Recent content by CNC101

Of course thanks for your help.

Okey doke here we go T = Tf + a e-t/τ. I can see what you mean when you say you have to use a figure inbetween the initial and final points T(4) = 10 + 40 e^-4/τ 24=10+40e^-4/τ solving for τ : τ= -4/ln(7/20) = 3.81 T(t)=10+40e^-t/3.81 T(20)= 10+40e^-20/3.81=10.21 (close to final...

Sorry I need some clarity here because it is getting too confusing, I need more methodology here and the correct syntax. According to my notes; T = Tf + a e^(-t/τ) T= output value τ= system time constant in seconds (a*0.368)?? i.e one time constant equals 36.8% temperature input a=...

10+50e^(-4/τ)=50 , τ=-4/(ln(4/5))=17.92 why t=4?? Im a bit lost now. I don't think this equation is working.

Im having trouble with the math: T=T∞+a.e^(-t/τ) T∞=10 a=50 T=50 at 0s 50=10+50e^(0/τ) is unsolvable T=10 at 20s 10=10+50e^-(20/τ) is unsolvable

Think I have it : T=T∞+a.e^(-t/τ) (where T∞ = limiting value which is 10°C) Choosing 2 points on the graph that are easily identifiable: (x=time,y=Temp) : (16,11) and (8,16) 11=10+a*e^-(16/τ) (1) 16=10+a*e^(-8/τ) (2) Simplifying by subtracting 10 from each equation and then...

"So how about T = Tie-t/τ + T∞ instead? Read Ti and T∞ from the graph to compute τ." Sorry but how would I do that, I don't understand the method,sorry.

Homework Statement The graph shows the response of a bare thermocouple which has been subjected to a step change in temperature from 50°C to 10°C. Assuming that the bare thermocouple behaves as a single transfer lag system, determine the mathematical relationship between the temperature (T) and...

Well I know a lot more now than when I started, Thanks!

Well I think inside the regulator is a forward biased zener diode which diverts excess current to its ground or "ADJ" in the diagram. Its own low resistance path means large changes of current have a small affect on its voltage and therefore protecting the load from excess current.

R1 is a sensing resistor and when Q1 senses a voltage drop at its base, it switches on and passes current around the regulator. When the voltage across R2 (error resistor) increases to a value just below Q1 max emitter-collector current limit, Q2 switches on and shunts R1 by creating a low...

Homework Statement Homework EquationsThe Attempt at a Solution I need to know if my understanding of this circuit is correct in terms of R1, R2, Q1, Q2. R1 is a sensing resistor and when Q1 senses a voltage drop at its base, it switches on and passes current around the regulator. I think...

Well my exact expression was -I(R1(Rf/R2)+1)+R)=V Which gave me 0.9991nA which on the simulation, Vo is shy by just 0.002mV compared to 0.09mV!

Could you explain what you mean by Numerical approximation simplification, cheers :)

With LTspice and a precision operational amplifier model, I have -100.088mV which I think is close enough. Simetrix required a voltage controlled voltage source to create ideal opamp conditions which just got confusing.