Hi all, I have a little question about my H bridge circuit with schematics attached as jpg file.
So, I have a circuit which is the same H Bridge as the image where the PNP Darlington are TIP127 and the NPN Darlington are TIP122 but without RX, the comparator and the zener diode.
The circuit I have attached is one that I will make new for a project, where of course, a 12VDC motor will be connected to the H Bridge.
I this new circuit I have to calculate the value for VX, which will be connected to the non inverting input of the comparator, so depending on the value for RX I will have a value for VX and than I'll can determine the value for the zener diode, but I am not totally sure about how to determine this value.
Testing the real H Bridge circuit, I connected it to 12V power supply, in point A e B I measured 10,6V (Voltage drop equals for two Darlington tranistor) and I measured the current consumed by the 12VDC motor connected to it and is about 70mA.
With these values, calculating the value for VX, so, treating the situation as a normal voltage drop formula, if I am not wrong, the formula is:
VX = VAB * RX / (RX + RMotor)
Where
VAB is the total voltage I have in output from the H Bridge and RMotor is the total resisteance of the motor, in my case 10,6V/70mA = 151,4 Ohm.
I am right, or I'm wrong and I am missing some detail?
unfortunately I don't have any potentiometer to determine it physically right now
Thank you for your help.
Simon
So, I have a circuit which is the same H Bridge as the image where the PNP Darlington are TIP127 and the NPN Darlington are TIP122 but without RX, the comparator and the zener diode.
The circuit I have attached is one that I will make new for a project, where of course, a 12VDC motor will be connected to the H Bridge.
I this new circuit I have to calculate the value for VX, which will be connected to the non inverting input of the comparator, so depending on the value for RX I will have a value for VX and than I'll can determine the value for the zener diode, but I am not totally sure about how to determine this value.
Testing the real H Bridge circuit, I connected it to 12V power supply, in point A e B I measured 10,6V (Voltage drop equals for two Darlington tranistor) and I measured the current consumed by the 12VDC motor connected to it and is about 70mA.
With these values, calculating the value for VX, so, treating the situation as a normal voltage drop formula, if I am not wrong, the formula is:
VX = VAB * RX / (RX + RMotor)
Where
VAB is the total voltage I have in output from the H Bridge and RMotor is the total resisteance of the motor, in my case 10,6V/70mA = 151,4 Ohm.
I am right, or I'm wrong and I am missing some detail?
unfortunately I don't have any potentiometer to determine it physically right now
Thank you for your help.
Simon
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