Hello,

But consider: a resistance change on the 200R resistor of page 6
will directly affect output voltage. 200R * 5mA = 1V.
A 100 ppm change of the 200R Resistor has a 1V/8.2V
= 12 ppm change on output voltage.
All other resistors have below 1ppm change with 100ppm resistance change.

So the 200 Ohms resistor will be the most
critical for tempco and ageing in the whole system.

So in my opinion its only worth to consider if you
really need to save the 20-30 mA for the heater.

With best regards

Andreas

Hello,

Please, look at the actual measurement results, maked by LTZ guru Lymex
Zhang/BG2VO (GIF attached).
And some information about tempco:
http://bbs.38hot.net/read.php?tid=5569
http://bbs.38hot.net/read.php?tid=2880

With best regards
Mickle T.

Friday, October 7, 2011, 1:37:38 AM, you wrote:

AJ> Hello,
AJ> But consider: a resistance change on the 200R resistor of page 6
AJ> will directly affect output voltage. 200R * 5mA = 1V.
AJ> A 100 ppm change of the 200R Resistor has a 1V/8.2V
AJ> = 12 ppm change on output voltage.
AJ> All other resistors have below 1ppm change with 100ppm resistance change.

AJ> So the 200 Ohms resistor will be the most
AJ> critical for tempco and ageing in the whole system.

AJ> So in my opinion its only worth to consider if you
AJ> really need to save the 20-30 mA for the heater.

AJ> With best regards
AJ> Andreas

Andreas Jahn wrote:

Except that in your case you should only need around 10% of 200 ohms
with a correspondingly reduced sensitivity to the tempco of this resistor.

Bruce

I would take the output voltage between pins 3 and 7 and leave the
resistor "outside". There is already a diode between the op-amp output
and the zener cathode and the diode must be much less time and
temperature stable than a resistor.

Or did I misunderstand something again?

Will

2011/10/7, Andreas Jahn Andreas_-_Jahn@t-online.de:

There's little point in adding the tempco compensation resistor if you
do that as your proposed connection essentially eliminates the
additional correction provided by the resistor.
There's no way of avoiding the sensitivity to the value of this resistor
if this method of tempco trimming is used.
However if a +50ppm/K LTZ1000 residual tempco (with the tempco trim
resistor = 0) is typical, then the sensitivity will only be about 10% or
so of that esstimated by Adreas.

Bruce

Will wrote:

Thanks Bruce.

Will wrote:

Yes I did. And I had the same mistake with my previous calculations,
which explains the strange results.

Since you are an expert, an "LTZ1000 circuit description for dummies"
would be great.

The heater control is straightforward, but all the interactions in the
reference circuit seem not to be clear to me. And most likely there
are other dummies too.

Will

2011/10/7, Bruce Griffiths bruce.griffiths@xtra.co.nz:

In the standard positive reference circuit depicted in the latest datasheet

Vref  = Vz +Vbe
Transistor collector current, Ic = Vz/R2 (This increases slightly with
temperature due to the positive tempco of Vz)

Zener current, Iz = Vbe/R1 (This decreases with temperature due to the
negative tempco of Vbe.)

where Vz is the zener voltage, Vbe is the base to emitter voltage of the
temperature compensation transistor.

Vz has a tempco of approximately +2mV/K and Vbe has an approximate
tempco of -2mV/K
Thus Vref is nominally compensated for temperature changes.
However the tempco matching isnt perfect with a "typical" (note sample
size = 1!!!!) residual of +50ppm/K (+350uV/K)
This can be reduced by adding a resistor (Rt) in series with the zener.

In this case (to a first order approximation)

Vref = Vz + (1+ Rt/R1)*Vbe
Ic = Vz/R2 +((Rt/R1)*Vbe)/R2
Iz = Vbe/R1
If the value of Rt is suitably selected the tempco can be adjusted to zero.
However the required value will vary for each LTZ1000 and the circuit
only works if the tempco with Rt =0 is positive.
The magnitude of the Vbe tempco can be increased by reducing the
collector current (requires increasing R2), however the required
collector current to achieve zero Tc will probably be 1uA or even less,
this increases the transistor noise and makes the circuit more
susceptible to leakage shunting R2.
The compensating transistor is operated at zero collector base voltage
which eliminates the variations in Vbe due to the Early effect.

Bruce

Will wrote:

I would not talk of a sample size = 1.
My LTZ #1 has  48ppm/K with 70k and 52ppm/K with 50K
My LTZ #2 has  54ppm/K with 70k
Both have a setpoint divider of 12K5/1K.

and:

from the datasheet you can calculate for the 13K/1K nominal temperature
setpoint
100 ppm in voltage divider change R4/R5 will give 1 ppm output change.
100 ppm change of 0,5V VBe will give around 50uV temperature setpoint
change.
2mV VBE/K or 50uV/2000uV/K = 0,025K temperature setpoint change.

So from datasheet with nominal divider you can calculate
40 ppm / K typical temperature gradient.

unheated reference. (0.95ppm output for 100ppm divider change)

So there seems to be less tempco (but faster ageing)
with higher temperature setpoints.

With best regards

Andreas

Andreas Jahn wrote:

Even a sample size of 2 is too small to be useful in predicting the
tempco of the entire population of LTZ1000s
There will be variations due to manufacturing tolerances.
Perhaps your 2 LTZ1000s came from the same batch/wafer?
Measuring the tempcos of a statistically significant sample would be
more useful.

The major problem with that analysis is that the sign of the tempco
cannot be deduced and this only gives an approximate upper bound for the
magnitude of the tempco.
The tempco could lie anywhere within the [-40, +40] ppm/K range.

Bruce

Hi,

Well that expert on 38hot also got a +40ppm tempco.

http://www.national.com/rap/Story/vbe.html
This page shows that he was right to say a much lower collector current would be required. A smaller zener current and a lower collector current will reduce a + tempco, but at the expense of higher noise, so a tradeoff is that we must increase the noise to reduce the required accuracy of the temp control loop.
It seems to me that  a lower set temp would reduce the sensitivity to orientation.