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I was wondering if I was correct in assuming these diodes were being used for ESD protection on the input of a power circuit. I'm a little bit confused what the point of the diodes are when there's a circuit breaker before them too if anyone has any idea.

The breaker is rated at 5 amps which is less than the maximum rating of the diodes. The only thing I can think of is the diodes are handling up to a certain level of transient and if the transient becomes too large the circuit breaker will kick in and protect the diodes and obviously the rest of the circuit as well.

I know usually there is a resistor in this type of network from what I've read online but there doesn't seem to be one in this setup which I was also confused about.

schematic

simulate this circuit – Schematic created using CircuitLab

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There is a mechanism that kicks the circuit breaker into action, and it relies on current (not voltage) in the traditional circuit breaker. When the current reaches a certain level a coil triggers the spring loaded switch to turn off. Some breakers also have a bimetallic that triggers when it gets to hot in response to current.

enter image description here
Source: https://electronics.howstuffworks.com/circuit-breaker2.htm

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Circuit breakers are mainly intended to prevent fires. They don't react quickly enough to block ESD pulses, and they don't react quickly enough to prevent many failure modes due to over-current in the protected equipment. What they do is prevent excess currents from flowing for minutes, hours, or days, which could overheat the protected wiring or equipment to the point of igniting nearby materials and burning a building down.

ESD protection diodes can react much more quickly, absorbing energy from an ESD pulse that might last only nanoseconds or microseconds.

Depending on the kind of power source, the diodes in your circuit might be protecting the load from a reversed power source. If the power source is reversed, a high current will flow through those diodes, which will trip the breaker. The diodes having higher current rating than the breaker threshold suggests this is more likely to be the function of the diodes, rather than ESD protection. (Yes, this would be using the breaker for a slightly different function than its usual one of fire prevention)

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  • \$\begingroup\$ What types of sources can experience reverses like that? Right now the source is 28 VDC from facility power. Thank you for the help! \$\endgroup\$ – Paul Sep 25 '19 at 16:29
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    \$\begingroup\$ @Paul, You'd most often use reverse protection for a battery powered device. You might also want it if the power is hooked up through terminal blocks or some other way where it could be miswired. \$\endgroup\$ – The Photon Sep 25 '19 at 16:43
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With the voltage source connect as shown in the schematic, the zener diodes are reverse biased and will (try to) clamp the input voltage to about the sum of their individual zener voltage.

If for example D1 and D2 are a 4.7V zener and D3 a 6.2V zener, the clamp voltage will be about (4.7+4.7+6.2)V = 15.6V.
If the input voltage is standard 15V, the zeners won't be clamping.
If the input voltage would become 30V, the zeners enter the break down, clamping the the input voltage at 15.6V or bit higher, depending how much current is running through the zeners.

Depending on the output impedance of the voltage source, a very high current can run through the zeners. Therefore, a circuit breaker is added to prevent the zeners will burn due to a too high current. If the zener would get damaged, they likely fail open and the circuit will still see the 30V of my example above.
That is the reason that the circuit breaker is rated less than the (lowest) maximum rating of the zener diodes, so it trips before the zener diodes get damaged.

The zeners also serve a function as reverse polarity protection (as The Photon already pointed out). Then, the zener diodes clamp the (negative) voltage at about the sum of their individual forward voltage. Again, this could cause a very high current to run and the circuit breaker should act before the zener diodes get damaged.

Why a circuit breaker instead of a resistor?
A resistor could also be used instead of shown circuit breaker. When the input voltages becomes higher than the clamping voltage, the zener diodes will try to clamp again, causing a current. This current also flows through the resistor which causes a voltage drop, such that the input voltage minus this voltage drop becomes about the clamping voltage again (or a bit higher depending on the current).

A draw back of using a resistor is that it will always cause a voltage drop (depending on the load current), even when the input voltage is below the clamping voltage.
For that reason, a circuit breaker may preveal over a resistor.

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  • \$\begingroup\$ This is what I thought originally! The Zener breakdown V total of the 3 diodes is around 10x higher than the ideal supply voltage in this case which is why I thought it was ESD protection as it would rarely be clamping with such a high avalanche region requirement. \$\endgroup\$ – Paul Sep 25 '19 at 18:25
  • \$\begingroup\$ Doublechecked diode datasheet and the diodes are actually clamping in total to only 3x the ideal input voltage which makes a lot more sense. \$\endgroup\$ – Paul Sep 25 '19 at 18:40
  • \$\begingroup\$ @Paul 3x the ideal input: I still would call that high. With a rated input of Y volts, having a tolerance of 10%, I would expect the clamping voltage at e.g. 1.5*Y volts. Is the maximum input voltage of your application defined? Would that voltage times about 1.5 match the claming voltage? \$\endgroup\$ – Huisman Sep 25 '19 at 19:58

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