skipping or other power saving mode. But that could introduce low frequency noise, and I would like this line to be quiet... So I sacrificed light load efficiency for light load noise performance, I hope it's worth it.
http://s14.postimg.org/hoje404gh/example_design_smps.jpg
I'm going to stick to the reference design as much as possible. It saves time on a part that is not so critical.
Output voltage
To set the voltage I'll take R1 as 31,6K and R2 as 10k.
Vout = 0,8*(1+(31,6/10)) = 3,33V
Softstart
Damn, it's gonna be emberrassing if I get this wrong. I'll just give it a shot anyway.
In the datasheet the softstart is discribed as a capacitor being charged by a 5uA current source till it reaches 0,8V.
Unlike charging with a voltage source, the voltage will increase in a linear fashion when charged with a current source.
C = (I(t)*t)/V(t)
I(t) is a constant, it won't change over time.
t is the softstart time in seconds. I'll go for 1msec.
V(t) I want it to be 0,8V at 1msec.
(5uA*1msec)/0,8V = 6,25nF (I'll go with 6,8nF as that is a more common value)
Input capacitor selection
I don't need a lot, there is quite a bit of capacitence on the 12V line on the board. So I'll take two of the previously used 22uF ceramic capacitors.
At that bias voltage I'll be lucky if I have 10uF left with the two of them in parallel.
Input ripple calculation, perhaps useless in this case, but I'll do it anyway.
Vripple_pp = (Iout/(freq*Cin))*(1-(Vout/Vin))*(Vout/Vin)
(2/(500000*(10*10^-6)))*(1-(3,3/12))*(3,3/12) = 80mV ripple
RMS current calculation
Icin_rms = Iout*sqrt((1-(Vout/Vin))*(Vout/Vin))
2*sqrt((1-(3,3/12))*(3,3/12)) = 0,89A
That shouldn't cause more than 1 degree C of temperature rise.
Part chosen: 2x 1210 X7R 25V 22uF TaiyoJuden TMK325B7226MM-TR 0,85 (1,70)
Inductor selection
The datasheet recommends a 4,7uH part, so lets check it out.
IL_rip_pp = (Vout/(freq*L))*(1-(Vout/Vin))
(3,3/(500000*(4,7*10^-6)))*(1-(3,3/12)) = 1,02A
That is about 50% ripple current, a bit to high in my opinion.
Lets try with 6,8uH.
(3,3/(500000*(6,8*10^-6)))*(1-(3,3/12)) = 0,7A
About 35%, acceptable.
Peak inductor current.
ILpeak = Iout+(IL_rip_pp/2)
2+(0,7/2) = 2,35A (the inductor will have to withstand that current without saturating)
Part chosen: Bourns SRU1048-6R8Y 6,8uH 13,6mOhm 4,1A 0,75
Output capacitor selection
I'll try to reuse the same 6,3V 22uF capacitors used on the 1,2V asic power supply.
At 3,3V three of these will have in parallel about 30uF of capacitence and at worst 2mOhm of ESR.
Output ripple calculation.
Vout_rip_pp = IL_rip_pp*(ESR+(1/(8*freq*Cout)))
0,7*(0,002+(1/(8*500000*(30*10^-6)))) = 7,2mV (acceptable)
Just to be sure lets check the RMS current.
Icout_rms = IL_rip_pp/sqrt(12)
0,7/sqrt(12) = 0,2A
As expected, nothing to worry about
Parts chosen: 3x 0805 X5R 6,3V 22uF Murata GRM21BR60J226ME39L 0,38 (1,14)
Feedback loop compensation
As said before this is a (peak)current mode regulator, it measures the current through the top mosfet and looks at the output voltage. Normal current mode will measure over a very small series resistor located after the LC filter. Or a more advanced scheme which measures over the inductor.
Voltage mode only looks at the output voltage. And needs to carry a lot of high frequency signal magnitude back to the feedback node. To react fast to transients.
That means quite a portion of the switching ripple is brought back aswell, which can create bothersome dutycycle jitter or at worst, instability.
One big benefit of (peak)currentmode is that the loop compensation can be simplified, in the voltage mode regulator we had to compensate for the two poles of the output LC filter.
Now the system is reduced to only a pole and a zero. But it's not all good, the pole is output load dependant. Meaning this pole will be all over the place when output current changes.
fp1= 1/(2*pi*Cout*Rload)
Cout = 30uF
A output of 2A equates a Rload of 1,65Ohm --> fp1= 3215Hz
1A -> 3,3Ohm --> fp1= 1607Hz
0,1A -> 33Ohm --> fp1= 161Hz
That sux but we'll deal with it.
The zero lies way up high due to the low ESR.
fz1= 1/(2*pi*Cout*ESR)
fz1= 1/(2*pi*(30*10^-6)*0,002) = 2,65MHz
And that will double if the ESR turns out to be half, that frequency will double.
Using same value components as used before in this design. The 6,8nF capacitor and the 10K resistor used in the output voltage setting.
Lets see if it will result in something acceptable.
It could be my lack of skill, or the guys making the datasheet didn't know either how to properly optimise the feedback loop.
The formullas they use to get the component values keep contradicting themselves, aswell as showing how different the requirements are for different output loads.
A futile exercise, unfortunatly.
So I grabbed the manufacturers spreadsheet to calculate the loop response.
It does not contain this specific part, but the AOZ1022 has an almost identical feedback loop.
So I used that to simulate it and it worked fine with a 6,8nF cap and a 10k resistor even over a wide range of loads.
@2A load
http://s9.postimg.org/c7v2upoxb/3_3_V_Loop_response_2_A.png
@1A load
http://s18.postimg.org/ml59fvwmx/3_3_V_Loop_response_1_A.png
@0,1A load
http://s13.postimg.org/q9z88fw07/3_3_V_Loop_response_0_1_A.png
So, yeah, I'll stick with those values.
Efficiency, for the most part above 90%, only at under 0,5A would you start to see lower efficiencies. I guesstimate.
Combined cost of the major parts: 1,14+0,75+1,70+0,69 = 4,28