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I am designing a piece of test equipment that has a DAC output. The test equipment might be used to troubleshoot UUTs that are damaged or malfunctioning. Therefore I want to add a protection circuit to the DAC output. The DAC circuit itself is already designed.

The question is what's the cheapest protection circuit for this test equipment?

  • I already looked at the Bourns TBUs but the smallest ones they had had a current limit of 100mA (which is too large). They also had rather large output impedance.
  • I already looked at PTC poly-fuses, but I couldn't find any that were rated for > 100V, had a hold current > 3mA, and a trip current < 10mA. Also poly-fuses with very low trip current have very high output impedance.
  • I came up with this circuit, using two depletion mode MOSFETs to form a bi-directional current limiter. I then reduced the apparent output impedance using an op-amp. This seems to work but it costs several dollars and I was hoping to do better price-wise.

enter image description here

These are the design goals:

1) The output range of the DAC is 0.0V to 5.0V.

2) The DAC output bandwidth is 1KHz max.

3) The protection circuit sits between the DAC output and the connector pin on the test box.

4) The voltage present on the output of the protection circuit should track the input to within ±2.5mV when the output current is within the range of ±3mA and the input voltage is between 0.1V to 4.9V.

5) When the input voltage is outside the range 0.1V to 4.9V then the protection circuit output voltage is not required to track the input voltage.

6) When the output current is more than ±3mA the output voltage is not required to track the input voltage.

7) The DAC output should survive a direct short to +33V DC indefinitely.

8) The DAC and protection circuits should survive intermittent voltage spike of ±100V occurring for less than 1 second not more than once every 10s at the UUT pins.

9) During a fault the protection circuit shall not allow more than 10mA to pass through the connector pin on the tester for shorts up to ±100V.

10) After an over-voltage or over-current fault is removed from the pin the protection circuit should be un-damaged and revert to normal operation within 10 seconds (faster is better).

11) The box is sealed and the user cannot open it to replace any parts or disposable fuses in the protection circuit.

12) ESD is handled separately and not considered here.

13) Protection circuit should not occupy an area larger than 20mm x 20mm on the PCB, and it shall not be more than 20mm tall.

I already have supplies 3.3V, +5.0V, and ±15V available for the protection circuit to use. For cost and space reasons I would like to avoid having to add extra power supply voltages in the tester.

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  • \$\begingroup\$ Why not just replace those FETS with a 1Kohm, 2 watt resistor? \$\endgroup\$ – analogsystemsrf May 3 at 4:11
  • \$\begingroup\$ "it costs several dollars" and your rather tough technical specification don't add up at all. It sounds as what you are looking for are "smart MOSFET drivers" (check Infineon), but those are relatively expensive. \$\endgroup\$ – Lundin May 3 at 7:02
  • \$\begingroup\$ Cost several dollars in what volume made where? (serious question). Any sort of volume made in China should be well under $1. Small volume made in US, yes. || To get 2.5 mV tracking at up to 3 mA a series resistor must be R=V/I <= 0.0025/0.003 <= 0.83 Ohm. Clamping the DAC end of this R to ground and 5V with schottky diodes would protect the DAC - but you'd need a P=V^2/R ~~= 33^2/.8 = 1.4 kW+ resistor :-). BUT that should give you time to accommodate a slower protection system. Sadly, perhaps, relays are still extremely cost effective in many switching applications. ,,, \$\endgroup\$ – Russell McMahon May 3 at 12:25
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    \$\begingroup\$ ... A circuit that senses out of range Vtest and opens a relay contact while the voltage remains may allow a cheaper than FET solution. As above, knowing volume and manufacturing location are pivotal to supplying a good answer. \$\endgroup\$ – Russell McMahon May 3 at 12:26
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    \$\begingroup\$ (past) Bedtime here - I'll perhaps expand on this in due course if nobody else does. The DAC is followed by a buffer amplifier and as large a resistor as will allow you to meet output spec. The "trick" is to sense Vout at the output end of the resistor so the buffer amp can swing as high or low as needed to bring the output to the desired voltage. The resistor max value seems to be be constrained by the need to deliver 3 mA with Vout anywhere in the 0-5V range. ... \$\endgroup\$ – Russell McMahon May 3 at 13:50
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Follow the DAC with a buffer amplifier and as large a resistor as will allow you to meet output spec.
The resistor protects the amplifier.
The "trick" is to sense Vout at the output end of the resistor so the buffer amp can swing as high or low as needed to bring the output to the desired voltage. The resistor max value seems to be be constrained by the need to deliver 3 mA with Vout anywhere in the 0-5V range.

So eg with 15V available and Vout at 3V,
Rseries max = V/I (15-3)V/3mA = 4K.
Say 3K3.
Now clamp the buffer amp output at about it's supply rails with diodes and the resistor must tolerate P = V^2/R = ((33-15)^2/3K3 = 100 mW!

100V transient output excursions can be tolerated.
(100-15)^2/3k3 ~= 2.2 W. Size resistor to suit max dissipation and duration.

Example

R1 + R2 protect DAC.
R1-R2 junction is clamped to supplies by D1 & D2 to prevent overswing from fault output.
Opamp input is clamped to supplies by D3 & D4 to prevent overswing.

R1 set to max load tolerable by opamp.
R1 + R2 total set to minimum acceptable under fault conditions.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ +1 I would have been wary about this circuit without the second set of protection diodes. As it stands it looks like a very cleaver solution if the specifications can be met. \$\endgroup\$ – KalleMP May 7 at 6:45
  • \$\begingroup\$ @RussellMcMahon This idea best answers my question, since it has the lowest cost. But I actually liked your idea of sensing the output voltage and just opening a relay. I came up with something based on that idea using a 400V SSR part that is already in the design. That method can withstand faults way beyond the original goal with very minimal power dissipation during a fault. \$\endgroup\$ – user4574 May 7 at 14:57

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