Operating mechanism of an SBD within a circuit (switch protection circuit, motor return circuit)

I'm just an amateur, but I'll do my bestLast time we covered the mechanism and characteristics of SBD itself.Was introduced.

This time, we will explain the characteristics of SBDs using an electric airsoft gun, which is a representative example of a simple DC motor electrical circuit."How can we utilize this and incorporate it into the circuitry of an electric airsoft gun?"I would like to consider this further.

Let's start by considering the circuit of a standard electric airsoft gun that doesn't have an SBD (Small Block Diode).

Simple circuit for a DC motor without an SBD

Let's start by considering the circuitry of a typical electric airsoft gun.

It's quite simple, consisting only of a battery for power, a mechanical switch, and a DC motor.

The trigger and switch are connected by a mechanism, and the switch is simply turned on or off. (DC motor refers to a motor that uses direct current.)

Next, let's consider what happens when the trigger is pulled, that is, when the switch is connected.

Switch on

When you pull the trigger, the switch engages and electricity flows to the DC motor.

This causes the DC motor to rotate, moving the gears and pistons, which expel air and fire the projectile. In other words, the gearbox moves.

Next, let's consider the case where the switch turns off when the trigger is released.

Switch off

When you release the trigger and switch it off, the power from the battery is cut off.

Here, the diagram is the same as the first one.The crucial difference is that the motor continues to rotate due to inertia.It's like how a car can't stop suddenly.

Now let's consider what happens when a DC motor rotates due to inertia.

The DC motor is rotating due to inertia (switch off).

What happens when a DC motor rotates due to inertia is that it acts as a generator, producing electricity.

The mechanism of a DC motor will be explained later, but when power is supplied, a DC motor rotates according to Fleming's left-hand rule.

Conversely, when a DC motor rotates due to some force, it generates electricity according to Fleming's right-hand rule.

The difference between the left and right hands lies in the direction in which electricity is generated.

When generating electricity, the direction of the current flow is opposite to the direction of the electricity supply, so the electricity will try to flow as shown in the diagram.

In other words, a DC motor can be described as an energy conversion device (its direction also changes). (Electricity → Kinetic energy, Kinetic energy → Electricity)

The power generated by a DC motor is called back electromotive force Vm [V]. I don't understand why the unit is V (volts) despite the name "power." Normally, power is measured in W (watts).

The fate of the back electromotive force of a DC motor

Now that we know that a DC motor generates a back electromotive force Vm, let's consider what happens to the generated power.

In the previous diagram, since the back electromotive force has no destination other than the switch after it is generated, the back electromotive force first goes to the switch.

Since the switch is physically separated, the electricity is essentially cut off. However, the generated power will try to be consumed in some way.

Whether it's kinetic energy or electrical energy, energy, much like me, tends to flow in the direction of ease—that's the law of physics in this world.

Furthermore, like the author, they have an exceptionally high ability to find easy solutions. And they try to be consumed as quickly as possible.

Returning to the circuit, the easiest way for power to flow is actually by jumping across the space between physically separated switches.

Compared to other methods of power consumption, the distance between the electrodes of a switch is short, making it one of the easiest ways for electricity to conduct power through it.

In particular, at the moment the switch is turned OFF, that is, immediately after the terminals are separated, the distance between the switches is short, making it relatively easy for electricity to flow.

Of course, not all of the back electromotive force is converted into electricity; some is converted into heat in the wiring and motor. However, the energy is so enormous that this alone is not enough to consume it, so the excess power is passed through the switches.

In that case, how do we think about the space between the switches in the circuit? We think of it as a resistor with an extremely large resistance value.

Because the resistance R is extremely large, even if the current flowing through it is somewhat small, the generated voltage will be extremely large.

This results in a phenomenon almost identical to lightning, which you are all familiar with. This is apparently called a spark discharge.

At that time, most of it is converted into heat and consumed (it becomes a spark). This heat (spark) burns dust and dirt in the air and sticks to the switch, and because the switch itself is made of metal, the heat accelerates its reaction with oxygen in the air, causing it to rust.

This double whammy causes the electrodes to deteriorate, eventually rendering them unable to function as a switch. If it progresses further, the electrodes (metal parts) of the switch will melt.

Furthermore, a very small amount of power that is not converted into heat is forcibly supplied to the battery. This is because the power is at an extremely high voltage due to the spark discharge, so the battery voltage of 8-12V is equivalent to 0V and flows through easily (to them, it's a voltage that's within the margin of error).

This can damage the battery, shortening its lifespan or causing it to break, whether it's a LiPo or NiMH battery.

I'm sure you're all wondering, "Can such a small DC motor really generate that much power?" So, I'll try a simple calculation, although it might be wrong.

Voltage across switches due to back electromotive force of a DC motor

We'll explain the mechanism of DC motors in detail elsewhere, so we'll skip over that for now.

First, let's consider an example setup: a standard electric airsoft gun with no recoil, using 0.2g BBs, with a muzzle velocity of around 90m/s and firing about 15 rounds per second (assuming a gearbox gear ratio of 18), powered by a 7.4V LiPo motor. (Assuming a standard rod motor from a major domestic manufacturer.)

It feels a bit strange, like I'm doing reverse engineering on a certain country's system, but I'll continue.

First, let's consider the kinetic energy W that the motor possesses when it is switched off.

This can be expressed in terms of angular kinetic energy.

Let I be the moment of inertia of the rotor of the DC motor, and ω be the motor's speed.

$ W =\frac{1}{2}Iω^2 $

Therefore, the motor rotor diameter is approximately 20mm, the height is 40mm, and the material is iron with a specific gravity of 7.85, resulting in a weight of about 30g.

Also, with a ratio of 18.72 and 15 shots per second, the motor's rotation speed is 280.1 revolutions per second, so ω is 561.6π (rad/s).

If we convert the units to the MKS system and perform the calculation from here...

$ W=11681[J] $

This is because each speed is per second, so it can be expressed as power per second W. If we assume that a semi-automatic firing rate of 15 rounds per second is possible, the total time for one round is 0.067 seconds, so the kinetic energy Ws of the motor is

Ws = 11681 × 0.067 = 782 [J]

Let's assume it's approximately 800 [J].

This work is measured in power (W) and can be calculated using the formula V (voltage) × I (current).

If we look at the large resistance of the spark discharge between the switches, conservatively speaking it is 0.1 MΩ, so from the relationship between voltage, current, and resistance...

$ V=RI=0.1×10^5I $

Since the DC motor has 800[W] of energy,

$ 800=V×\frac{V}{0.1×10^5}, V=2830[V] $

2,830Vになります。

This doesn't have an absolute value meaning, but I think the order of magnitude is correct. Even with this conservative estimate, it will still be a sufficiently high voltage.

However, it is consumed instantly and converted into heat.

You'll see that a battery voltage difference of around 12-13V is within the margin of error.

From here, we will consider a circuit with an SBD (Screen Block Diode).

Circuit with SBD

First, let's review the circuit symbol for an SBD.

Let's use this to design a circuit.

The problem with the circuit is degradation caused by spark discharge between switches due to the back EMF of the DC motor, so we will create a circuit that returns the back EMF to the motor.

This results in a circuit that looks like this.

Please pay attention to the orientation of the SBD.

Please be careful during the actual installation. While making a mistake won't cause major problems (only the SBD might break), it will render the entire system useless.

Next, let's consider how the SBD works when the problematic electromotive force is generated.

The fate of the back EMF of a DC motor with an SBD

I explained that energy flows wherever it's easiest.

As explained in the previous post, an SBD conducts electricity when a voltage higher than the voltage drop VF is applied (VF is usually less than 1V).

On the other hand, the resistance of a switch that generates a spark discharge is extremely high, so the generated back electromotive force goes to the SBD rather than the switch (electricity flows much more easily through the SBD than through the switch when it's off). When electricity flows in the direction of the SBD, it returns to the motor.

This is called recirculation to the motor, and since electricity basically does not flow to the switch, the switch will not produce spark discharges.

However, a small current will flow when the back electromotive force is lower than the voltage drop VF of the SBD. This small current can cause a spark discharge, but the amount of energy is overwhelmingly lower than without an SBD, so the amount of spark is greatly reduced. This extends the life of the switch.

On the other hand, the electricity generated by the DC motor is returned to the motor by the SBD, so the DC motor itself consumes power and stops rotating.

This is the basic concept behind switch protection using SBDs.

Next, let's consider what the SBD does when the switch is turned on.

How the SBD works when switched on

So, what does the SBD do when the electric airsoft gun is operating and the switch is on? Well, one of the undesirable characteristics of the SBD is that leakage current (IR) flows.

Moreover, most of this is converted into heat.

In other words, the power obtained by multiplying this leakage current IR by the voltage V applied to the DC motor (the voltages are the same because it's a parallel circuit) is converted into heat.

A simple calculation shows that a fully charged 7.4V LiPo battery is 8.4V, and if all of that voltage is applied to the DC motor, the same voltage (in parallel) is also applied to the SBD. Therefore, calculating with 100% heat conversion, the heat energy Q = 8.4 × IR [J]. (This depends on the LiPo's discharge capacity and the SBD's heat conversion rate.)

The leakage current IR depends on the voltage applied to the SBD, but it seems to be around 0.01 [A]. If a voltage of 8.4V per second is applied, the heat generation Q will be 0.084 [J/s]. For reference, cal = 4.18J, and 1 cal is the amount of heat required to raise the temperature of 1g of water by 1°C.

While this may not seem like a big deal at first glance, the heat generated by the DC motor, the heat generated by the SBD itself, and the increased leakage current IR (a characteristic of the SBD) due to the heat can create a vicious cycle that causes the temperature to rise even further, ultimately leading to the SBD being destroyed by heat.

Please refer to the previous explanation for details on these characteristics (such as the increase in leakage current due to heat).

In other words, you should be careful about situations where there is an excessive and continuous flow of electricity, such as prolonged firing in full auto mode with an electric airsoft gun.

Furthermore, when the load on the DC motor temporarily increases due to some factor (such as piston failure or a jam), the LiPo battery will exhibit its maximum discharge capacity.

For example, a 7.4V 30C 2000mAh LiPo battery (mini size equivalent) can supply 30[A/h] x 2000[mAh] = 60[A].

The internal resistance of a DC motor in the class of an electric airsoft gun is usually around 2-3 ohms, so if 60A flows through it, a voltage of 120-180V could be applied to the SBD (estimated value for a typical motor from a major domestic manufacturer).

Therefore, if you don't select an SBD that can withstand voltages of 120-180V, the SBD will break (SBD reverse voltage tolerance characteristic). In fact, when you look into the voltage tolerance characteristics of SBDs, it seems that there are SBDs with a reverse voltage tolerance of less than 200V that are of good quality.

However, be careful when selecting an SBD with unnecessarily high reverse power tolerance, as this can increase the leakage current (IR) and voltage drop (VF) due to the SBD's characteristics, generating heat.

Furthermore, if the SBD has high reverse power tolerance, the current recirculation reaction rate when the circuit is switched off will be sacrificed. If the recirculation reaction rate is slow, the current will reach the switch before it can flow through the SBD, causing a spark discharge, which defeats the purpose of the SBD, so care must be taken.

It appears that there is a trade-off between performance characteristics such as reverse power resistance, leakage current, and recirculation reaction rate.

Therefore, since SBDs are designed for recirculation, selecting an SBD based solely on the back EMF scenario can lead to damage. Consequently, it seems that the firing conditions of the electric gun (circuit switch on) must also be carefully considered when selecting an SBD; otherwise, it's pointless.

Heat generation in SBD due to back electromotive force

Just as an SBD generates heat when switched on, it also generates heat due to the back electromotive force when switched off.

Especially with ultra-high-performance motors, the high performance generates more kinetic energy, resulting in a larger back electromotive force and an increase in the current passing through the SBD.

Even in the forward direction, an SBD loses a voltage drop VF, and this loss is converted into heat. This means that it generates heat, and if a large current flows through it, there is a high probability that it will fail due to the large amount of heat.

Even with a normal motor, excessive semi-automatic rapid firing will continuously generate heat due to back electromotive force, so I think it would be quite stressful.

In short, SBDs generate heat and can break down regardless of whether they are switched on or off, so caution is advised. In the case of electric airsoft guns, excessive semi-automatic firing or prolonged full-automatic firing should be avoided.

Summary

While installing SBDs is easy, choosing the right SBD is a rather complex topic.

Click here for the SBD installation article.

If you're using an electric airsoft gun and the grip (motor) feels hot, the SBD (Speed ​​Bullet Diode) may already be broken.

In that case, countermeasures such as installing a high-capacity MOSFET, reducing the motor load (by using a weaker spring), or switching to a more efficient motor may be necessary.

I've written about MOSFETs here, so if you're interested, please take a look →Considering MOSFETs

In any case, the temperature is easily discernible, so taking precautions such as stopping use when it gets hot may help save the SBD.

Furthermore, excessive continuous use in semi-automatic or fully automatic modes can damage not only the SBD but the entire mechanism, so it might be wise to be careful.

Fortunately, even if an SBD breaks, it won't cause the airsoft gun to remain switched on like an FET does. However, this makes it difficult to determine if it's still functioning, which is a drawback.

Thank you for your time.