USB to UART current consumption.

[electricity86]

Hey.

I'm not sure what is the current consumption of the CP2102 USB to UART bridge that I use, according to datasheet.

https://www.silabs.com/Support%20Documents/TechnicalDocs/cp2102.pdf

This Supply Current under this table means to the current that comes from the USB connecter (from PC) to the CP2102?

https://www.electro-tech-online.com/attachments/cp2102-jpg.27118/

 

[dknguyen]

It's the current the IC consumes. WHere it comes from depends on how you built the thing. So if you built it to be powered from the PC USB, that's where it comes from.

 

[electricity86]

It's the current the IC consumes. WHere it comes from depends on how you built the thing. So if you built it to be powered from the PC USB, that's where it comes from.

But why is the supply voltage mentioned there is 3.3V and not 5V which comes from USB? (called VBUS)

The 26mA is not only what comes for the 3.3V regulator (if enabled), but for all the CP2102?

 

[electricity86]

help please fellow.

 

[blueroomelectronics]

It can supply 20ma (26max) from the internal 3.3V regulator.

 

[electricity86]

It can supply 20ma (26max) from the internal 3.3V regulator.

I think that you might have a mistake.
Take a look please at this table:

 

[blueroomelectronics]

My mistake, it can supply from 1 to 100ma plus it needs 20ma to 26ma for itself. I don't use that chip I've used the FT232R which is similar.

 

[electricity86]

My mistake, it can supply from 1 to 100ma plus it needs 20ma to 26ma for itself. I don't use that chip I've used the FT232R which is similar.

Thank you very much.
So the 3.3V Regulator consumes 20mA (Typically) in order to output 3.3V.

How do i know then how much the whole CP2102 consumes?
It cant be that a manufacturer doesnt tell the consumer how much its product consumes.

I cant understand it from datasheet.

 

[electricity86]

anyone know maybe?

 

[blueroomelectronics]

Your USB port can easily supply 100ma and up to 500ma if the device requests it. Why not build one and measure it? Is it critical?

 

[dknguyen]

Thank you very much.
So the 3.3V Regulator consumes 20mA (Typically) in order to output 3.3V.

No no. Think of the USB and regulator as two separate pieces of silicon in a single IC package. The USB part of the chip needs 26mA to run and is powered by the regulator part, but the regulator part can output 100mA. That means that the regulator has an extra 100mA-26mA = 74mA that can be used to power other things.

This extra 74mA is NOT being consumed by the USB IC- it's being consumed by the other devices being that are also being powered from the internal regulator and you can account for this by adding up the current from those other devices. So the IC itself only uses 26mA. If you power a second IC from the internal regulator and it consumes 50mA, then the USB IC still consumes 26mA while the second IC consumes 50mA. The USB IC *DOES NOT*/B] consume 26mA+50mA. THe 50mA is being passed straight through the linear regulator (with the extra voltage regulated by burning it off as heat energy) to power the second IC and neither the USB part or regulator part consumes this current.

It's like saying a 1A linear regulator consumes 1A...it does not. THe things it is powering are consuming 1A. The regulator itself is only consuming something in the uA range to keep itself running, and this current is separate from the 1A output current.

 

[electricity86]

No no. Think of the USB and regulator as two separate pieces of silicon in a single IC package. The USB part of the chip needs 26mA to run and is powered by the regulator part, but the regulator part can output 100mA. That means that the regulator has an extra 100mA-26mA = 74mA that can be used to power other things.

This extra 74mA is NOT being consumed by the USB IC- it's being consumed by the other devices being that are also being powered from the internal regulator and you can account for this by adding up the current from those other devices. So the IC itself only uses 26mA. If you power a second IC from the internal regulator and it consumes 50mA, then the USB IC still consumes 26mA while the second IC consumes 50mA. The USB IC *DOES NOT*/B] consume 26mA+50mA. THe 50mA is being passed straight through the linear regulator (with the extra voltage regulated by burning it off as heat energy) to power the second IC and neither the USB part or regulator part consumes this current.

Hi, I very appreciate your answer.

What do you consider as the USB part?
The USB supplies voltage for the regulator, meaning the USB operates the regulator, so how come the regulator runs the USB part?

Isnt it true that the USB is the power supply of the whole chip?
And therefore the current that the regulator sources is coming from the USB?

It's like saying a 1A linear regulator consumes 1A...it does not. THe things it is powering are consuming 1A. The regulator itself is only consuming something in the uA range to keep itself running, and this current is separate from the 1A output current.

Assuming that it is a linear regulator, then IIN =~ IOUT in the regulator, so how come the regulator sources 26mA but sinks only µA?

 

[dknguyen]

Hi, I very appreciate your answer.

What do you consider as the USB part?

The circuitry inside the IC that deals with the USB signals, not the power supplies (ie. internal regulator). The manufacturers could have completely removed the regulator from the IC and it would still have all the USB functions, except you would have to provide a 3.3V regulator separately. See Figure 1 in your datasheet. The USB part is every block that is not the internal regulator (well technically the UART is a separate part too because it serves a different function but it is also powered from the internal regulator).

The USB supplies voltage for the regulator, meaning the USB operates the regulator, so how come the regulator runs the USB part?
Isnt it true that the USB is the power supply of the whole chip?
And therefore the current that the regulator sources is coming from the USB?

You are getting confused about two separate "parts" being housed in a single IC pacakge. The difference is as meaningless as to whether or not your PC power supply was mounted inside your tower, or sitting outside beside it. Like I said imagine your one IC as TWO separate ICs (a USB IC and a regulator IC). The one with all the USB functionality and a linear regulator. The manufacturers just threw in the regulator with the USB part so you don't have to juggle two ICs in order to make things more convenient for you. The PC's USB powers the regulator which in turn powers the USB part of the chip. So yes, the PC's USB powers the "whole chip" BUT the whole chip consists of the regulator part and USB part. The PC's USB directly powers the internal regulator, and the internal regulator directly powers the USB part, therefore the PC's USB only indirectly powers the USB part. Understand? Remember think of them as two separate parts.

Now, the reason the regulator part of the IC runs the USB part of the IC Is because the PC USB outputs 5V, but the USB part of the IC needs 3.3V. Look at Figure 1 in your data sheet, you can see that the regulator block accepts 5V and turns it into 3.3V which powers everything else EXCEPT the USB transceiver which has it's own separate connection to 5V. So internally, the IC processes the information using 3.3V (because it lets them uses lower power or faster transistors, and lets the IC communicate over the UART with 3.3V microcontrollers), but when the IC needs to send data to the PC it converts the signal back to 5V (because USB signals are 5V). So, in fact a tiny section of the USB part is powered directly from the PC's USB, but most if it is powered indirectly from the PC's USB (powered directly from the internal regulator and indirectly through the PC's USB).

I'm just trying to get you to visualize the functional components of the chip. Visualizing the chip as a whole is meaningless because you can stuff as many components as you want into a single IC package.

Assuming that it is a linear regulator, then IIN =~ IOUT in the regulator, so how come the regulator sources 26mA but sinks only µA?

26mA + XXXuA enter the IC, but the 26mA is not really consumed (sink may the wrong word to use here, sorry about that). Some of the extra voltage is removed to bring it down to 3.3V, but it exits the IC at 3.3V (not at 0V) with enough energy left to power lots of stuff (the energy was not consumed by the internal regulator).

 

[electricity86]

help please.

 

[blueroomelectronics]

help please.

What didn't you understand? Why is it critical? What are you building?

 

[dknguyen]

Think of a regulator a bit like a switch or more accurately an intelligent diode that is able to adjust it's forward voltage drop. Just because 1A is entering the it, does not mean that the switch is using 1A of current because it's being passed out the other end with pretty much all of it's useful energy (which can then be used by another device). The intelligent diode just happens to always be able to adjust it's voltage drop according to the voltage on the input side so that the voltage on the output end is always the same.

If you can't wrap your head how a regulator inside an IC is no different than a regulator sitting outside the IC, then don't use the internal regulator to power any external devices. Forget what voltages the IC uses internally and just assume the IC runs off 5V and consumes 26mA Forget that the internal regulator is even there. Forget that some parts of the IC run off 5V and other parts run off 3.3V, and that the regulator is there so you do not need to feed the IC both 5V and 3.3V. If you need to convert the PC USB's 5V to 3.3V to power something else, use a separate regulator IC.

A lot of microprocessors and microcontrollers do this actually. Most of the devices they must communicate with are 3V (or 5V), but the silicon technology that is fastest, smallest, and lowest power is 1.8V (or lower). So they make most of the brains of the chip run off 1.8V but there is hardware between the brains and input/output to change 1.8V and 3V signals back and forth. Then the manufacturers stick a little voltage regulator into the chip so that you can completely forget that the chip uses 1.8V for most of the it's internal operation. You just feed it the 3V supply and forget about it! The 3V powers both the input/output hardware of the chip and the 1.8V regulator which provides power to the brains of the chip- you never need to know what voltage the chip actually uses inside.

Maybe after working with an external regulator you will understand how a regulator integrated with other components is no more than a normal regulator except that a few wires have already been connected for you (to power other parts of the chip, whether you want it to or not).