seebeck coefficient Measurement

[becidea]

Hi,

No : I was suggesting to increase slowly the gradient, say for one or two minutes, and measure from time to time, both dV and dT. For high temperature you may need 50 to 100 points, for low temperature 20 to 50 will suffice 

..............................

What abour temperature gradient between two ends of the sample!. The temperature gradient should be less than 5K, so I need to keep gradient of temperature constant, needen't I ?

[nitad54448]

Hi

You need to measure the gradient every second or so T1, T2 and dV, at one end of the sample you put a small heater (from 1 W to about 50 W, depending on the temperature range you want to work)... You apply a small power to this heater (like a chain saw or sinus form..,) until the gradient is few K (selectable by the user).

I use typically 2K gradient at low temp (10K to 250K) and 3-5K at higher temperature (room temperature to 1000 K). I have built four different instruments of this type and I am routinely measuring S and (simultaneously, resistivity) from 15 K to 1200 K.. So, believe me, this is the best option. This is stated also in the previous paper I listed. Read it first then post back if you have questions

N

[becidea]

This was exactly what I wanted and I did so. One heater on one end of the sample, up to 550K temperature increased,but temperature gradient was so high. I put a heater on one end of the sample but gradient rises up to 25K, thats why I will use two heater from both end.

I have read this review. It is very nice review for measurements of Seebeck and Resistivity.It was combined of measurement methods papers. But some inofrmations about thermocouples, thermal contact are thing good to know. Do you have mns, icq, skype to talk about it, also do you have some time to spend for me.

Best...

[nitad54448]

---

@becidea wrote:

This was exactly what I wanted and I did so. One heater on one end of the sample, up to 550K temperature increased,but temperature gradient was so high. I put a heater on one end of the sample but gradient rises up to 25K, thats why I will use two heater from both end.

---

the sample should be in a furnace of some sort (or cryostat) and one or two heaters inside, attached to the sample. the amount of gradient is controlled by how much power you give to this heater. I am using a PID control to heat the small heater attached to sample in order to make a smooth and a slow gradient from equilibrium up to 2-5K difference then back to equillibrium in 1 to 5 minutes. I measure something like 50 to 500 points T1, T2 dV. You will need more than the 182 (which is good but quite slow), or you need a switch with multiplexer or matrix card. One of my devices uses CDAQ units, it's cheaper and faster than the nanovoltemeters albeit not as precise. depending upon the application these may be enough. PXI is another (better, more expensive) alternative.

Thermocouples are selected by the temperature you use (see the omega site for a ton of information). thermal contact is seen in the form of the dV vs dT points (if it s linear it s ok)

send me your email address, I can send you some of my papers

regards

[nitad54448]

hi again,

I re read your post (see also my previous). If I understand correctly you use two heaters connected somehow to the sample. Indeed you need then to use these heaters to make gradient ; you can do something like

1. make T1=T2 for the the two heaters (which is the measurment temperature)

2. make T1-x and T2+x where x is a function of time (from -3 to +3 for instance)

3. measure during this period of time T1, T2 and dV, (you can calculate the dt, dv then S)

4. don t forget to correct for the absolute S of the wire -unless you use Nb which has almost zero value of S)

5. go to the next temperature

N

[becidea]

Hi,

I have a problem. in my vi, I am reading two different temperatures via thermocouples. The problem is that one of the temperatures show 0.00, but there is no pbolem with the system bacuse display show that its normal. I don't know why but, when I use highlight execution lamp there is no problem; however, if I don't use one of them is not working. what can be the problem?

[nitad54448]

it is hard to answer without seeing more. show parts of your code. you need also to start a new thread...this question is different that Seebeck measurement system

[becidea]

I started a new thread with png

[becidea]

hi,

I have a vi to measure Seebeck Coefficient. Would you please check for any mistake. Actually program is working but I am not sure that I wired correctly.

Seebeck coefficient = (Th-Tc) / microVolt

plot Seebeck coefficient vs Avg Temp.

Th=hot side of temperature,

Tc=cold side of temp.

Best.

[nitad54448]

hi

S is the other way around is voltage/(T1-T2) and needs to be obtained from the slope of dV versus dT

also S needs to be corrected with the absolute value of the thermopower of the wire. some people use Nb wire which has low thermopower (almost nul); copper has 1 to 2 microV/K from room down to low Temp.

I personnaly use either copper for LT or platinum wire for HT (platinum has about -5 microV/K at RT larger value at HT)

in order to see if you did it correctly you need to measure a couple of known compounds : the easiest way is to measure two samples (one p and one n), for instance a Pt wire and a Pt/Rh sample

The difference of these will give you the S or R thermocouple (following the percentage of Rh) which are very well known (see for instance NIST webpage for absolute values)

N