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I need a soft-start circuit for an electronic load to avoid sparks at the input terminals when a power source is connected (in this example when SW1 is closed).

Below circuit diagram is working fine in the simulator but in reality OA1 output (NODE1) voltage doesn't go all the way down to negative rail (-10 V) and outputs ~3 V:

schematic

simulate this circuit – Schematic created using CircuitLab

Any idea why OA1 output doesn't clip to negative rail?

It's obvious that OA1 inputs both are connected to ground when OA2 activates M2 gate.

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    \$\begingroup\$ Supply rails? Supply rail connections? \$\endgroup\$
    – Andy aka
    Commented Sep 4, 2021 at 19:22
  • \$\begingroup\$ @Andyaka ±10 V independent of the 'source' with a common ground. \$\endgroup\$ Commented Sep 4, 2021 at 19:23
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    \$\begingroup\$ They need to be grounded to the source. Please don't make this hard i.e. show all power connections. \$\endgroup\$
    – Andy aka
    Commented Sep 4, 2021 at 19:24
  • \$\begingroup\$ @Andyaka Done.. \$\endgroup\$ Commented Sep 4, 2021 at 19:30
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    \$\begingroup\$ What is Vsource? What is Vref? \$\endgroup\$
    – Andy aka
    Commented Sep 4, 2021 at 19:42

3 Answers 3

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in reality OA1 output (NODE1) voltage doesn't go all the way down to negative rail (-10 V) and outputs ~3 V

When the voltage reference is bypassed to 0 volts via M2, the op-amp is then trying to regulate the output voltage to 0 volts. That's the first thing to remember. So, if the output of MOSFET M1 is purposefully controlled to be 0 volts, the gate voltage will naturally be a little higher than 0 volts because the MOSFET is a source follower. 3 volts isn't unexpected in these situations.

But, it wouldn't be unexpected to find that the gate voltage would be a little negative either; it all depends on M1's characteristics and circuit leakages. In other words, if you built several of these you might find a range of gate voltages from a full negative 10 volts to something around 3 volts positive.

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  • \$\begingroup\$ The second solution in my answer to the question, is it any good? \$\endgroup\$ Commented Sep 5, 2021 at 9:03
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    \$\begingroup\$ If OA2 has a negative 10 volt supply rail then it looks likely to work just as well as the first solution. \$\endgroup\$
    – Andy aka
    Commented Sep 5, 2021 at 9:06
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You designed a delay circuit with OA2, not soft start. If you add a capacitor across M2 source and drain, you will have a soft start capability plus delay. The capacitor will also reduce noise at that node in a real circuit.

I assume you're trying to make a 1 V regulated supply. You'll probably need a capacitor (on the order of 100pF to 1nF) across R4 to stabilize the loop due to wiring and opamp input capacitance.

As for your output at Node 1, you are trying to make a zero volt regulator. The opamp OA1 will produce a voltage to maintain approximately zero volts which depends on the opamp input offsets.Thus, seeing 3V at Node 1 may be perfectly normal for the offsets in your actual part (it could be trying to maintain a 10mV output voltage for zero volts).

You can change the -10V supply to zero volts if you use an opamp where the input voltage range extends below ground in a single supply application. An AD823 has this capability.

In a real circuit, you'll probably need an input capacitor at the drain of M1 to ground which will create an arc across your switch. An RC snubber across your switch will help.

As a linear regulator, you are dropping 29V across M1. At 10A, that's 290W of dissipation. Very low efficiency, i.e, not practical. For a low voltage, high current supply, you are better off using a switching supply. You'll end up with other difficulties, depending on topology, with a switching supply since your input voltage is quite a bit larger than your output voltage.

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    \$\begingroup\$ "As a linear regulator, you are dropping 29V across M1. At 10A, that's 290W of dissipation." - That's by design. The soft start circuit controls an 'electronic load' which is a constant current sink. Must be a big FET to handle that power! \$\endgroup\$ Commented Sep 4, 2021 at 22:56
  • \$\begingroup\$ How adding a capacitor across drain and source avoids sparks at SW1? it will increase the spark instead! it's a constant current source not a voltage regulator! \$\endgroup\$ Commented Sep 5, 2021 at 5:39
  • \$\begingroup\$ @BruceAbbott "Must be a big FET to handle that power!" Eight parallel MOSFETs on a big wind tunnel. \$\endgroup\$ Commented Sep 5, 2021 at 5:45
  • \$\begingroup\$ @ElectronSurf I simulated your circuit in LTspice and it did not have the desired soft turn on. I then moved C1 from the Gate of M2 to the Drain, and it did have a soft turn on, taking ~50ms to reach full current. \$\endgroup\$ Commented Sep 5, 2021 at 6:26
  • \$\begingroup\$ @BruceAbbott Did the OA1 output clip to negative rail? \$\endgroup\$ Commented Sep 5, 2021 at 6:27
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I solved the problem by pulling down the reference voltage to a negative voltage:

schematic

simulate this circuit – Schematic created using CircuitLab

Another simple solution:

schematic

simulate this circuit

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