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

 

Hi simo_x,

Start by picking an arbitrary voltage drop you can live with across RX. Usually you will want this to be as low as possible since it would subract from the motor voltage and therefore torque.

Example: if you want to keep the voltage drop less than 1 Volt then try RX=8.2Ω. Therefore, Vrx= motor current X RX = 0.07 X 8.2 = 0.574 Volts.

Use a 4.7V zener and get the detection level (0.574V) with a resistor divider.

Using this approach, you can re-calculate the values to suit your requirements.

MOSFETs would be a better choice for H-Bridge drivers as they would have much less voltage drop across them.

Regards,
Ifixit

 

Hi ifixit, thank you very much. Now is all clear!