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From seeing a few schematics where the flyback or snubber diode has been placed across the transistor C-E terminals (Right Configuration), instead of what I typically seen as the the flyback being placed across the coil terminals (Left Configuration).

Which of these are "correct"? Or does each have a separate purpose?

As a note, the diodes are normally listed as external 1N400x type diodes (on TIP120 Darlingtons), not the internal body diode of the BJT or Mosfet.

Final note, I have seen a few schematics that have both diodes, one across the coil and another across the CE terminals. I assume that one is just redundant without really affecting the circuit in that case, is that a wrong assumption?

schematic

simulate this circuit – Schematic created using CircuitLab

The answer to When/why would you use a Zener diode as a flywheel diode (on the coil of a relay)? touches on this slightly, by showing a regular Diode in the above left configuration, while showing a Zener Diode in the right configuration. It doesn't say that the opposite isn't true (or why) So as a second part, can a Zener work in the left configuration, and a regular diode in the right configuration? If so, how does it change how it operates?

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    \$\begingroup\$ Woo Gold Badge 10k views. \$\endgroup\$ – Passerby Nov 9 '15 at 22:17
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Consider the operation of the circuit.

When the transistor is on current is flowing in the coil from top to bottom as the circuit is drawn we now switch the transistor off. The current in the coil still wants to flow.

For the circuit on the left this current can now flow back to Vcc via the diode the voltage across the coil has reversed direction and is limited by the diode the current can decay to zero safely.

For the circuit on the right the diode does not help. The current flowing in the coil will force the voltage on the collector to rise to the point where the transistor (or possibly the diode) breaks down and starts to conduct. At this point the current can start to decay in the coil but the energy in the broke down transistor (or less likely diode) will be excessive and may well result in the transistors death. Note a zener diode here will work because you allow the voltage on the coil to reverse so the current can decay to zero while limiting the voltage across the transistor to a safe value.

It should be noted the allowing the voltage across the coil to reverse to an higher voltage means the current can decay more quickly which is why you sometimes see a zener in the right hand circuit or more than one diode in series in the left hand one.

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A zener can work in both but a diode wouldn't

A zener.

For the left it would just function a diode (with some supply clamping..) For the right it will rapidly discharge the coil (if rated correctly - tvs)

A diode

For the left it will be a normal chopper with a free wheel path. For the right you have a dead transistor

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The latter can't possibly be correct. The induced current flows in the same direction that the original current did, and a reverse-biased junction diode won't help. The voltage that builds up from such a current across the now near-infinite resistance is what damages the transistor in the first place (the Zener works by allowing the current to flow past once the voltage reaches a given maximum). That the transistor is still operational after switch-off in such a configuration is dumb luck.

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The inductor is causing a high voltage peak because it the current path is interrupted. The current will try to find the a new path and until it does, it increases its voltage. Best alternative

The left circuit is the best of the two, it suppresses the voltage spike at the source. If the voltage across the inductor rises, the diode starts conducting until the energy is all dissipated in the circuit.

The right circuit attempts to do the same thing, but relies on power supply having a low impedance path. This is not always true and some voltage regulators don't like reverse current to be fed into its output. Bad alternative.

The zener or MOV alternative suffers from the same issue as the right circuit, it relies on a low impedance path through the power supply. Bad alternative.

I personally don't like 1N400x for this use because it is rather slow. For small currents (<100mA) I prefer a 1N4148 which is much faster. For larger currents I would check one of the various selection guides on Internet.

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    \$\begingroup\$ Can you elaborate a bit more why a zener rated below the Vce breakdown placed across the collector and emitter is not a good approach ? When the relay turns on, current passes through the collector, and when the relay turns off, the collector voltage builds up until the zener breaks down and current passes through the zener to ground. Problem ? \$\endgroup\$ – efox29 Jul 14 '15 at 16:21
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    \$\begingroup\$ @efox29: Using a diode across the coil will cause flyback energy to be dissipated slowly, largely in the resistance of the coil windings. Using a zener will cause flyback energy to be dissipated more quickly, but most of it will go into the zener. Whether this is a good or a bad thing depends upon the thermal characteristics of the coil and the zener. \$\endgroup\$ – supercat Nov 11 '15 at 21:04
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You want to use the circuit on the left (whether you're using a standard diode or a diode+Zener combo) for two reasons:

  1. Some power supplies (just about all linear regulated supplies, actually) cannot sink current, which the second circuit asks them to do. If you try to ask the supply to sink when it can't, the output voltage will rise in an uncontrolled fashion, potentially damaging the supply and anything else connected to it.

  2. Even if the supply can sink current, the left-hand circuit is still superior because the loop area for the dI/dt turn off transient is kept far smaller, keeping it from emitting as much EMI as it would if it went all the way to the power supply and back. This is especially important if you're clamping the back-EMF at a significant value because the resulting EMI will be larger in that case.

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    \$\begingroup\$ I really ought to take a relay, a diode, and a long length of cable and measure how much more EMI is emitted when the snubber network is at the switch instead of the load. \$\endgroup\$ – ThreePhaseEel Oct 5 '16 at 1:05
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The back-EMF incurred by a fast turn-off of coil energizing current causes a rapid collapse of the coil's magnetic field, thus inducing a reverse current equal and opposite the current the coil had charged or saturated to. This negative current will take the resistive paths through which a negative voltage will result.

That danger presented to the switching element is best dealt with quickly and decisively with an anti-parallel, free-wheeling diode across the coil.

This reduces the EMI radiating path length, and simplifies the analysis by keeping the problem between the coil and the diode. That alone avoids any unnecessary reverse voltage breakdown stress on the junction of the driving transistor, as well as avoiding fancy zener selection to try and match the transistor's breakdown threshold, or worrying about spreading the power incurred between a coil and a zener, all of this being dependent upon switching characteristics, duty-cycle, saturation current etc., etc.

With a freewheeling diode, the only power you need to worry about is the power dissipated given the coil/core's maximum saturation current times the forward-biased diode's drop. Secondly, if the coil is going to heat-up by being snubbed, it will be heated at least as much, typically more by its being energized; the snubbing cannot dissipate more energy than the power it dissipated across the time it was energized.

The diode PIV can only matter in the perverse case of a very high supply voltage and a very lengthy, highly resistive coil.

If power dissipation in the diode is a concern at all, the duty-cycle can also be considered, as that may avoid heat-sinking or a constant Pd rating at least as high as the max Pd calculated.

In general, simple is beautiful; additional snubber complexities are generally incurred when trying to minimize switching losses and matching components as closely as possible to get the most out of the most expensive component in the switched loop -- generally the switch itself -- while minimizing the cost of all other, less expensive components in the switched loop, and maintaining EMC.

A more detailed snubber analysis is generally a DFM (design for manufacture) refinement for maximizing a cost-effective, mass-produced product, which invariably puts reliability in the balance, as thermal management defines the rate of long-term breakdown in semiconductor devices.

For prototyping, the freewheeling diode involves the least number of terms in its selection, and is the most direct approach.

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The capacitance presented through VCC means that from an AC standpoint, the cathode of the diode in the left hand diagram is effectively connected to the emitter of the transistor. It would appear, therefore, that there is little difference in the protection provided in both the left hand and right hand diagrams. Mr Dorian Stonehouse.

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    \$\begingroup\$ Except for when the coil generates 10's of volts when the inductor is switched off. In one the overvoltage current goes only through the diode, in the other through VCC and then to ground and back. \$\endgroup\$ – laptop2d Jan 24 '17 at 20:02

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