Part Number: TPS54160
I wrote a C++ Program to try to calculate the equation 47 the data sheet
Can't paste the data sheet equation in but I will give you the C++ code and Output to show what the computer calculates.
Below are inputs to the equation:
// set RLOAD to 10 ohmsR_LOAD = 3.3 / .33; // tenth the currentcout << "R_LOAD = " << R_LOAD << endl;cout << "fc = " << fc << endl;cout << "COUT = " << COUT << endl;cout << "pi = " << pi << endl;COUT_Resr = 10.0e-3;cout << "COUT_Resr = " << COUT_Resr << endl;
values printed out by the C Program:
COUT = 4.7e-05 after deratingfp_MOD = 1539.22pi = 3.14159COUT = 4.7e-05fz_MOD = 338628fc_max = 45353.6fc_min = 7696.08fc = 45000R_LOAD = 10fc = 45000COUT = 4.7e-05pi = 3.14159COUT_Resr = 0.01
G_MOD_fc = 67.773 value come back
G_MOD_fc_numerator = gm_ps * R_LOAD * ( (2.0 * pi * fc * COUT * COUT_Resr ) + 1.0);G_MOD_fc_denominator = ( 2.0 * pi * COUT * (R_LOAD + COUT_Resr ) ) + 1.0;G_MOD_fc = G_MOD_fc_numerator / G_MOD_fc_denominator;//cout << "G_MOD_fc = " << G_MOD_fc << endl;
G_MOD_fc = 67.773 // Calculated output from C++ Program
Expected answer in the data sheet is
For the example problem, the gain of the modulator at the crossover frequency is 0.542. Next
This is on page 37 of the data sheet.
Thanks for any help on this
TI Wide Vin Buck Converter & Controller Applications Engineer
In reply to YangZhang:
Thank You Yang,
I would like to go to another equation in the app note:
Equation 51 and 48 calculate a R_C value. When I plug in the values into the equation the numbers for Equation 51 seem to be wrong here is the
C code I am using:
double R_C;double gm_EA;double VREF;fz_MOD = 338.0e3;gm_EA = 97.0e-6;VREF = 0.8;G_MOD_fc = 0.542;cout << "fz_MOD = " << fz_MOD << endl;cout << "gm_EA = " << gm_EA << endl;cout << "VREF = " << VREF << endl;cout << "VOUT = " << VOUT << endl;cout << "G_MOD_fc = " << G_MOD_fc << endl;//R_C = ( VOUT ) / ( G_MOD_fc * fc * gm_EA * VREF ); // 51 fz_MOD//cout << ( G_MOD_fc * fz_MOD * gm_EA * VREF ) << endl;R_C = VOUT / ( G_MOD_fc * fz_MOD * gm_EA * VREF ); // 51 fz_MODcout << "R_C = " << R_C << endl;R_C = ( VOUT ) / ( G_MOD_fc * gm_EA * VREF ); // 48cout << "R_C = " << R_C << endl;
And the output or values I used in the equations:
fz_MOD = 338000gm_EA = 9.7e-05VREF = 0.8VOUT = 3.3G_MOD_fc = 0.542R_C = 0.232133R_C = 78460.8
expected value is
• RC = 76.2 kΩ
Not that for equation 51 it is very small:
Equation 48 seems reasonable. The app note suggests need to use equation 51 for this example:
For the example problem, the ESR zero is located at a higher frequency compared to the crossover frequency soEquation 50 through Equation 53 are used to calculate the compensation components. In this example, thecalculated components values are:• RC = 76.2 kΩ• CC = 2710 pF• Cƒ = 6.17 pF
This is on page 38 of TPS54160.pdf data sheet.
In reply to Gary Olson:
App note SLVS9492.pdf (heading at the top TPS54620. Entitled
TPS54620 4.5 to 17V input 6-A Synchronout Ste Downt Swift Converter
Calculating the Type 2 compensation resistor R4 in this app note
had an equation on page 28 equation number 35
R4_num = 2.0 * pi * fc * VOUT * COUT;R4_denom = ( gm_ps * gm_EA * VREF );
R4 = R4_num / R4_denom;
I can get a resistance in the ball park especially if I change the fc = 36500
leaving all the other parameters the same.
double R4_num; double R4_denom; //double R4;
//fc = 0.1 * f_sw_hz; fc = 36500.0;
R4_num = 2.0 * pi * fc * VOUT * COUT; R4_denom = ( gm_ps * gm_EA * VREF );
cout << endl << endl << endl;
cout << "fc = " << fc << endl; cout << "VOUT = " << VOUT << endl; cout << "COUT = " << COUT << endl; cout << "gm_ps = " << gm_ps << endl; cout << "gm_EA = " << gm_EA << endl; cout << "VREF = " << VREF << endl; cout << endl;
cout << "R4_num = " << R4_num << endl; cout << "R4_denom = " << R4_denom << endl; cout << "R4 = " << R4 << endl;
Results of C++ output
fc = 36500VOUT = 3.3COUT = 4.7e-05gm_ps = 6gm_EA = 9.7e-05VREF = 0.8R4_num = 35.5701R4_denom = 0.0004656R4 = 76396.2
when fc is left at 45000
R4_num = 43.8535R4_denom = 0.0004656R4 = 94187.1
I agree with this as the answer to the first question. I was hoping to hear more about the next question.
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