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I would like to do a simple undervolatge lockout. I thought a BJT as a switch like this: enter image description here

In normal conditions I would have 30V (and -30 and the othre rail) and 25 mA to 50 mA. These power rails will be used to make the symetric supply of an OpAmp. It's rated for 10 to 45 volts for its +vcc pin (and -10 to -45 and the -vcc pin). A great plus would be to have a rail openned if a fault is detected on the other one.

The idea is: when the voltage "In" dropes, the current in Rsense will drop too and thus the BJT will pass in "OFF" mode because the current in its base is too low. Is my thinking correct?

I'm looking for a solution that can be implemented on a postiive voltage rail and on a negative one for a symetric voltage supply. If you have a better idea (that works for postive and negative voltages). It doesn"t have to be only with BJTs. I'm all ears!

EDIT 20/04: so I did what Math Keeps Me Busy suggested. And it works:

enter image description here green : Vin , blue: Vout

I was wondering if it was possible to have a more snappy response when the 10V are reached?

Here is my complete design: enter image description here

The goal of this thing is to make a split (+/-30V) supply for an OpAmp. The OpAmp is rated for 10 V to 45 V on +Vcc pin and -45 to -10 V on -Vcc pin. It has to be able to handle votlages has high as +/-60V. Q1 is a pre regulator. LM317 is the voltage regualtor. Q2 is for current regualtion in case of a short. Q6 and Q5 are for the Undervoltage Lockout (if that's the correct name). The 1200 Ohm resistor acts like a 25mA load.

EDIT: just to clarify. In the end, I don't need this UVLO for my circuit. I missunderstood the opamp's specs. Voltage under 10 V isn't going to damage it. But the solution proposed by Math Keep Me Busy does work.

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    \$\begingroup\$ Please describe in more detail what you are trying to do (specifically the voltages and currents) as well as the exact application (with schematics if possible) and what the motivation is. UVLO would typically consist of a voltage reference, a comparator and a switch. \$\endgroup\$
    – Spehro Pefhany
    Apr 19, 2021 at 14:49
  • \$\begingroup\$ Your base current will be enough to keep the thing open even at lower voltages. Put enough diodes in series with base to lift minimum voltage. Potential problem happens if your transistor's base is already conducting, but the load wants to pull more current than BJT can pass through. This may or may not be a relevant problem, depending on load current. This is a very simplistic approach anyway, it's simple and cheap, but can't be perfect. Better try P-MOSFET+NPN, where conducting NPN would bring PMOS's gate to ground and pass power through PMOS \$\endgroup\$
    – Ilya
    Apr 19, 2021 at 14:53
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    \$\begingroup\$ That's too simple for most applications. Anything involving just one transistor is going to have too soft a transition for most purposes. You can get a nice snap action with two transistors a zener, and some resistors, but (A) it'll still not be precise (expect 1/4 to 1/2 a volt over temperature) and (B) it'll be bigger than voltage monitor ICs. Tell us what you really need (i.e., what voltage range, do you need hysteresis, etc.), and we'll help. \$\endgroup\$
    – TimWescott
    Apr 19, 2021 at 15:01
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    \$\begingroup\$ Is my thinking correct? No, this circuit is going to behave extremely similar to a diode. \$\endgroup\$
    – Bimpelrekkie
    Apr 19, 2021 at 15:20
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    \$\begingroup\$ I think you want to lockout both rails if either input is low or in other words detect Vin on both before enabling outputs. Some LDO’s have this enable function or shutdown but nothing with memory to lockout, so be clear about your in/out specs. \$\endgroup\$
    – Tony Stewart EE75
    Apr 19, 2021 at 15:34

1 Answer 1

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This is the simplest circuit I can think of using BJTs that will provide under-voltage lockout.

schematic

simulate this circuit – Schematic created using CircuitLab

How it works:

If Vin is higher than the the Zener voltage of D1 plus 1 diode drop (Vbe of Q1) then D1 will conduct, providing base current for Q1. Q1 will conduct providing base current for Q2, and Vin - Vce(Q2) will appear at Vout.

If however, Vin is lower than the Zener voltage of D1 plus 1 diode drop, then D1 will not conduct appreciably, Q1 will not provide base current for Q2, and Vout will float.

R1 and R2 serve to limit the current through the bases of the BJTs.

The Zener and BJTs should be specified to have low leakage current. If the Zener does not, you may need a resistor from the base of Q1 to ground (to divert leakage current).

Note, this circuit has no hysteresis, so if Vin remains close to the threshold voltage, the output may/will bounce on and off.

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  • \$\begingroup\$ This is a nice solution, and it could be adapted to suit something I asked about a couple months ago: How to improve this battery-status (voltage monitor) circuit? \$\endgroup\$
    – Theodore
    Apr 19, 2021 at 15:26
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    \$\begingroup\$ A lockout implies a latch which requires positive feedback from at least 2 inversions \$\endgroup\$
    – Tony Stewart EE75
    Apr 19, 2021 at 15:35
  • \$\begingroup\$ Thanks for your answer. It's working but how can i tweek it so it's more "snapy" when in goes from not passing to passing? For now I have an exponential curve \$\endgroup\$
    – Neeko
    Apr 20, 2021 at 12:01
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    \$\begingroup\$ As mentioned before in case of leakage current, a resistor from base of Q1 to ground can help to shorten the transition. \$\endgroup\$
    – Bravale
    Apr 20, 2021 at 17:04
  • \$\begingroup\$ Also an additional resistor from base to emitter of Q2 will make even faster. \$\endgroup\$
    – Bravale
    Apr 20, 2021 at 17:56

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