Can I level shift to 1V like this?

[Coming]

Hi,

I want to use DIO of USB6008 to give some digital incentive to some other IC chip. However, the max input is this IC is only 1V and the digital output of 6008 is about 5V. So I want to level shift signal "1" from 5V to 1V. Can I do this convertion like the pic below? I just use RI around 1.1k, then the 5V supply of USB6008 will voltage divide between Rp(4.7K inside 6008) and RI(1.1). The output (node A) will [5V/(4.7k+1.1k)]*1.1k around 1V. Can I do like this? I think the input of the IC chip is the gate of an inverter. Is there any other attentions I shall pay to?

BTW, is there any model of connectors that I can put on my PCB to easy connect to USB 6008?

Thank you~

Kangmin

[K C]

Hi Kangmin,

What levels does the IC accept. Are the levels OK when you calculate the levels with all the tolerances ?

[Coming]

Hi,
The IC can tolerate max 1.2V, otherwise it may break down. So I assume 1V for safety.
I am not sure whether I can level shift the 5V of USB6008 to 1V like this...

Kangmin

[ryan_d]

Hi Kangmin,

If I understand you correctly, you are trying to shift the output voltage of a digital output line of a USB-6008 from 5V to 1V. You are wondering if you can accomplish this by dividing down the voltage using the on board pull-up resistor and your RI resistor on you PCB.

You are correct in that you cannot divide the voltage down as in your diagram. You can, however, treat the USB-6008 as a black box, and add a resistance from the P0.0 output to A on your PCB to create a voltage divider. If you keep your current configuration where RI is 1.1 kOhms you would want to place a resistance that is near 4.7 kOhms.

Please post if you have any questions. Have a great day!

[Coming]

Hi,Ryan,

Oh, do you mean I shall voltage divide like the pic below? Use R2=4.7kohm, R1=1.1kohm, and foreget Rp, and then I can get 1V at node B?

However, I just curious: because the manual of 6008 says its digital output is open drain, so in this configuration what if the signal is "0"? This will turn down the open drain transistor at the output of the 6008 chip, and the supply voltage(+5V) will be divided among Rp(inside 6008, with value of 4.7kohm), R2 and R1. Then the voltage of B will be [5V/(4.7k+4.7k+1.1k)]*1.1k=0.52V...  Am I wrong somewhere?

Kangmin

[K C]

Hi Kangmin,

With the voltage divider you have to depend on what's inside your USB device. Seeing it as a blackbox and use a voltage divider at the outside with a pullup at the inside wont work. Like you said...

I think this is a better solution 1 transistor (Any NPN will do) and a few resistors. You replace the 4k7 with a smaller resistor and a potmeter to adjust the output value if you like. This way it does not matter what is inside the USB device as long there is 0 to 5V on the output.

Message Edited by K C on 11-27-2007 08:32 AM

[ryan_d]

Hello everyone,

KC is correct. I neglected to consider the low output case.

[Coming]

Thanks KC, that make sense. It is safe to do the way you draw or use some level shifter. However, I just want to keep fewest off-chip components on my PCB...

Hi, Ryan. What do you mean by "low output case"? For my application, the output will just drive gate capacitor of the inverter inside the chip. I know KC's method is safe. I want to make sure the schematic I draw on my first post is correct or not?...


Thanks all~
Kangmin

[K C]

Hi,

Ryan, is the pull-up resistor (4k7) at the output of this USB device a plain pull-up resistor or some kind of other electronic pull-up ???

Kangmin, if the pull-up (4k7) is just a plain pull-up resistor your first method will work. Like you said my last suggestion is more safe.

If it is not a plain pull-up resistor you better use the last suggestion with the transistor.

[ryan_d]

Hello everyone,

The pull-up resistor is just a plain pull-up resistor. Either method will work.