Thread: need help in solving equation usin matlab

Zthjsl(timel)=(Rjs1*(1-exp(-timel/taujs1)))+(Rjs2*(1-exp(-timel/taujs2)))+(Rjs3*(1-exp(-timel/taujs3)))+(Rjs4*(1-exp(-timel/taujs4)))+(Rjs5*(1-exp(-timel/taujs5)))+(Rjs6*(1-exp(-timel/taujs6)));

Zthjsh(timel)= Rjs1*(1-exp(-(T-timel)/taujs1)) + Rjs2*(1-exp(-(T-timel)/taujs2)) + Rjs3*(1-exp(-(T-timel)/taujs3))+ Rjs4*(1-exp(-(T-timel)/taujs4)) + Rjs5*(1-exp(-(T-timel)/taujs5)) + Rjs6*(1-exp(-(T-timel)/taujs6)) ;

Zthsal(timel)= Rsa*(1-exp(-timel/tausa));

Zthsah (timel)= Rsa*(1-exp(-(T-timel)/tausa));


Thetal(timel) = plossl*(Zthjsl(timel) + Zthsal(timel));

Thetah(timel) = plossh*(Zthjsh(timel)+ Zthsah(timel));

Thetal(timel)=Thetah(timel);

only one unknown is there timel.i tried but i could not come up with a solution how to get the values of timel if all other parameters are specified

Originally Posted by electra

Zthjsl(timel)=(Rjs1*(1-exp(-timel/taujs1)))+(Rjs2*(1-exp(-timel/taujs2)))+(Rjs3*(1-exp(-timel/taujs3)))+(Rjs4*(1-exp(-timel/taujs4)))+(Rjs5*(1-exp(-timel/taujs5)))+(Rjs6*(1-exp(-timel/taujs6)));

Zthjsh(timel)= Rjs1*(1-exp(-(T-timel)/taujs1)) + Rjs2*(1-exp(-(T-timel)/taujs2)) + Rjs3*(1-exp(-(T-timel)/taujs3))+ Rjs4*(1-exp(-(T-timel)/taujs4)) + Rjs5*(1-exp(-(T-timel)/taujs5)) + Rjs6*(1-exp(-(T-timel)/taujs6)) ;

Zthsal(timel)= Rsa*(1-exp(-timel/tausa));

Zthsah (timel)= Rsa*(1-exp(-(T-timel)/tausa));


Thetal(timel) = plossl*(Zthjsl(timel) + Zthsal(timel));

Thetah(timel) = plossh*(Zthjsh(timel)+ Zthsah(timel));

Thetal(timel)=Thetah(timel);

only one unknown is there timel.i tried but i could not come up with a solution how to get the values of timel if all other parameters are specified

1. Where has this problem come from, may-be you are trying to solve the wrong problem, that is there may be an easier way to solve the original problem than solving these equations.

2. Why Matlab? Are you seeking a symbolic solution or do you have numerical values for all these constants and would a numerical solution be acceptable.

RonL

In your PM you wrote:

i have TO CALCULATE power loss in rectifiers ...

vt0d =0.7 ;
rtd = 2.3;

Rjs1 = 0.1765; % components of transient thermal impedance, junction to sink
Rjs2 = 0.206;
Rjs3 = 0.06481;
Rjs4 = 0.04177;
Rjs5 = 0.01089;
Rjs6 = 0;

taujs1 = 2.965;
taujs2 = 0.56393;
taujs3 = 0.09829;
taujs4 = 0.00902;
taujs5 = 0.0008;
taujs6 = 1;

Rsa= 0.04;
tausa=2;

il = 0.84*50; % phase current RMS value, low power (2pi/3)
ih = 0.84*100; % phase current RMS value, high power

plossdl = (vt0d + rtd*abs(il))*abs(il); % power losses of one diode
plossl = 6*(plossdl);
plossdh = (vt0d + rtd*abs(ih))*abs(ih); % power losses of one diode, high load
plossh = 6*(plossdh ); % total losses for rectifier bridge, high load

T=85;
timel=75;

% thermal impedances (junction to sink and sink to ambient) for lower and higher power
Zthjsl=(Rjs1*(1-exp(-timel/taujs1)))+(Rjs2*(1-exp(-timel/taujs2)))+(Rjs3*(1-exp(-timel/taujs3)))+(Rjs4*(1-exp(-timel/taujs4)))+(Rjs5*(1-exp(-timel/taujs5)))+(Rjs6*(1-exp(-timel/taujs6)));

Zthjsh = Rjs1*(1-exp(-(T-timel)/taujs1)) + Rjs2*(1-exp(-(T-timel)/taujs2)) + Rjs3*(1-exp(-(T-timel)/taujs3))+ Rjs4*(1-exp(-(T-timel)/taujs4)) + Rjs5*(1-exp(-(T-timel)/taujs5)) + Rjs6*(1-exp(-(T-timel)/taujs6)) ;

Zthsal = Rsa*(1-exp(-timel/tausa));

Zthsah = Rsa*(1-exp(-(T-timel)/tausa));

% temperature decrease during lower power, initial calculation
Thetal = plossl*(Zthjsl + Zthsal);
% temperature rise during high power pulse, initial calculation
Thetah = plossh*(Zthjsh+ Zthsah);
%here Thetal=Thetah, and only one variable is there 'timel'.how can i find timel

So this is a numerical problem.

It is easy to get lost in the detail when the problem is written like this, so I will rewrite it as:






That is make all the constants and intermediate calculation that are independent of globals. Then put each of the functions into a Matlab function of its own (or into one vector valued function)

Now you are going to use one of the numerical solvers, use the Matlab help to look up which is most appropriate.

Also, if you have some idea where the solution is, you might look at plotting:



which if you are lucky will have a zero that you can spot on the graph (note that this will not be differentiable at the zero so only use this for exploratory work).

RonL