3
\$\begingroup\$

I have an application where I need to switch a contactor on and off using a spare output of an off the shelf Battery Management System (BMS)

The input is the signal coming from the BMS, while the output is the signal going to the contactor

Input/Output

  • When the input is FLOATING, the output should be OFF
  • When the input is GROUNDED, the output should be ON

The manufacturer of the BMS provided the following circuit diagram to follow: manufacturer example

Constraints

  • BMS output must not sink more than 175mA
  • BMS output must not have more than 24V across it

The main supply of this circuit is a battery which will have a maximum operating voltage of 35V. Looking through the P MOSFETS available in my country, I can only find a max Vgs of -20V. I have looked at gate voltage protection circuits and tried to adapt the following circuit to work: Adapted Circuit

The circuit I have derived is the following:

Zener diode: BZX85C22V PMOS: Vishay IRF9540PB

My Circuit

My logic is that the 10K resistor acts as the pull-up when the BMS input is floating. When the input is grounded, then the Zener starts conducting and clamps the gate to 22V. Therefore Vgs is -13V (well under the -20V maximum). Iz looks to be about 22mA for the zener to conduct, so I assume let 25mA flow. so then

R2 = (35-22)/0.025 = 520R. Choose 470R, which gives 28mA. Resistor power dissipation is P = I^2 * R = 0.36W (Choosing 0.6W resistors)

Zener dissipation = V * I = 22 * 0.025 = 0.55W (Under the maximum of 1.3W).

My question is that will this circuit work reliably? I am not experienced with circuit design, so your feedback would be very appreciated.

\$\endgroup\$
7
  • \$\begingroup\$ Why the 470Ω pullup resistor? \$\endgroup\$
    – Hearth
    May 13, 2018 at 23:57
  • \$\begingroup\$ @FelthryThe zenor needs 22mA to breakdown. I calculated that 520R is the minimum that will provide this. \$\endgroup\$
    – Russell
    May 14, 2018 at 0:03
  • \$\begingroup\$ You realize this is directly in parallel with your 10k pullup, and thus effectively your pullup resistor is just 470Ω? \$\endgroup\$
    – Hearth
    May 14, 2018 at 0:11
  • \$\begingroup\$ Typically one would use a common emitter with a collector R to limit current with R pullup to act as a Voltage divider to say 15V across source gate so no Zener needed or emitter resistor. \$\endgroup\$ May 14, 2018 at 0:16
  • \$\begingroup\$ @TonyStewartolderthandirt Yes I saw that from the circuit I adapted, but I was not sure how to make it work with the input signals for my application. \$\endgroup\$
    – Russell
    May 14, 2018 at 0:18

2 Answers 2

1
\$\begingroup\$

Your resistors R1 and R2 are in parallel, essentially a single resistor of value slightly less than 470Ω. To be honest I don't really see why this has to be so low, considering you are only driving a MOSFET gate at very low frequency.

I assume when you say that the "voltage across the BMS output" shouldn't exceed 24V, that you really mean the potential of the BMS output shouldn't exceed 24V.

I don't know how you intend to "pull up" the the BMS open drain output using a 35V source, but it will also need protecting:

schematic

simulate this circuit – Schematic created using CircuitLab

Whichever way I look at this, the constraint on the maximum BMS output potential limits that output to 0V and 24V levels, and this whole thing becomes an interfacing dilemma. How do you switch a P-channel MOSFET whose source is +35V using a 0V/+24V signal?

Your own design actually addresses this problem, using a zener diode:

schematic

simulate this circuit

D3 effectively "adds" 18V to BMS_OUT. When BMS_OUT is low, Q1's gate potential is +18V, low enough to switch on Q1. As BMS_OUT rises, Q1's gate rises too, always 18V above BMS_OUT, until it clips at 35V. At that point Q1's \$V_{GS}\$ is zero, and the transistor is off.

Crucially, this means that \$V_{GS}\$ never exceeds \$35 - 18 = 17V\$.

\$\endgroup\$
0
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
2
  • 1
    \$\begingroup\$ How does Q1 behave when a floating input is present? \$\endgroup\$
    – Russell
    May 14, 2018 at 1:40
  • \$\begingroup\$ If the wire is like an induction wire it can be sensitive to turning on with 10uA , but any CMOS driver is low impedance \$\endgroup\$ May 14, 2018 at 1:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.