In article <8436b95b-c9fa-4b36-911a-4dd625c904b2@googlegroups.com>, whit3rd@gmail.com says...> > On Friday, May 30, 2014 2:21:01 PM UTC-7, Phil Hobbs wrote: > > On 5/30/2014 12:50 PM, whit3rd wrote: > > > > > On Friday, May 30, 2014 6:59:20 AM UTC-7, Phil Hobbs wrote: > > > > >> On 05/29/2014 01:44 PM, whit3rd wrote: > > > > > > > > > >>> There's one voltage source that has very low ripple, good filtering, and > > >>> the same noise output as a low-value resistor. That's a thermopile. > > > >> Ultrareliable isn't how I'd describe it. Also its tempco would be on > > >> the order of 3000 ppm/K. > > > > Not so. That's the tempco of the series resistance, not of the voltage > > > output! > > > > A thermopile is a bunch of thermocouples in series. Their output > > voltage is approximately proportional to the temperature drop across the > > junction pairs. That means that V = K(T_hot -T_cold). > > > Even if T_cold = 0 kelvin, that scheme can't do better than PTAT > > That is a tremendously improbable case! Thermopiles operated on > HEAT SOURCES stabilize to a fixed gradient (established by heat conduction) > and one expects (T_hot - T_cold) to be a constant. It is NEVER > proportional to absolute temperature in normal circumstances. > > It's true, by the third law of thermodynamics, that thermocouple coefficients > (and other things) do vanish at absolute zero, but at room temperature a thermocouple > has no capability to give an absolute temperature reading, and that's > because the Seebeck equation contains no absolute temperature sensitivity.Which is why you need a cold junction reference to measure against. Jamie
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Started by ●May 7, 2014
Reply by ●May 31, 20142014-05-31
Reply by ●May 31, 20142014-05-31
On Saturday, May 31, 2014 8:32:07 AM UTC-7, Phil Hobbs wrote:> On 5/31/2014 2:49 AM, whit3rd wrote:> >>>>> There's one voltage source that has very low ripple, good filtering, and > >>>>> the same noise output as a low-value resistor. That's a thermopile.> >>>> ... its tempco would be on > >>>> the order of 3000 ppm/K.> >>> Not so. That's the tempco of the series resistance, not of the voltage > >>> output!> >> A thermopile is a bunch of thermocouples in series. ... That means that V = K(T_hot -T_cold).> >> Even if T_cold = 0 kelvin, that scheme can't do better than PTAT> > That is a tremendously improbable case! Thermopiles operated on > > HEAT SOURCES stabilize to a fixed gradient (established by heat conduction) > > and one expects (T_hot - T_cold) to be a constant. It is NEVER> You miss my point. The output voltage is proportional to the > temperature drop across the junction, and so the tempco goes as > 1/(Thot-Tcold), which is huge--generally _much bigger_ than PTAT. > (Tcold being zero kelvin was a fictitious best case for your argument, > with zero physicsy stuff in it--see the linear approximation above.)Actually, you need the physicsy stuff to find a regime where PTAT is the case. Take 100x100mm of flexible circuit board, put quarter-millimeter wires with quarter-millimeter spacings all over it. That's 200 thermocouples. Roll it into cigarette-size, glue a base on one end, and a heater resistor on the other. Insulate everything but the base. Send enough current through the resistor to raise the hot end temperature 100 degrees above the ambient-temperature end (i.e. set the power at the thermal conductance of the wires times 100 degrees). Thot - Tambient = heat_input / thermal_conductance = 100 Now look at dependence on ambient temperature Ta dV/dTa = d/dTa (K) * (Thot - Ta) + K * d/dTa ( Thot - Ta) but we know that the sum Thot - Ta is set by the heat input, that sum is constant! So the temperature dependence is PTAT whenever the thermocouple has a Seebeck coefficient that is proportional to temperature. That happens at/near absolute zero (third law of thermodynamics). The principle is not at all impractical; accurate RF current meters have worked this way for decades, inside the box where one rarely sees it happening.
Reply by ●May 31, 20142014-05-31
On 5/31/2014 3:33 PM, whit3rd wrote:> On Saturday, May 31, 2014 8:32:07 AM UTC-7, Phil Hobbs wrote: >> On 5/31/2014 2:49 AM, whit3rd wrote: > >>>>>>> There's one voltage source that has very low ripple, good filtering, and >>>>>>> the same noise output as a low-value resistor. That's a thermopile. > >>>>>> ... its tempco would be on >>>>>> the order of 3000 ppm/K. > >>>>> Not so. That's the tempco of the series resistance, not of the voltage >>>>> output! > >>>> A thermopile is a bunch of thermocouples in series. ... That means that V = K(T_hot -T_cold). > >>>> Even if T_cold = 0 kelvin, that scheme can't do better than PTAT > >>> That is a tremendously improbable case! Thermopiles operated on >>> HEAT SOURCES stabilize to a fixed gradient (established by heat conduction) >>> and one expects (T_hot - T_cold) to be a constant. It is NEVER > >> You miss my point. The output voltage is proportional to the >> temperature drop across the junction, and so the tempco goes as >> 1/(Thot-Tcold), which is huge--generally _much bigger_ than PTAT. >> (Tcold being zero kelvin was a fictitious best case for your argument, >> with zero physicsy stuff in it--see the linear approximation above.) > > Actually, you need the physicsy stuff to find a regime where PTAT is > the case. Take 100x100mm of flexible circuit board, put > quarter-millimeter wires with quarter-millimeter spacings all over it. > That's 200 thermocouples. Roll it into cigarette-size, glue a base > on one end, and a heater resistor on the other. > Insulate everything but the base. > > Send enough current through the resistor to raise the hot end temperature > 100 degrees above the ambient-temperature end (i.e. set the > power at the thermal conductance of the wires times 100 degrees). > > Thot - Tambient = heat_input / thermal_conductance = 100 > > Now look at dependence on ambient temperature Ta > > dV/dTa = d/dTa (K) * (Thot - Ta) + K * d/dTa ( Thot - Ta) > > but we know that the sum Thot - Ta is set by the heat input, that sum is constant! > > So the temperature dependence is PTAT whenever the thermocouple has > a Seebeck coefficient that is proportional to temperature. That happens > at/near absolute zero (third law of thermodynamics). > > The principle is not at all impractical; accurate RF current meters > have worked this way for decades, inside the box where one rarely > sees it happening. >You're just blowing smoke. A very garden-variety voltage reference can do 30 ppm/K, and good ones are below 5 ppm/K (e.g. the LT1021). Good luck getting anywhere near that with a thermopile, especially one in SOT-23. ;) Cheers Phil Hobbs -- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
Reply by ●May 31, 20142014-05-31
On Saturday, May 31, 2014 12:44:23 PM UTC-7, Phil Hobbs wrote:> ... A very garden-variety voltage reference can > do 30 ppm/K, and good ones are below 5 ppm/K (e.g. the LT1021).Ah, if only life were that simple. To get best errors from a biased photodiode, you need the bias voltage NOT to be temperature independent (because the diode capacitance, and stored charge, isn't). There's a square root of T in the formula...
Reply by ●May 31, 20142014-05-31
On 5/31/2014 5:30 PM, whit3rd wrote:> On Saturday, May 31, 2014 12:44:23 PM UTC-7, Phil Hobbs wrote: > >> ... A very garden-variety voltage reference can >> do 30 ppm/K, and good ones are below 5 ppm/K (e.g. the LT1021). > > Ah, if only life were that simple. To get best errors from a biased > photodiode, you need the bias voltage NOT to be temperature > independent (because the diode capacitance, and stored charge, isn't). > There's a square root of T in the formula... >What formula are you talking about, exactly? The PD bias is there to reduce its capacitance (by as much as 7 times), and therefore to reduce the high frequency noise of the TIA, which is i_N = e_NTIA * 2 pi f C_diode . As long as the PD bias doesn't have significant noise in the bandwidth of interest, you don't care about small variations. At one point, we were talking about voltage regulators, and the noise rejection thereof. Cheers Phil Hobbs -- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
Reply by ●May 31, 20142014-05-31
On Sat, 31 May 2014 12:33:52 -0700 (PDT), whit3rd <whit3rd@gmail.com> wrote:>On Saturday, May 31, 2014 8:32:07 AM UTC-7, Phil Hobbs wrote: >> On 5/31/2014 2:49 AM, whit3rd wrote: > >> >>>>> There's one voltage source that has very low ripple, good filtering, and >> >>>>> the same noise output as a low-value resistor. That's a thermopile. > >> >>>> ... its tempco would be on >> >>>> the order of 3000 ppm/K. > >> >>> Not so. That's the tempco of the series resistance, not of the voltage >> >>> output! > >> >> A thermopile is a bunch of thermocouples in series. ... That means that V = K(T_hot -T_cold). > >> >> Even if T_cold = 0 kelvin, that scheme can't do better than PTAT > >> > That is a tremendously improbable case! Thermopiles operated on >> > HEAT SOURCES stabilize to a fixed gradient (established by heat conduction) >> > and one expects (T_hot - T_cold) to be a constant. It is NEVER > >> You miss my point. The output voltage is proportional to the >> temperature drop across the junction, and so the tempco goes as >> 1/(Thot-Tcold), which is huge--generally _much bigger_ than PTAT. >> (Tcold being zero kelvin was a fictitious best case for your argument, >> with zero physicsy stuff in it--see the linear approximation above.) > >Actually, you need the physicsy stuff to find a regime where PTAT is >the case. Take 100x100mm of flexible circuit board, put >quarter-millimeter wires with quarter-millimeter spacings all over it. >That's 200 thermocouples. Roll it into cigarette-size, glue a base >on one end, and a heater resistor on the other. >Insulate everything but the base. > >Send enough current through the resistor to raise the hot end temperature >100 degrees above the ambient-temperature end (i.e. set the >power at the thermal conductance of the wires times 100 degrees). > >Thot - Tambient = heat_input / thermal_conductance = 100 > >Now look at dependence on ambient temperature Ta > >dV/dTa = d/dTa (K) * (Thot - Ta) + K * d/dTa ( Thot - Ta) > >but we know that the sum Thot - Ta is set by the heat input, that sum is constant! > >So the temperature dependence is PTAT whenever the thermocouple has >a Seebeck coefficient that is proportional to temperature. That happens >at/near absolute zero (third law of thermodynamics). > >The principle is not at all impractical; accurate RF current meters >have worked this way for decades, inside the box where one rarely >sees it happening.My wife gave me this as a birthday present. Can't imagine where she got it. https://dl.dropboxusercontent.com/u/53724080/Gear/HF_Ammeter.JPG -- John Larkin Highland Technology Inc www.highlandtechnology.com jlarkin at highlandtechnology dot com Precision electronic instrumentation
Reply by ●May 31, 20142014-05-31
On Saturday, May 31, 2014 2:41:02 PM UTC-7, Phil Hobbs wrote:> On 5/31/2014 5:30 PM, whit3rd wrote: > > > On Saturday, May 31, 2014 12:44:23 PM UTC-7, Phil Hobbs wrote:> > >> ... A very garden-variety voltage reference can > >> do 30 ppm/K, and good ones are below 5 ppm/K (e.g. the LT1021).> > Ah, if only life were that simple. To get best errors from a biased > > photodiode, you need the bias voltage NOT to be temperature > > independent (because the diode capacitance, and stored charge, isn't).> What formula are you talking about, exactly?The discussion started with stabilizing a bias voltage for a photodiode; the current from a photodiode is I = Iphoto + Isat + d/dt(C * Vbias) one usually ignores (or calibrates out) the constant Isat That last term isn't just one error, it's the sum of two terms dC/dt * Vbias + C * d/dt(Vbias) Thus, in a temperature-varying-with-time case, change 't' time for 'T', temperature, and optimum photocurrent measurement requires 0 = dC/dT * Vbias + C * d/dT(Vbias) d/dT( Vbias) = -( dC/dT * Vbias) / C It's not worth worrying about unless low-frequency response is important, or unless the amplifier could saturate, because temperature is a slow-changing variable. Sze (_Physics of Semiconductor Devices_) gives a formula for abrupt-junction diode capacitance that seems appropriate C = [constant] * sqrt(Vbi - V - 2kT/q) where Vbi is the builtin voltage, V is the bias...> The PD bias is there to reduce its capacitance (by as much as 7 times), > and therefore to reduce the high frequency noise of the TIA, which is > > i_N = e_NTIA * 2 pi f C_diode . > > As long as the PD bias doesn't have significant noise in the bandwidth > of interest, you don't care about small variations.
Reply by ●May 31, 20142014-05-31
On 5/31/2014 7:21 PM, whit3rd wrote:> On Saturday, May 31, 2014 2:41:02 PM UTC-7, Phil Hobbs wrote: >> On 5/31/2014 5:30 PM, whit3rd wrote: >> >>> On Saturday, May 31, 2014 12:44:23 PM UTC-7, Phil Hobbs wrote: > >> >>>> ... A very garden-variety voltage reference can >>>> do 30 ppm/K, and good ones are below 5 ppm/K (e.g. the LT1021). > >>> Ah, if only life were that simple. To get best errors from a biased >>> photodiode, you need the bias voltage NOT to be temperature >>> independent (because the diode capacitance, and stored charge, isn't). > >> What formula are you talking about, exactly? > > The discussion started with stabilizing a bias voltage for > a photodiode; the current from a photodiode is > > I = Iphoto + Isat + d/dt(C * Vbias) > > one usually ignores (or calibrates out) the constant Isat > That last term isn't just one error, it's the sum of two terms > > dC/dt * Vbias + C * d/dt(Vbias)There's no error term in the external circuit from dC/dt. It's caused by the spreading of the shielding regions on the edges of the depletion region, and doesn't have to be charged up from zero volts.> > Thus, in a temperature-varying-with-time case, change 't' time for 'T', > temperature, and optimum photocurrent measurement requires > > 0 = dC/dT * Vbias + C * d/dT(Vbias) > > d/dT( Vbias) = -( dC/dT * Vbias) / CHave you ever actually designed a photoreceiver? Doesn't sound like it. You're just blowing smoke.> > It's not worth worrying about unless low-frequency response is important, > or unless the amplifier could saturate, because temperature is a slow-changing > variable. > > Sze (_Physics of Semiconductor Devices_) gives a formula for abrupt-junction > diode capacitance that seems appropriate > > C = [constant] * sqrt(Vbi - V - 2kT/q)The temperature of the photodiode changes on a timescale of minutes to hours. If you have a capacitance of 10 pF at 25 C, it might change by a picofarad or so between there and 100 C. Typical parameters might be a 10V reverse bias, and a very generous estimate of the thermal TC for the assembly might be 30 seconds. For an extreme example, say you dunked the whole thing into boiling oil at 200 C, i.e. a 175K transient. The current error resulting from your mechanism would be delta I = (dT/dt)*dC/dT*V_bias = (175 K)/(30s) * (1 pF/100 K) * 10 V = 0.6 pA. Not very impressive numbers, even for an extreme case, and of course it would almost all be well below 1 Hz.> where Vbi is the builtin voltage, V is the bias... > >> The PD bias is there to reduce its capacitance (by as much as 7 times), >> and therefore to reduce the high frequency noise of the TIA, which is >> >> i_N = e_NTIA * 2 pi f C_diode . >> >> As long as the PD bias doesn't have significant noise in the bandwidth >> of interest, you don't care about small variations.How many sensitive photoreceivers have you actually designed? Cheers Phil Hobbs -- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot nethttp://electrooptical.net
Reply by ●May 31, 20142014-05-31
In article <qdkko95v4baiom25edjh6tr3497vcuklgg@4ax.com>, jjlarkin@highNOTlandTHIStechnologyPART.com says...> The principle is not at all impractical; accurate RF current meters > >have worked this way for decades, inside the box where one rarely > >sees it happening. > > My wife gave me this as a birthday present. Can't imagine where she got it. > > https://dl.dropboxusercontent.com/u/53724080/Gear/HF_Ammeter.JPG > >I haven't seen one of those around in years. Jamie
Reply by ●May 31, 20142014-05-31
On Saturday, May 31, 2014 4:41:21 PM UTC-7, Phil Hobbs wrote:> On 5/31/2014 7:21 PM, whit3rd wrote:[about error current in a photodiode]> > dC/dt * Vbias + C * d/dt(Vbias)> There's no error term in the external circuit from dC/dt.Of course there is! The charge on the capacitor doesn't move through the depletion region, it goes in the external wiring.> The temperature of the photodiode changes on a timescale of minutes to > hours. If you have a capacitance of 10 pF at 25 C... > The current error resulting from your mechanism would be> delta I = (dT/dt)*dC/dT*V_bias> = (175 K)/(30s) * (1 pF/100 K) * 10 V = 0.6 pA.> How many sensitive photoreceivers have you actually designed?One, in particular, was for fluorescence spectroscopy on microscopic samples. It had a ~1 cm photodiode to cover the monochromator's exit slit, and took a minute or so to scan the spectrum with a stepping motor moving the grating. The output was picoamps. So was leakage. Capacitance was high, seconds-to-minutes response was what worked best with a chart recorder. It mismatches by a few orders of magnitude the numbers you've plugged in.
