Hello again,LM335 V's are 420mV(Starts working 42 degrees Celsius) and 380mV (stops working at 38 degress Celsius)(Because of 10mV per degree celsiu)
But how will i find the Vref,i don't know R10 neither R11 (and i can't find a way to calculate them)
Assuming Vref is measured at the node of R3, R10 and D1, I think that Vref = VD1 .But how will i find the Vref,i don't know R10 neither R11 (and i can't find a way to calculate them)
Hello,I have understand the philosophy of the circuit but i can't analyze it so i can't do that.
If u can show me a way to start analyzing it.
So, I gave it a shot. For Part 2 I think it is:Ok, so we have an expression for VL which is the voltage at the non inverting input when the output is low, so now we need the expression for the voltage at the non inverting input when the output of the comparator is 12v. Can you find this expression? It's a voltage divider again, but this time we have a non zero voltage at both ends of the divider. That means you may want to use superposition to find the expression.
Can you at least attempt to get this?
[Hint: use the voltage divider formula twice, killing just one source for each solution, then add the results].
Hi,So, I gave it a shot. For Part 2 I think it is:
vH1 = vLM * (R11/(R10+R11) killed the 12V of the comparator output
vH2 = 12 * (R10/(R10+R11) killed vLM
So, overall vH should be the the algebraic sum of vH1 and vH2, which should equal vH = vH2 - vH1 = ((12*R10)-(vLM*R11)) / (R10+R11).
At this point there are two equations and three unknowns. Should we assume vLM=380mV at Part 1 and vLM=420mV at Part 2 ?
Hi,I don't want to rain on your parade, but, here is a picture from the LM335 datasheet. You might want to rethink your voltage levels.
View attachment 150436
Why do we get the sum of VH1 and VH2?So, I gave it a shot. For Part 2 I think it is:
vH1 = vLM * (R11/(R10+R11) killed the 12V of the comparator output
vH2 = 12 * (R10/(R10+R11) killed vLM
So, overall vH should be the the algebraic sum of vH1 and vH2, which should equal vH = vH2 - vH1 = ((12*R10)-(vLM*R11)) / (R10+R11).
At this point there are two equations and three unknowns. Should we assume vLM=380mV at Part 1 and vLM=420mV at Part 2 ?
Our teacher gave us this output for LM335 10mV per degree celsiu(see attached file)Hi,
Yes thanks for pointing that out.
I love rain in the summer time
Thread starter | Similar threads | Forum | Replies | Date |
---|---|---|---|---|
N | In need for help for my homework | Homework Help | 0 | |
W | Need guidance on this homework problem | Homework Help | 32 | |
M | Need help with a homework! | Homework Help | 11 | |
M | Hi, I need help with my homework on circuit diagrams! | Homework Help | 12 | |
I need homework help | Homework Help | 4 |