5
\$\begingroup\$

I have an analog output that I need to protect from continuous overvoltage in case of incorrect connection by amateur installers. For those who are interested, I am currently working on a part of the BMS for a professional lithium-ion system.

The output range is 0V to 15V with an expected load resistance ≥10kΩ that is tied to ground in most cases. The output is not necessarily required to achieve true 0V but it should come as close as possible. It is mostly a question of balancing the effort of an additional negative supply and the expected benefit. The input signal is pure DC. The output's dynamic behavior is not important, except for it being stable.

The analog output consists of a digital to analog converter followed by an amplifier for range adjustment and buffering. The operational amplifier is supplied by the battery voltage (6V to 15.4V minus two schottky diode drops) with the option of doubling the supply voltage through a charge pump if required to achieve the desired output voltage. The circuit is obviously battery powered so the quiescent current should be low. I therefore opted to use an operational amplifier with shutdown and a charge pump instead of a boost converter for supply voltage increase. The protection circuits additional quiescent current should be no more than an additional 100µA when the amplifier is enabled and no more than 10µA when the amplifier is disabled.

The expected overvoltage is continuous and in the range of -16V to 32V (when the output is accidentally connected to a battery pole in a connection of up to two batteries in series).

Although I researched the internet extensively, I did not find a solution for my requirements. In fact, I did not even find something mildly related. So I had to come up with a solution of my own and I would like some feedback from developers with similar problems.

Here is the circuit that I came up with. It works in simulation but I did not test it in practise yet.

schematic

simulate this circuit – Schematic created using CircuitLab

Datasheets: OPA, BJT, FET

The components have the following functions:

  • R1 & C1 are the input filter for DAC filtering and loop shaping
  • OA1 is the operational amplifier with internal short circuit and over-temperature protection. The output has ESD protection diodes towards negative and positive supply
  • R2, R3 & C3 are the feedback network for steady state gain setting and loop shaping
  • D1 & D2 block reverse current into the supply and are part of said charge pump for supply voltage doubling
  • C2 is the local supply capacitor and the charge pump's output capacitor
  • M1 blocks current due to negative output overvoltage. R5 biases M1 so that it is active under normal operating conditions. If the output falls more than a diode drop below ground, Q1 disables M1 and therefore blocks excessive current through OA1's clamping diode. R4 can limit the current if Q1 reacts too slow in practice. R6, R7 & C4 are the biasing network for Q1. Since the failure condition may stay indefinitely, the DC resistance must be large to avoid excessive leakage current (large R7). The dynamic base resistance must be small though, to allow fast reaction (small R6 and C4).
  • M2 limits the positive overvoltage to the supply voltage minus M2's threshold voltage. The gate network performs the same function as Q1's. The OPA will sink the short circuit current in this case and go into thermal limitation. It will not die but likely age significantly. Maybe I will use an additional BJT to get rid of this problem.

Please criticize my circuit. I am open for all types of improvements or alternative solutions. I was looking for an integrated solution, either integrated in an OPA or as a separate IC, but could not find any.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ What is the maximum allowable added resistance to the op-amp output? Why can't you just use a resistor and a TVS? \$\endgroup\$ Commented May 3 at 14:24
  • \$\begingroup\$ @Spehro'speff'Pefhany Fair point. I wanted the output to have as low internal resistance as possible to maximise the available voltage swing without needing a negative bias generator and using the charge pump as little as possible. I did not mention it specifically but the circuit should be really low power since it is a battery powered application, obviously. I will reflect that in the question. \$\endgroup\$ Commented May 3 at 16:33

1 Answer 1

4
\$\begingroup\$

There are no "rail to rail output" op-amps when the output is loaded. A reasonable control voltage output should be able to supply say +/-5mA while maintaining voltage control. So, you will need supplies that go past the 0..+15V output range, no way around that. A small switcher will gladly generate the supplies you need - both positive and negative with single inductor. There's lots of such circuits out there to be inspired from.

I'll leave the critique of your circuit for other answers. It looks like it should work.

To show what else could be done about it, we can investigate several other approaches to this problem.

Simple Output Clamp

When the output voltage goes below -0.7V, or above +15V, the unipolar TVS1 starts to conduct and keeps the op-amp output within safe voltage range. PTC1 initially has low resistance. As soon as the TVS conducts, PTC1 heats up, its resistance rises - since its tempco is positive - and the PTC limits the current flowing through the TVS.

schematic

simulate this circuit – Schematic created using CircuitLab

Load Powers the Op-Amp

Since the op-amp can take 40V between supply rails, we can use the output overvoltage to power the op-amp. In the circuit below, the op-amp's supply limits won't be exceeded in either over-voltage condition on the output: 32V+5V<40V and 20V+16V<40V.

schematic

simulate this circuit

Current Steering Bridge

A diode bridge can steer current away from the op-amp in an overvoltage condition. The current sources Q1,Q2 and Q3,Q4 limit the output current to about 6mA. Once the bridge output forced beyond the output range of the op-amp, the bridge disconnects the op-amp from the output, and one of the current sources takes over driving the output current.

schematic

simulate this circuit

Current Limiter #1

The op-amp's output current is split between two current mirrors, whose output voltage range is larger than the op-amp's output range. R5 and R6 load the op-amp and limit the op-amp - and thus output - current to about 5mA. The output of the current mirror is loaded with 50kΩ, and sets the maximum output impedance.

The output stage is inverting. C4 should be zero, not the value shown.

schematic

simulate this circuit

Current Limiter #2

An absolute current limiter can protect the op-amp, ensuring that the op-amp always has enough capacity to keep its output within the supply rails, no matter the load voltage.

schematic

simulate this circuit

Q12's B-C junction protects Q11's B-E junction from excessive reverse voltage. R5 shunts Q12;s B-E junction and thus limits the B-E reverse voltage as well. In typical transistors, the B-E junction can take about 5V of reverse bias before it breaks down.

\$\endgroup\$
1
  • \$\begingroup\$ You have some good recommendations. Thanks you very much for the effort. The active solutions are pretty fancy and there function is well explained. Solutions 3 and 4 come at the cost of high quiescent power as far as I can see. Solution 5 is very nice. I already thought about implementing a bidirectional current limitation but could not come up with such a sleek solution. I will certainly keep this one in my repertoire. Solutions 1 and 2 seem to be the most practical for my solution though since I am looking for the lowest quiescent power possible. \$\endgroup\$ Commented May 4 at 10:50

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.