Sciencemadness Discussion Board

Measuring pressure a la Pirani's Vacuum Meter

Alek - 22-3-2005 at 15:07

My name is Alek Zawada. I'm polish. My english is very fatal:(. I'm so sorry:(
That's why I beg you:
don't write a lot of very hard words.

I have the vacuum pump. It isn't very good pump, because I have
70 torr vacuum (760 torr=760mmHg=1013,25hPa). However I haven't
very high vacuum it's posibble to make the tubes with cold cathode.
I can to do it.

I measure my vacuum by mechanical vacuum meter, but it isn't good idea.
Why? Because this vacuum meter is big. I tried to do better vacuum meter-
Pirani's vacuum meter. (I reconstructed ligtbulb-I united the glass tube with bulb).
Metal wire in the bulb must be hot (mabye 400 celsus degrees it's good -then wolframe
couldn't burn). That's why you must to connect the serial resistor to reduce current.
Now, If you reduce the pressure inside the bulb, resistance of metal wire in bulb will be
grow up (because gas cool down the wire worse nad worse). It's no problem to compute the resitance to presure and display it on LCD.
But I don't know if displayed pressure (vacuum) depend on atmospheric pressure.
What do you think about it?

If you want, you can see my glass tubes here (there is in the file *.jpg:)

Edited by chemoleo: Changed title to something more appropriate.

[Edited on 23-3-2005 by chemoleo]

chemoleo - 22-3-2005 at 17:42

It's a good question.
Could you please point out which picture you mean, there are several.

Hmm, so essentially you say resistance in Ohms changes within the wire depending on the pressure (concentration of gas) the wire is surrounded by? Essentially, a higher pressure would mean more rapid cooling, while a cooler wire would mean a smaller resistance (a red glowing wire has a greater resistance than a cool one)- which results in a resistance being somewhat proportional to the pressure? Proportional...uhm. I doubt it.

You'd need to derive the equations for that. The system needs to be calibrated naturally, if the lightbulb i.e. had the size of a room, the cooling would be more efficient (and thus the resistance proportional to the pressure, taking some gas/electric laws into account?) because the heat generated by it would not be able to feed back into the system so quickly.
So there are lots of factors to take account of.
Essentially, yes, the resistance depends to some degree on the pressure; the higher the pressure, the smaller the resistance.
So a vacuum insulates perfectly, causing the resistance to increase (because it can heat up, without the heat being carried away by gasses).

But unless you have a perfect system, I think you'd need calibration (of known pressures) to make this work.
I like the idea though :)

Do correct me if I misunderstood or something.

[Edited on 23-3-2005 by chemoleo]

Twospoons - 22-3-2005 at 18:44

The displayed pressure will depend less on atmospheric pressure, and more on atmospheric temperature.

A better way to run a Pirani gauge is to run it so that its temperature (resistance) stays constant and then measure the power needed to keep that temperature.

Have a look here for a thermistor gauge that works on the same principle

Alek - 23-3-2005 at 00:16

Good morning!
Thank you for answeres. I watch Twospoons's link. It i'snt exactly my idea, because I haven't thermocouple in my bulb.

My sample schematic diagram is here:

(the file called "wakuometr idea.jpg".

In my circuit, the metal wire have resistance:
when k-constant
Ro-resistance in temperature=0 celsus degrees
T-temperature (depend on presure:) )

If pressure fall from ca.760 Torr to ca.76 Torr the current in microamperometer change of 100 microamperes. Firstly, I couldn't belive that it's possible- I though that Pirani's vacuum meter is good for low pressure (for example 10^-2 torr) only.

I'm going to reconstruct my circuit - I will have to add operational amplifier, microcontroller, LCD display etc.

I'm afraid that use thermocopule is better idea- we haven't errors from temperature of atmosphere.

Twospoons - 23-3-2005 at 14:45

This is closer to what you are doing. In this circuit the bridge is kept in balance by the amplifier, and the bridge excitation voltage gives you a measure of the pressure.

Since you are using a filament instead of an NTC thermistor, you would need change the circuit - put your filament in the upper arm of the bridge in place of R1.

Geomancer - 24-3-2005 at 19:09

That's an impressive bit of glassblowing. Scientific American had a column describing a Pirani gauge in the November 1996 issue. The sensor was a "glow plug" for some sort of model engine, but I assume the calibration and electronics will be similar. I don't have a copy right now, so that is all I can tell you. I can probably give some more details on Monday.

Alek - 26-3-2005 at 11:53

Having some time I'm making the system with LCD 2*16, microcontroller ATMEGA 8 and precision amplifier MAX 430. Writing the program I used polynomial (5 stage) to change voltage to pressure. I'm happy, because it's working good!

Geomancer-I will be waiting for your answer.