You need a P-channel logic-level mosfet to switch it, like IRLML6402 for example.

Hi Kubeek,

Thanks a lot for the diagramm. I see my mistake now

regards Rainer

 

Kubeek is correct that you need a P-type if you are choosing to interrupt the power line. My comments about N-type would apply to switching the path to ground for the load. Sorry for any confusion.

 

I am not too sure what the consequences might be if you interrupt the gnd line, it could leave the pullups on the data lines active and leave the usb device still posing as connected, so I think it is better to stay as close to the standard as possible and iterrupt the power line.

 

Yup, no argument.

 

I am not sure what you mean. Only usb host is allowed to put voltage on the +5V line, a usb device is allowed to use it as power supply but should not be putting any voltage by itself.
No idea what you mean with short and long lines..
The mosfet in the picture I posted is interrupting the +5V line, with the host on the left side and the device on the right side.

Hi Kubeek,

Sorry that was misinterpretation from my side. I deleted those 2 nonsense paragraphs. After looking a bit deeper into the diagramm I realised that you are interruting the +5V line in order not to have the 5V voltage potential in the camera.

With long and short lines I referred to the battery or power symbol and misinterpreted the positive long line of the symbol being negative.

Everything is now more or less well understood

Now if I want to use an optocoupler i can use foe example a TLP521-1 and connect the pins 3 and 4 as the switch needed for switching the gate right ?

The activation of the TLP521-1 would happen applying a 5V to max. 20V voltage on pins 1 and 2 ¿ correct ?

 

Between pins 1 and 2 is an LED, so you need a current limiting resistor set tou roughly 5mA. The LED has Vf of 1.3V, so if you feed it from 5V source then the resistor should be (5-1.3)/0.005=740ohm, but anything between 560ohm and 2k would work.

 

Between pins 1 and 2 is an LED, so you need a current limiting resistor set tou roughly 5mA. The LED has Vf of 1.3V, so if you feed it from 5V source then the resistor should be (5-1.3)/0.005=740ohm, but anything between 560ohm and 2k would work.

Hi Kubeek,

Thanks. Using 12V would then mean 12V -1.3V / 0.005mA = 2140ohm ~ 2200ohm ¿ right ?

Does it matter where the current limiting resistor is ? I mean ... before anode or after cathode ?

 

Yes an yes, the position doesn´t matter. Just make sure the polarity of the diode is correct.

 

Hi Kubeek,

For my better understanding.

How does the 10K R1 pull down the voltage charge of the Gate being connected to the +5V line ? ¿ Slowly through the Drain > Source line of the MOSFET ?

Thanks

 

The 10k resistor is connected to the source, so that when nothing else is connected the gate is kept at the same voltage as the source, thus turning the mosfet off.

As for the book, try the allaboutcircuits e-book: http://www.allaboutcircuits.com/

 

Hi,

I got now the parts and when I have time I will do the tests.

I see that the SMD MOSFET I bought have a very short pin on the Drain. Do I assume correctly that I can use the plate at the back for connecting the drain in the same way I would connect a cable on the very short Drain pin ?

Another question. I also bought some 2SJ 598 MOSFET, maybe a mistake, but I assume I can use them too ¿?

Thanks and regards Rainer

 

I quess you have the mosfet in this package http://media.digikey.com/photos/National Semi Photos/14-TO-252.jpg the short middle pin is connected to the back tab, and the tab is meant to be the connection to drain.

 

I quess you have the mosfet in this package http://media.digikey.com/photos/National Semi Photos/14-TO-252.jpg the short middle pin is connected to the back tab, and the tab is meant to be the connection to drain.

Yes, they look like little bugs

OK, understood.