Hey all. I have small electronics project with the following specs.

- I am given a K-type thermocouple and I have to deisgn an amplifier using MOSFET BS170. The ouput should be 10mV/Kelvin (10 mV per 1 kelvin change).

Now I found out that the thermocouple gives 40uV/Kelvin and this means I have to design an amplifier voltage gain of 250.

-Decided to built a common source amplifier. The problem is that MOSFETS give small voltage gains, generaly less than 100, so if anyone knows a way to increase the gain to 250 or maybe a circuit configuration that can do that.

We are told that we should not use any other devices except the thermocouple, MOSFET and resistor

 

Are you permitted to use two of the BS170 in a two stage amplifer? That would give you more than enough gain.

hgmjr

 

Hi,

unfortunately we are not permitted to use more than one MOSFET.
Any other tips?

 

Now I found out that the thermocouple gives 40uV/Kelvin and this means I have to design an amplifier voltage gain of 250.

Look at this link to see if it will give you some inspiration.

Designing and Building Your Own MOSFET Boosters

By biasing the MOSFET gate just over the Vg(th), the overall small signal gain can be very high. The author mentioned 320!!!

However, one must tweak the biasing of the circuit to fit each MOSFET individually as the Vg(th) varies a lot even with the same part number.

Added: I did a quick & dirty simulation with LTSpice and pick a MOSFET. I will let you work out the gain in this case.

 

How do you get to the resistor value (R = 100k) in the circuit above? Also please can someone explain this paragraph in the link pasted above... http://www.geofex.com/Article_Folders/mosboost/mosboost.htm

"It would do that, except we're choking it off with our measurement method. Rdrain is 1K ohm, remember? That's 1V per milliampere. At 9ma, the entire battery voltage is dropped across the drain resistor and there isn't any more voltage to push current through the MOSFET. In fact, the current in Rdrain rises from zero milliamperes to 9ma as the Vbias rises from 2.0V to 2.028V. That additional 28 millivolts is enough to cause the MOSFET to go from completely cut off to completely saturated. (If we had used a 1 ohm measuring resistor, it would have limited the Vsupply current to 9V / 1 ohm = 9A, and we'd have seen the drain current go as high as the gate voltage told it to - until the MOSFET burned up, that is.)"

Still don't get this...