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Hi everyone, for quite some time I've been interested in building my own ltz1000 reference, but no matter how long i stare at the circuit and how many simulations i do in LTSpice, there are a few things i still don't understand.
1: how does it start? I know it's silly but i really can't understand how the opamp A2 starts sourcing current to bias the zener and the compensating transistor. It seems that once we power up the opamp, the two opamp inputs are already at an equal potential, so why would it start to source current? I thought it may be due to the input offset, but the offset can be in both ways, so startup is not guaranteed.
2: what is the purpose of the diode at the output of the opamp? I did simulations with it removed and it seems to not make a difference. Why is it there?
I understand the presence of the diode at the heater negative, which is to raise it above ground to avoid forward biasing the substrate diode between heater negative and zener anode.
3:How does the 400k resistor for TC compensation works exactly? My thought is that is slightly changes the current through the zener. But why would it compensate the TC? Does the zener varies the TC with current?
And why we don't need it for the LTZ1000A, why the higher thermal resistance means we don't neet to put that resistor?
I swear i've read everything for months, watched many videos (thanks Marco Reps and Xdevs), simulated everything, but there is something i'm missing here, and i'm asking your help to understand what it might be.
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Hello,
1. what prevents you to use a pull up resistor like in "real life" applications (see 3458A reference)?
2. probably to prevent forward biasing the substrate diodes (substrate is pin 4 of the reference)
3. The influence is from the heater voltage which is the amplified temperature. I use it also with the LTZ1000A to trim the T.C. further down.
with best regards
Andreas
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For the startup the circuit may have an added resistor to provide some extra current to the reference voltage.
The other point that help with start up is, that the LT1013 is not rail to rail output. This especially helps if the diode at the OP-amps output is not there.
For the purpose of the diode at the A2 output - it is a bit strange and circuits work without it. It could be a "left-over" from an emitter follower.
The residual TC compensation uses the heater signal to add a tiny bit to the reference. The heater signal kind of senses the circuit temperature and does a little correction.
It is not so clear how the residual TC comes in. One part can be the thermal gradient cause by not perfectly symmetric heating or the thermal EMF from the heat flow through the pins. In both cases it makes sense to use the heater power for correction, not just an added sensor for the external temperature. The compensation is by no means perfect, as it is nonlinear (more like square root of the heater power), while the residual termal drift is more linear. For most purposes the compensation is still good enough - maybe unless used with a very large temperature swing, which is unusual for highest precision. -
Hello,
1. what prevents you to use a pull up resistor like in "real life" applications (see 3458A reference)?
2. probably to prevent forward biasing the substrate diodes (substrate is pin 4 of the reference)
3. The influence is from the heater voltage which is the amplified temperature. I use it also with the LTZ1000A to trim the T.C. further down.
with best regards
Andreas
Well, the pullup resistor is not necessary to ensure startup, and many circuits don't use it. With the pullup it's easy to explain how the startup works, but i would like to understand how does it work without any pull up -
For the startup the circuit may have an added resistor to provide some extra current to the reference voltage.
The other point that help with start up is, that the LT1013 is not rail to rail output. This especially helps if the diode at the OP-amps output is not there.
For the purpose of the diode at the A2 output - it is a bit strange and circuits work without it. It could be a "left-over" from an emitter follower.
The residual TC compensation uses the heater signal to add a tiny bit to the reference. The heater signal kind of senses the circuit temperature and does a little correction.
It is not so clear how the residual TC comes in. One part can be the thermal gradient cause by not perfectly symmetric heating or the thermal EMF from the heat flow through the pins. In both cases it makes sense to use the heater power for correction, not just an added sensor for the external temperature. The compensation is by no means perfect, as it is nonlinear (more like square root of the heater power), while the residual termal drift is more linear. For most purposes the compensation is still good enough - maybe unless used with a very large temperature swing, which is unusual for highest precision.
The added pull up resistor can ensure startup (like in the lm399 bootstrap circuit), but it is not necessary, and i am really struggling to understand why. With the lm399 it's not necessary because, as you said, the opamp has it's low output swing limited by the negative rail, so the output is always positive and sourcing a small current. This would explain startup of the LTZ1000 only if there was no diode, but the diode makes the little positive output voltage of the opamp practically useless for this purpose. -
What you are looking for is a dirt effect. The LT1013 datasheet shows pnp input transistors. A2 positive input base current (12 .. 30 nA) may be enough to init the reference circuit. When using a different type of OpAmp one may need some explicit startup circuit, possibly using some kind of zener/diode gate as seen in the HP 3456A reference. The diode in the LTZ1000 has a similar function. As Andreas wrote, it becomes necessary when the negative supply of A2 is more negative than Gnd.
For the temperature compensation: Heater voltage is about proportional to the square root of the difference between LTZ1000 chip temperature and board temperature. As the first one is near constant, the voltage measures board temperature (in a nonlinear fashion). So it can be used as a feed forward control to simulate infinite oven gain - which means zero residual TC in the end. LTZ1000A oven gain is already higher due to the improved thermal insulation of its reference chip.
Regards, Dieter -
The pullup resistor is necessary if the servo op amp's output can reach ground because that will also be a stable operating point where its inverting and non-inverting terminals will be at the same voltage. In the ADR1000 datasheet they actually mention that the op amp in their eval board (the ADA4084) can't get its output all the way to ground, so it will start up. I believe the same is true of the LT1013, so it won't actually be able to maintain that stable operating point at 0 V. As Andreas said, the diode is there to prevent forward biasing the Zener. Forward biasing tends to shift the breakdown voltage of Zener diodes, so if it were allowed to forward bias that would be detrimental to stability.
A common complaint I have seen about this circuit is that the residual temperature coefficient (i.e., with the oven off) is not insignificant, which is to say that the zero tempco point (as the temperature coefficient is a parabola with negative curvature) is too high to be useful. I am sure there are people who would know better than me what that residual tempco is around ambient temperature, but I think it's in the range of +20 to +50 ppm/K. Note that when the heater is powered, the chip will only move a small fraction of the ambient temperature change, so the tempco will appear linear as others have mentioned. That the residual tempco is a parabola is most useful when it comes to selecting a heater temperature - ideally it will be near zero around the heater setpoint. I don't know if the tempco of the zener itself is parabolic or if the curvature comes mainly from the compensating BJT, which will have a tempco curvature of about -1 uV/K^2 (https://web.mit.edu/klund/www/Dphysics.pdf). Another way of compensating this is to add a resistor at the cathode of the Zener, which was done in the Wavetek 7000, which operates at a lower temperature set point than the datasheet circuit. The collector resistors also influence tempco as increasing the collector current of a BJT makes the Vbe tempco less negative. The ADR1000 residual temperature coefficient is closer to zero than that of the LTZ1000, so if you pop one of those in it will probably push the temperature coefficient negative.
One of the reasons that people don't really like that the residual tempco is high is that if there is a residual tempco, the temperature setting divider R4/R5 needs to be more stable with respect to temperature. There aren't any really good 13k/1k or 12k/1k dividers in stock, if you want a BMF one, for instance, you are stuck waiting whatever the lead time is and possibly dealing with MOQs. Discrete BMF resistors just aren't as good, even though the datasheet typical specs suggest they should be. So a clever strategy of compensating the residual tempco loosens the requirements around this divider. An advantage of the cathode resistor strategy here is that you can characterize the tempco over a range and select a heater setpoint where it is minimal, but the resistor is itself influential, which is a tradeoff one must weigh in their own design. Others have incorporated a source follower JFET to provide the zener current, which has the advantage of reducing the current draw (and heating) of the servo amp. The CERN DVM circuit does this, and I have used it in all the references I have made.
To design your own doesn't require knowing the device physics - the things you are looking out for are much more mundane. You don't want heater current flowing in the same path as your sense- voltage, thermocouples with the kovar leads are important, so efforts must be taken to keep them at the same temperature (as mentioned in the datasheet), and be aware of the thermal gradients caused by the reference itself and the heater transistor to avoid unwanted effects from parasitic thermocouples. -
The amplifier A2 comes with positive or negative input offset voltage.
So there is already a 50 % chance for safe startup.
In the other case the diode at A2 output prevents it from shorting out the bias currents of the pnp input stages of the two amplifiers. Leakage current of the diode is less than bias current of the pnp-inputs. In this case startup will also depend on the reference sense line load.
Regards, Dieter -
1: how does it start? I know it's silly but i really can't understand how the opamp A2 starts sourcing current to bias the zener and the compensating transistor. It seems that once we power up the opamp, the two opamp inputs are already at an equal potential, so why would it start to source current? I thought it may be due to the input offset, but the offset can be in both ways, so startup is not guaranteed.
https://www.eevblog.com/forum/projects/does-anyone-know-what-this-diode-is-for/msg161194/#msg161194
2: what is the purpose of the diode at the output of the opamp? I did simulations with it removed and it seems to not make a difference. Why is it there?
I understand the presence of the diode at the heater negative, which is to raise it above ground to avoid forward biasing the substrate diode between heater negative and zener anode.