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bsburns2506
modified 7 years ago

Temp sensor

1
26
186
01:22:19
Turning centigrade to F. The mV input represents a temp sensor that puts out 1mV for every degree centigrade. It's not 100% accurate, anyone that can offer solutions would be appreciated. The parameters are the temp sensor, converting c to f and having the output in mV be 100 times the temp in F, and the only power supplies allowed are the ones for the op amps. It's a work in progress *Update: I added a freezer alarm to the circuit. When the output goes above 32 degrees F the red led lights and won't go back until the power is cycled or the temp goes to below -32 F
published 7 years ago
bsburns2506
7 years ago
I had it working, but now it seems like it cant find a solution for some reason
zorgo
7 years ago
What exactly is your desired outcome? And I've found out that if you set the input voltage as greater than 80mV, then the led shines. However, it doesn't stop shining for any value you give the input even uptil 1KV
bsburns2506
7 years ago
The leds are just my effort to add 3.2v at the end
bsburns2506
7 years ago
My project is to simulate a temp sensor that will take in C as a mV value and convert it to F. For example 21.1 C = 21.1 mV then convert that to 70 F. But the mV output has to bee 100 times the temp in F so 70F would be 7000mV or 7V
bsburns2506
7 years ago
The conversion from C to F is F= 1.8C +32. I have the C times 1.8 part, but i cant figure out a way to add 3.2V without adding another power supply
crake
7 years ago
Like this? http://everycircuit.com/circuit/5275238278627328
bsburns2506
7 years ago
Yea i like it. It is a lot simpler than the solution my book proposed. They used three op amps, one to amplify, one to invert and than one to sum the 3.2v. But I guess he did say to only use the 15v power supply that supplies the op amp. I'm going to update it his suggested solution
bsburns2506
7 years ago
That is, if i can get it figured out
crake
7 years ago
If you wan't to use the circuit I linked here in chat, you may need to add a voltage follower to the input, and add a voltage follower at the DC offset section (assuming you derive 3.2V from 15V using a voltage divider, and use resistors above 10K).
bsburns2506
7 years ago
Thank you sir
crake
7 years ago
You're welcome, bsburns2506.
2ctiby
7 years ago
Here is one variation: http://everycircuit.com/circuit/6443919830417408
bsburns2506
7 years ago
Very nice solution
2ctiby
7 years ago
Thanks ...Here is a full F to C and vice versa version: http://everycircuit.com/circuit/5362322511233024 And yet a different variation here:  http://everycircuit.com/circuit/5785433417187328
crake
7 years ago
@bsburns http://everycircuit.com/circuit/5275238278627328
crake
7 years ago
@bsburns this is what I was talking about before. Since we don't know out output resistance of your sensor, this may be an improvement.
bsburns2506
7 years ago
@crake yes that is the exact solution he talked about. Its hard for me because its an audio lecture and i cant see his results. I can only hear his description of them and this is spot on
2ctiby
7 years ago
That looks good ... a little more playing on that crake version above here: a 111 Ohm calibration resistor to enable higher C input? http://everycircuit.com/circuit/5765474435727360
crake
7 years ago
@bsburn2506 Glad this helped. If you haven't already, look into op amp characteristics like input and output resistance, and negative feedback etc. Mess with the most basic amp topology first and then increase the complexity once you're sure you have a grasp on characteristics. The amp here has a lot going on. It has signal gain, addition, buffering etc.
bsburns2506
7 years ago
@crake The series ive been listening to has done a few lectures on opamps. I think that no current goes through the input and that with feedback, both the pos and neg inputs are almost the same. But they are still pretty new to me. Trying to decipher your circuit is helping me
bsburns2506
7 years ago
@crake i was able to remove the summing amp and its feedback resistor without any effect on the output. What is the purpose of that amp?
crake
7 years ago
@bsburns2506 This amp was a buffer providing a low output resistance and a DC offset voltage. The low output resistance seen by the 180kohm resistor helps maintain the balance of the difference amp at the center, as well as maintains the DC offset voltage provided by the voltage divider. Since you chose low resistor values for your voltage divider, this potential problem isn't obvious. Make a spreadsheet with pairs of resistor values for your voltage divider. Start with 3kohm and 817ohm and then increase their values. Take the pairs of resistor values from the spreadsheet and then try them in the simulator. You will see that the output resistance from your voltage divider begins to interact with the 180kohm resistor for ever increasing values of the resistor pairs.
crake
7 years ago
Check out this source. Analog devices rocks and I would encourage anyone to check out their articles, as well as Ti and others but, here you go... scroll down about half way. You will see something very similar to what you're working with here. http://www.analog.com/en/analog-dialogue/articles/common-problems-when-designing-amplifier-circuits.html
bsburns2506
7 years ago
Thank you sir. I will dive into that recommendation
bsburns2506
7 years ago
Thanks to everyone involved in this. Its been an awesome learning experience
crake
7 years ago
@bsburns2506 I am glad we could help. Keep at it. After sometime you will get to where you want to be, and then some :D

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