LSF0108: :

Onur Kusakoglu

Intellectual 960 points

Part Number: LSF0108
Other Parts Discussed in Thread: SN74AHC541

Good morning,

I have a question about pull up resistor values and how to limit the current.

I see the table in the datasheet and the formula Rpu=(Vpu-0.35)/0.015A to limit it to 15mA.

My questions,

Let's assume, I have a signal which is an Input to B1 pin with a 5V pull up and it is an output to port A1 pin. And VREFB is 5V and VREFA is 3.3V.

1-) Do I need to add any pull up on A pins or the only pull up which is added to B pins would be enough? Example: To set the current to 15mA, then Pup B1 pin with 310 ohm with a 5V to set the current to 15mA. Correct?

2-) What if I need an initial state to on port A and I also want to add a pull up there. Assume I have 10k pull ups on both port A and port B. Signal is going through from Port B to Port A. Vref B is 5V and Vref A is 3.3V. What would be the current that the driver needs to drive(set current)?

I am a bit confused, happy to hear your feedback.

Best regards

Onur

over 2 years ago

0 Clemens Ladisch over 2 years ago

Guru 242930 points

Higher pull-up currents allow higher speed, but use more power and increase V_OL. You have to choose a current that fits your requirements.

1. If the A pin is only an output, it might not need a pull-up resistor. See Voltage-Level Translation With the LSF Family.

2. The driver must be able to sink the current flowing through both pull-up resistors.

0 Onur Kusakoglu over 2 years ago in reply to Clemens Ladisch

Intellectual 960 points

In my case, A pin is always output and B pin is always input. Driver ICs are all at 5V level and I want to shift the level to 3V3.

VREFB is 5V and VREFA is 3V3.

All ICs, which drive pin Bs of LSF device, have drive capabilities up to 4mA and I want to limit the current to at most 1mA.

(On B pin) I have a 10k pull up on 5V side which sets the current to 0.465mA->(5-0.35)/10k

I don't fully understand what would be the impact of the pull up on A side. I also have a 10k pull up to 3V3 on A side of the shifter.

If driver sends a low signal, this signal shall have the strength to pull down both pull ups on B pin and A pin.

0 Clemens Ladisch over 2 years ago in reply to Onur Kusakoglu

Guru 242930 points

The LSF is a passive switch; for a low signal, the A and B pins are connected, and the driver must be able to hande the current through both resistors.

For unidirectional level shifting to a lower voltage, a more simple and robust solution is a plain buffer with overvoltage-tolerant inputs, such as the SN74AHC541.

0 Onur Kusakoglu over 2 years ago in reply to Clemens Ladisch

Intellectual 960 points

Hi Clemens,

Thanks for the new part information, but I have already the design ready and looking for the best solution with what I have in hand.

If I have pull ups on both side then driver shall be able to drive

(On B pin) I have a 10k pull up on 5V side which sets the current to 0.465mA->(5-0.35)/10k

(On A pin) I have a 10k pull up on 3V3 side and to pull it down 0.295mA->(3.3-0.35)/10k

Correct?

+1 Clemens Ladisch over 2 years ago in reply to Onur Kusakoglu

Guru 242930 points

Yes; the total current is about 0.75 mA.

0 Onur Kusakoglu over 2 years ago in reply to Clemens Ladisch

Intellectual 960 points

Thanks a lot.

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LSF0108: :