If he is using this LED voltmeter the zeroing isn´t needed. This pannel is zeroing itself and can be used with the same powersource. Just the ground needs to be attached directly to "-" but not the zeroing pot.
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FrozenGate by Avery
DIY Thermal LPM for under $50
- Thread starter MarioMaster
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If he is using this LED voltmeter the zeroing isn´t needed. This pannel is zeroing itself and can be used with the same powersource. Just the ground needs to be attached directly to "-" but not the zeroing pot.
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Some opamps have slight offsets that may need to be compensated for. Not all, but some.
Yes I know and my opamp is having it, so I need the pot when attaching a DMM or one of those LCD panels that need a separate powersource. Just this panel has a build in auto-zeroing to compensate such offsets and this is working really well.
That´s why I replaced the previous panel I was using before with this one.
That´s why I replaced the previous panel I was using before with this one.
Im doing some brainstorming. please excuse the bad schematic :3
Question: Why not this way?
I dont understand why you measure the TEC voltage without a load resistor.
Without it the TEC integrates the energy. Which adds a lot of error especially if you measure high power lasers.
Or is it impossible to deal with low power lasers, without integration?
Question: Why not this way?
I dont understand why you measure the TEC voltage without a load resistor.
Without it the TEC integrates the energy. Which adds a lot of error especially if you measure high power lasers.
Or is it impossible to deal with low power lasers, without integration?
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Linearity would probably be better with a load resistor, however this adds an extra step in calibration and a layer of complexity that most people who want to build this circuit likely will not be able to perform easily.
The circuit was designed for measurement of lasers up to 2 watts.
The circuit was designed for measurement of lasers up to 2 watts.
but doesn't the circuit output rise as long as you point the laser on the TEC? How do you know when it will show you the correct value? Sure, you can wait till you reach thermal equilibrium, but a load resistor would speed that process up by a huge amount, because you would reach the equilibrium almost instantly.
besides, what happens >2W? Just an error to big problem?
*im not criticizing your schematic. Im totally new to this. im just trying to understand*
besides, what happens >2W? Just an error to big problem?
*im not criticizing your schematic. Im totally new to this. im just trying to understand*
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No, the TEC itself has thermal mass and thermal conductivity from face to face, even when unpowered. The TEC outputs a voltage proportional to the temperature difference from face to face. The laser provides heat to one side, this energy input combined with the thermal resistance of the TEC face to the heatsink gives a way to quantify the heat flow (thus, laser power).
So response time is basically limited to the thermal mass of the TEC and its thermal resistance. I just tested an old TEC LPM sensor I had laying around and response time showed no significant improvements with a 1K and 100 ohm load resistor.
So response time is basically limited to the thermal mass of the TEC and its thermal resistance. I just tested an old TEC LPM sensor I had laying around and response time showed no significant improvements with a 1K and 100 ohm load resistor.
thats strange. if you short the TEC, it should be actively trying to equalize the temperature on both sides, the counter effect is the flowing current.
Shorting the TEC should lower the temperature resistance between the faces.
With a high load the current cant flow, and the temperature has to equalize by normal means
Either something is off, or i totally miss understand something :/
edit:
on second thought. i THINK, your statement cant be right. If i put a fixed amount of energy on one side, i can use this energy to create a current across a resistor. The heat will be transferred equally to the other side. If i increase the electric load, more current will flow. The heat will then faster equalize. If that is not the case it would imply i can suck more energy out of the heat if i increase the load, which cant be right.
Shorting the TEC should lower the temperature resistance between the faces.
With a high load the current cant flow, and the temperature has to equalize by normal means
Either something is off, or i totally miss understand something :/
edit:
on second thought. i THINK, your statement cant be right. If i put a fixed amount of energy on one side, i can use this energy to create a current across a resistor. The heat will be transferred equally to the other side. If i increase the electric load, more current will flow. The heat will then faster equalize. If that is not the case it would imply i can suck more energy out of the heat if i increase the load, which cant be right.
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That effect will become greater as the laser power increases. At the relatively low powers we're dealing with the TEC is only generating tens of millivolts.
)
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TEC is chosen because it gives an output based on a temperature difference, instead of an absolute temperature reading.
TEC generates the same signal regardless of the ambient conditions (assuming the ambient is stable), so the data acquisition system does not need to compensate for external temperature, beam shape, beam size, etc
TEC generates the same signal regardless of the ambient conditions (assuming the ambient is stable), so the data acquisition system does not need to compensate for external temperature, beam shape, beam size, etc
The TEC design and this circuit has been proven to work correctly, so I don´t get where the problem is. You are messuring voltage not current, so the voltage drop will be the same regardless you are soldering a resistor in parallel or not. To reduce the noise there is a capacitor.
All you need is a linear outputting TEC and a calibrated circuit. For calibration you can use SMD resistors and other lasers with a known output.
I´ve build a calibration tool for my LPM and calibrated it up to 4W but already measured up to 6W with combined beams. But also measuring them separately and adding the outputs gives me the same results. So the whole LPM is working as expected, even if the heatsink on the back of the TEC is warming up.
For a diy LPM this is a really good working circuit but if you need better response time, you better go with a thermophile sensor rather than with a TEC.
All you need is a linear outputting TEC and a calibrated circuit. For calibration you can use SMD resistors and other lasers with a known output.
I´ve build a calibration tool for my LPM and calibrated it up to 4W but already measured up to 6W with combined beams. But also measuring them separately and adding the outputs gives me the same results. So the whole LPM is working as expected, even if the heatsink on the back of the TEC is warming up.
For a diy LPM this is a really good working circuit but if you need better response time, you better go with a thermophile sensor rather than with a TEC.
Yeas, but there is no absolute linearity. Even photons are not traveling absolute linear through the spacetime. You only reach almost absolute linearity for a limited section.
The same for this design. When you are using a TEC with a linear output, there is nothing to worry about but when using a TEC with a non linear output, you will need a more compicated circuit to compenstate the drift.
However if you reach the saturation point of the TEC, even a more complicated circuit will be not able to compensate it.
The same for this design. When you are using a TEC with a linear output, there is nothing to worry about but when using a TEC with a non linear output, you will need a more compicated circuit to compenstate the drift.
However if you reach the saturation point of the TEC, even a more complicated circuit will be not able to compensate it.
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You could use the peltier actively to keep both sides equally warm, then measure the necessary current. I think this should give you a very good linearity.
>Even photons are not traveling absolute linear through the spacetime.
im not sure about which fluctuations you are speaking, but yes they should fly perfectly linear, if they dont then external effects are at fault. c is a constant.
>Even photons are not traveling absolute linear through the spacetime.
im not sure about which fluctuations you are speaking, but yes they should fly perfectly linear, if they dont then external effects are at fault. c is a constant.
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