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I think you understand it well. They are basically telling you that without an added connection (made by you, somehow), their power supply will be ON by default. They then add that if you add a connection to over-ride their default behavior, which they anticipate you may want to do, then you need to be aware of the fact that to over-ride it you have to "sink" (or pull-down) enough current to achieve it.

It's a very modest amount of current. But since they don't entirely know exactly what \$V_{IN}\$ you will be providing, they instead tell you that they have a \$100\:\textrm{k}\Omega\$ resistor there and also that there may be some added current you have to sink (beyond that resistor's requirements) for their added protection circuitry. The upshot of this is that you should be prepared to sink at least \$20\:\mu\textrm{A}\$ per volt and that you should consider the input to look like this:

schematic

simulate this circuit – Schematic created using CircuitLab

(Here, you can easily compute \$R_1=\frac{1\:\textrm{V}}{20\:\mu\textrm{A}}=50\:\textrm{k}\Omega\$.)

If you want to disable the module, you need to make sure that the ENABLE line is pulled below \$600\:\textrm{mV}\$. Just grounding it works fine (of course.) That pretty much guarantees it. But they know that no one is really just going to ground that input. That would be pointless. So they also want to make sure you understand the details enough that you can make sure that whatever you use will be able to "pull down" enough.

Suppose you didn't just ground it but instead hooked up a \$10\:\textrm{k}\Omega\$ resistor from ENABLE to ground. And suppose \$V_{IN}=35\:\textrm{V}\$. Then it's pretty clear that the voltage would be perhaps as high as \$35\:\textrm{V}\cdot\frac{10\:\textrm{k}\Omega}{10\:\textrm{k}\Omega+50\:\textrm{k}\Omega}\approx 5.8\:\textrm{V}\$, which obviously won't work. In this case, you'd need to use a resistor of less than \$872\:\Omega\$, which will sink near \$690\:\mu\textrm{A}\$, to meet the specifications. 

This is what they are trying to get across to you. You might need to sink almost a milliamp in some cases and they want you to be aware of it. They also want you to know enough so that issueif you do want to control it, you can successfully design a reasoned circuit arrangement.

I think you understand it well. They are basically telling you that without an added connection (made by you, somehow), their power supply will be ON by default. They then add that if you add a connection to over-ride their default behavior, which they anticipate you may want to do, then you need to be aware of the fact that to over-ride it you have to "sink" (or pull-down) enough current to achieve it.

It's a very modest amount of current. But since they don't entirely know exactly what \$V_{IN}\$ you will be providing, they instead tell you that they have a \$100\:\textrm{k}\Omega\$ resistor there and also that there may be some added current you have to sink (beyond that resistor's requirements) for their added protection circuitry. The upshot of this is that you should be prepared to sink at least \$20\:\mu\textrm{A}\$ per volt and that you should consider the input to look like this:

schematic

simulate this circuit – Schematic created using CircuitLab

(Here, you can easily compute \$R_1=\frac{1\:\textrm{V}}{20\:\mu\textrm{A}}=50\:\textrm{k}\Omega\$.)

If you want to disable the module, you need to make sure that the ENABLE line is pulled below \$600\:\textrm{mV}\$. Just grounding it works fine (of course.) That pretty much guarantees it. But they know that no one is really just going to ground that input. That would be pointless. So they also want to make sure you understand the details enough that you can make sure that whatever you use will be able to "pull down" enough.

Suppose you didn't just ground it but instead hooked up a \$10\:\textrm{k}\Omega\$ resistor from ENABLE to ground. And suppose \$V_{IN}=35\:\textrm{V}\$. Then it's pretty clear that the voltage would be perhaps as high as \$35\:\textrm{V}\cdot\frac{10\:\textrm{k}\Omega}{10\:\textrm{k}\Omega+50\:\textrm{k}\Omega}\approx 5.8\:\textrm{V}\$, which obviously won't work. In this case, you'd need to use a resistor of less than \$872\:\Omega\$, which will sink near \$690\:\mu\textrm{A}\$, to meet the specifications. This is what they are trying to get across to you. You might need to sink almost a milliamp in some cases and they want you to be aware of that issue.

I think you understand it well. They are basically telling you that without an added connection (made by you, somehow), their power supply will be ON by default. They then add that if you add a connection to over-ride their default behavior, which they anticipate you may want to do, then you need to be aware of the fact that to over-ride it you have to "sink" (or pull-down) enough current to achieve it.

It's a very modest amount of current. But since they don't entirely know exactly what \$V_{IN}\$ you will be providing, they instead tell you that they have a \$100\:\textrm{k}\Omega\$ resistor there and also that there may be some added current you have to sink (beyond that resistor's requirements) for their added protection circuitry. The upshot of this is that you should be prepared to sink at least \$20\:\mu\textrm{A}\$ per volt and that you should consider the input to look like this:

schematic

simulate this circuit – Schematic created using CircuitLab

(Here, you can easily compute \$R_1=\frac{1\:\textrm{V}}{20\:\mu\textrm{A}}=50\:\textrm{k}\Omega\$.)

If you want to disable the module, you need to make sure that the ENABLE line is pulled below \$600\:\textrm{mV}\$. Just grounding it works fine (of course.) That pretty much guarantees it. But they know that no one is really just going to ground that input. That would be pointless. So they also want to make sure you understand the details enough that you can make sure that whatever you use will be able to "pull down" enough.

Suppose you didn't just ground it but instead hooked up a \$10\:\textrm{k}\Omega\$ resistor from ENABLE to ground. And suppose \$V_{IN}=35\:\textrm{V}\$. Then it's pretty clear that the voltage would be perhaps as high as \$35\:\textrm{V}\cdot\frac{10\:\textrm{k}\Omega}{10\:\textrm{k}\Omega+50\:\textrm{k}\Omega}\approx 5.8\:\textrm{V}\$, which obviously won't work. In this case, you'd need to use a resistor of less than \$872\:\Omega\$, which will sink near \$690\:\mu\textrm{A}\$, to meet the specifications. 

This is what they are trying to get across to you. You might need to sink almost a milliamp in some cases and they want you to be aware of it. They also want you to know enough so that if you do want to control it, you can successfully design a reasoned circuit arrangement.

1
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I think you understand it well. They are basically telling you that without an added connection (made by you, somehow), their power supply will be ON by default. They then add that if you add a connection to over-ride their default behavior, which they anticipate you may want to do, then you need to be aware of the fact that to over-ride it you have to "sink" (or pull-down) enough current to achieve it.

It's a very modest amount of current. But since they don't entirely know exactly what \$V_{IN}\$ you will be providing, they instead tell you that they have a \$100\:\textrm{k}\Omega\$ resistor there and also that there may be some added current you have to sink (beyond that resistor's requirements) for their added protection circuitry. The upshot of this is that you should be prepared to sink at least \$20\:\mu\textrm{A}\$ per volt and that you should consider the input to look like this:

schematic

simulate this circuit – Schematic created using CircuitLab

(Here, you can easily compute \$R_1=\frac{1\:\textrm{V}}{20\:\mu\textrm{A}}=50\:\textrm{k}\Omega\$.)

If you want to disable the module, you need to make sure that the ENABLE line is pulled below \$600\:\textrm{mV}\$. Just grounding it works fine (of course.) That pretty much guarantees it. But they know that no one is really just going to ground that input. That would be pointless. So they also want to make sure you understand the details enough that you can make sure that whatever you use will be able to "pull down" enough.

Suppose you didn't just ground it but instead hooked up a \$10\:\textrm{k}\Omega\$ resistor from ENABLE to ground. And suppose \$V_{IN}=35\:\textrm{V}\$. Then it's pretty clear that the voltage would be perhaps as high as \$35\:\textrm{V}\cdot\frac{10\:\textrm{k}\Omega}{10\:\textrm{k}\Omega+50\:\textrm{k}\Omega}\approx 5.8\:\textrm{V}\$, which obviously won't work. In this case, you'd need to use a resistor of less than \$872\:\Omega\$, which will sink near \$690\:\mu\textrm{A}\$, to meet the specifications. This is what they are trying to get across to you. You might need to sink almost a milliamp in some cases and they want you to be aware of that issue.