I am using a low side driver from Infenion BTS3080EJ. Details here. The input from the microcontroller will drive the IC BTS3080EJ which will drive a contactor (12V, 1.3A). I want to know whether I should use an optoisolator in between the microcontroller and BTS3080EJ . Is it needed? The circuit provides features like overvoltage prrotection, short circuit detection and current limiting, thermal shutdown. But signal isolation is not provided by BTS3080EJ. If I have to use an isolation, what should I use? Will Digital isolators be good? Please help enter image description here

  • \$\begingroup\$ From what do you want to protect the microprocessor? Or from what needs the microprocessor to be isolated? \$\endgroup\$ – Huisman Apr 17 at 20:33
  • \$\begingroup\$ I want ground isolation between DC-DC,7805,microcontroller, relay and battery. Since relay is using battery ground and microcontroller is using ground of DC-DC converter, Do I need to use an optocoupler to isolate these two ? I am not well versed with the concept of ground isolation and keep getting confused. \$\endgroup\$ – Senorita Apr 19 at 3:36
  • \$\begingroup\$ Is "relay" which you're commenting about the same thing as "contactor" in the block diagram? \$\endgroup\$ – Nick Alexeev Apr 19 at 21:34

If you needed galvanic isolation, you could use e.g. an optocoupler. However, because your microprocessor and contactor share the same power source, there will not be any galvanic isolation.

You probably want to prevent ground loops.
Wires and PCB traces have an inductance and resistance. You could represent this impedance in every line wire in your diagram between two wire connections. In the schematic you provided it would look like this (I only drew the impedances in the main current paths)

enter image description here

Every impedance contributes to a voltage drop. The return current of the contactor flows through Z3, Z4, Z5 and Z6 and causes for each a voltage drop.
Now, if your low side driver were close to the actuator and both would be far away from the optocoupler, Z3 would be big. When a big current would flow through the Contactor, it could cause a not to be neglected voltage drop across Z3.
So, while you would think you would drive the low side driver with \$V_B-V_A\$ you actually drive it with a lower voltage being \$V_B-V_A-V_{Z3}\$ (with \$V_{Z3}\$ being the voltage drop across Z3).
Do note the use of an optocoupler will not prevent this.

Luckily, the current through the contactor is relative small (1.3A) and moreover, the BTS3080EJ has an internal gate driver, so the input of the BTS3080EJ is probably not sensitive to variation in voltage \$V_B-V_C\$.

If you still want to avoid the discussed voltage drops, you should be using one (or more) starpoint(s).
In the picture above the return current of the contactor flows through Z3, Z4, Z5 and Z6, each causing a voltage drop. So, each impedance lifts the ground locally with respect of the 'real' ground defined in the left lower corner. E.g. for the local 'ground' voltages \$V_C\$ and \$V_A\$ applies \$V_C > V_A > 0V\$.

By using a starpoint you can get rid off Z3, Z5 and Z6 as shown in the picture below. Z4 does not affect the controlling signals coming from the microprocessor as it has the same (local) ground with the BTS3080EJ. (I removed the optocoupler, but as discussed above).

enter image description here

Schematics are just for visualisation. Of course, the real starpoints should be considered in the PCB layout.

Hint/additional note
Try to keep trace Z4 as small as possible. Z4 could lift the ground of the microprocessor. If the microprocessor gets other (analog) signals wrt 'real' ground as well, Z4 can become quite disturbing.
I could have flipped around the DC/DC converter and the 7805 in the schematic such it is closer to the starpoint, but then the schematic probably becomes unreadable.
You should consider the use of starpoints and take care of trace lengths in the PCB layout anyway.

  • \$\begingroup\$ Thank you for thedetailed explanation. It was very helpful. \$\endgroup\$ – Senorita Apr 29 at 4:38
  • \$\begingroup\$ I however have another doubt. I am using BTS3035EJ. Its output is same as its cooling pad. There is no separate pin for putput. How should I take the output from it? \$\endgroup\$ – Senorita Apr 29 at 4:40
  • \$\begingroup\$ @Senorita You should treat the cooling pad as pin. So, just connect the cooling pad with a wide trace. \$\endgroup\$ – Huisman Apr 29 at 19:23
  • \$\begingroup\$ BTW, note that Figure 46 of the datasheet also shows the starpoint connection such that the high current holding wire (wire is drawn thicker) leaving the BTS3035EJ is as short as possible. \$\endgroup\$ – Huisman Apr 29 at 19:23
  • \$\begingroup\$ Thank you. This was really helpful \$\endgroup\$ – Senorita May 1 at 2:08

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