| Electronics > Beginners |
| uSupply Problems |
| << < (4/4) |
| JS:
--- Quote from: VEGETA on May 07, 2018, 07:57:13 pm ---You would need a negative rail to get it to 0v, which is doable. I saw TL431 or so used to achieve this... make it -5v and it will be enough. I don't understand why digital pots are a problem of stability. They are simply resistors. Current limit will have a drop voltage of around 0.43v between pin 5 and 2, so it won't be a big problem. However, the bench psu circuit shows more details and more resistances with an op-amp... but it won't get worse. getting 10k for voltage control and 100k for current will guarantee 30v 2A, but I don't fully understand the current limit control loop. I would like a nice explanation for it (the bench psu config.) --- End quote --- Current limit will have a small voltage across it but will be floating high, dpots can't do that, you'd need an external power source and isolation, bit messy. I'm finishing a PCB design for the µCurrent and I used just a diode drop to get 0V in case my results are different than David's but it's just a bridge away of taking it out, I also added differential remote voltage sensing. I forgot to add OCP and OVP but I wanted to, pity the PCB is nice and ready, self etch friendly single sided, for those who want it, I'd probably be posting it at any time now. The board I designed doesn't contain the µC but a pin-header to connect to a control board. PWM filters are included in the PCB and two different voltage sensing outputs, one taken directly from the output and other with a differential amp for remote sensing. I'd like to build it, maybe tomorrow, and fully test it on the week when my new scope arrives, then I start a topic for troubleshooting or sharing (depends on the results) JS |
| mikerj:
--- Quote from: daubmaso on April 30, 2018, 01:26:00 am ---By 4x10k matched he means that the 10k resistors should be hand picked and measured to make sure each value is identical across all resistors. This is so that the differential amplifier for the current shunt resistor has as close to a 1:1 gain as possible to ensure an accurate current limit. --- End quote --- FWIW the resistor matching is to ensure the common mode rejection ratio (CMRR) is as high as possible. With poorly matched resistors, changes in the input supply voltage would cause the output of the op-amp to vary, which would be seen as changes in load current. |
| JS:
--- Quote from: mikerj on June 03, 2018, 01:23:24 pm --- --- Quote from: daubmaso on April 30, 2018, 01:26:00 am ---By 4x10k matched he means that the 10k resistors should be hand picked and measured to make sure each value is identical across all resistors. This is so that the differential amplifier for the current shunt resistor has as close to a 1:1 gain as possible to ensure an accurate current limit. --- End quote --- FWIW the resistor matching is to ensure the common mode rejection ratio (CMRR) is as high as possible. With poorly matched resistors, changes in the input supply voltage would cause the output of the op-amp to vary, which would be seen as changes in load current. --- End quote --- Luckily the source impedance difference is low (the 1Ω resistor, at most, and pretty much fixed) which makes not to need the two sides to be matched, just ratio. You could hand trim a resistor with a file to get as high as you can measure if you want so. Effects in CMRR are quite hard in this case (with the original pre regulation) as it can change 14V in the original design and 2mV per bit, the error introduce by those 14V will make about 80dB needed to be less than 1LSB, or 2mA. I'm not counting the INA current sensing as it has 120dB typ CMRR. The other caveat, aiming for precision and accuracy is that about 2% of the current to the load is going to the Vcontrol pin of the LT3080 which is not being sensed, so there's a gain error in that as well. JS |
| VEGETA:
--- Quote from: JS on June 03, 2018, 02:16:14 am --- --- Quote from: VEGETA on May 07, 2018, 07:57:13 pm ---You would need a negative rail to get it to 0v, which is doable. I saw TL431 or so used to achieve this... make it -5v and it will be enough. I don't understand why digital pots are a problem of stability. They are simply resistors. Current limit will have a drop voltage of around 0.43v between pin 5 and 2, so it won't be a big problem. However, the bench psu circuit shows more details and more resistances with an op-amp... but it won't get worse. getting 10k for voltage control and 100k for current will guarantee 30v 2A, but I don't fully understand the current limit control loop. I would like a nice explanation for it (the bench psu config.) --- End quote --- Current limit will have a small voltage across it but will be floating high, dpots can't do that, you'd need an external power source and isolation, bit messy. I'm finishing a PCB design for the µCurrent and I used just a diode drop to get 0V in case my results are different than David's but it's just a bridge away of taking it out, I also added differential remote voltage sensing. I forgot to add OCP and OVP but I wanted to, pity the PCB is nice and ready, self etch friendly single sided, for those who want it, I'd probably be posting it at any time now. The board I designed doesn't contain the µC but a pin-header to connect to a control board. PWM filters are included in the PCB and two different voltage sensing outputs, one taken directly from the output and other with a differential amp for remote sensing. I'd like to build it, maybe tomorrow, and fully test it on the week when my new scope arrives, then I start a topic for troubleshooting or sharing (depends on the results) JS --- End quote --- EPOT is a resistor network eventually so it replaces the resistor... what is the problem? the IC's power source is independent of the resistor array in it so the resistors are somehow isolated like having a separate resistor. Dave used EPOTs in his uSupply so I guess they are suitable somehow, you just have to find the way. Also, if 100K epot is 256 taps that is equal to 8-bit DAC or similar... So if we put 10k epot + 10k pot with each 256 taps in series (instead of the 100k in l200 circuit, but how? how to modify the values) would that be like 16 bit resolution? About the negative rail... Put ICL7660 IC which gives you -5v from your +5v or if you have something like +12v it will output -12v (or maybe limited to -10v) so put TL431 after it (with resistor feedback network) to get your desired negative rail, this will be stable enough. |
| JS:
--- Quote from: VEGETA on June 03, 2018, 07:09:56 pm ---EPOT is a resistor network eventually so it replaces the resistor... what is the problem? the IC's power source is independent of the resistor array in it so the resistors are somehow isolated like having a separate resistor. Dave used EPOTs in his uSupply so I guess they are suitable somehow, you just have to find the way. Also, if 100K epot is 256 taps that is equal to 8-bit DAC or similar... So if we put 10k epot + 10k pot with each 256 taps in series (instead of the 100k in l200 circuit, but how? how to modify the values) would that be like 16 bit resolution? About the negative rail... Put ICL7660 IC which gives you -5v from your +5v or if you have something like +12v it will output -12v (or maybe limited to -10v) so put TL431 after it (with resistor feedback network) to get your desired negative rail, this will be stable enough. --- End quote --- No, the potential at resistor terminals must be within digital rails, take a look at the datasheet. It won't behave as 16 bits. Many things there, first you load the first one with the second one so the first one won't behave linearly. This might be desirable as you get more definition in the lower range. Using a 10k pot first ans a 100k pot later will diminish this effect. If you just need more definition ibln the lower range (or upper range) and you are fine with 256 steps you can load the pot with a passive resistor and have a non linear response, this can be tweaked to a steep or shallow curve depending on the ratio of the pot and resistor, widely used, at least in audio were non linear tapers are the usual prefered and not alwaya available. Still, when both pots are on the top, a step in anyone of them would attenuate the same ammount. You could use 3 pots, the first two in feeding the extremes of the last one and track them a step or two appart, behaving like a KVD. Even then, you do have a ton of steps to output now, maybe more than 16 bits, but linearity would be horrible as the first two are ment to be 8 bits anyway and build to a cost with that in mind, even if they error is a fraction of a LSB you can only get that much bettrr out of them for 3 times the cost. JS Enviado desde mi LG-M250 mediante Tapatalk |
| Navigation |
| Message Index |
| Previous page |