For Isolation related questions:
Please post a question on the new Digital Isolators Forum.
Part Number: TCAN1043-Q1
My customer asks for confirmation of the resistance value connected to the WAKE PIN of TCAN1043.
The using formula calculated is shown below.
VSUP = 26.5V.
R_serise = Vsup/Io(wake) = 26.5V/3mA =8.4kohm.
R_serise tries to apply 10 kohm.(More than 8.4 kohm)
R_bias < ((VSUP – Vih)/Iih) – R_series = ((26.5 – (26.5-1.9V))/Iih) – Rseries = 1.9V/25uA – 10KΩ = 76KΩ
R_bias tries to apply 10 kohm.(Less than 76 kohm)
Question : Is it OK to apply both R_serise and R_bias at 10KOHM?
Are you doing this to create a resistor divider on the WAKE pin for threshold crossing as shown in the datasheet?
If so, then putting 10k at both will create a 50% voltage divider, leaving the high voltage on the WAKE pin at 13.25V and never allowing it to cross the high threshold. The calculation for the series resistance is correct, but the combination of R_bias and R_series should limit the current into the pin at 3mA (absolute maximum), not just the one resistor, and I would recommend limiting it to something much lower than 3mA.
Because the high level input voltage has to be a minimum of VSUP - 1.9, to cross the high logic threshold, the voltage seen by the WAKE pin will need to be at least 24.6V in this case, and the low level input will need to be at most 23.1V.
If my understanding of your WAKE pin configuration was incorrect, can you please share a schematic?
Texas Instruments | INT-TRX| Applications Engineer
We are glad that we were able to resolve this issue, and will now proceed to close this thread.
If you have further questions related to this thread, you may click "Ask a related question" below. The newly created question will be automatically linked to this question.
In reply to Eric Hackett:
I can not understand being a voltage distribution.
Customer will configure same as above TI.(Low side switch input)
When the switch is open, the internal current source (25uA) will flow to R_Series.
Since the switch is open, I think the voltage of the Wake Pin is ((R_bias+R_series) * 25uA)+Vsup.
So, V_wake pin_switch open = ((10k+10k)*25uA)+26.5V = 27V
When the switch is closed, internal current will flow through R_S and switch to GND.
So, V_wake pin_switch closed = 10k*25uA = 250mV (If Switch impedance is 0ohm)
For the above reason, I think that R_bias and R_series both use 10k.
If you think I'm wrong, please tell me.
And I have additional questions.
In the datasheet, we use the following formula to calculate Rb for a high side switch configuration.
RBIAS < ((VSUP - VIH) / IIH) - RSERIES
The customer wants to design the circuit in a Low Side Switch configuration.
When configuring with a low side switch, is it right to change the formula as shown below?
RBIAS < ((VSUP - VIL) / IIL) - RSERIES
All content and materials on this site are provided "as is". TI and its respective suppliers and providers of content make no representations about the suitability of these materials for any purpose and disclaim all warranties and conditions with regard to these materials, including but not limited to all implied warranties and conditions of merchantability, fitness for a particular purpose, title and non-infringement of any third party intellectual property right. TI and its respective suppliers and providers of content make no representations about the suitability of these materials for any purpose and disclaim all warranties and conditions with respect to these materials. No license, either express or implied, by estoppel or otherwise, is granted by TI. Use of the information on this site may require a license from a third party, or a license from TI.
TI is a global semiconductor design and manufacturing company. Innovate with 100,000+ analog ICs andembedded processors, along with software, tools and the industry’s largest sales/support staff.