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Part Number: SN74LVC1G08
i've been asked by Customer below.
1.Is there any C value limitation customer can place capacitor between Y-GND pin? if so how much is it?
what customer tries to use is 0.47uF. is this no problem?
2.if customer has the load (40uA) connected with each above logic, would it have any problem ? should customer not care about it?
When the output is high and the capacitor is charged, and the output then switches low, the capacitor discharges through the Y pin. At VCC = 3.3 V, the output impedance is about 12 Ω, so this would result in a current of about 275 mA, which would exceed the absolute maximum ratings.
To make this safe, add a series resistor (between Y and C) to limit the current to less than 50 mA, i.e., 56 Ω or more.
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In reply to Clemens Ladisch:
Besides what Clemens has already mentioned, the larger the C value, the longer the capacitor will take to discharge. This impacts output frequency. As long as the outputs of the devices are not being switched too fast, there should be no problem.
In reply to Karan Kotadia:
may I get your answer for below?
In reply to Masaharu Takahashi:
hi, One more thing,
as Clemens-san mentioned , where we find the output impedance ? does it mean the impedance how much C has where connects between Y-GND ?
When the output is high and the capacitor is charged, and the output then switches low, the capacitor discharges through the Y pin. At VCC = 3.3 V, the output impedance is about 12 Ω, so this would result in a current of about 275 mA, which would exceed the absolute maximum ratings
Even with a series resistor, the output will have no problem driving a 40 µA load.
The output impedance can be computed from the typical VOL/IOL values. (The datasheet does not list the typical VOL; I looked into SZZA010.)
We have some great FAQ's that typically answer your questions.
This is your first question on how much capacitive load can be driven.
This is your second question on how to calculate the output impedance ROH and ROL of a device.
Is there any equation to know how much output cap IC can be placed like Clemens mentioned?
ex: Vol : 0.1V , Iol ...16mA(in case of Vcc:3V)
output impedance is 6.25ohm
if the cap customer will use has 0.1ohm as ESR. (total R is ≒ 6.3ohm)
output voltage is 3.3V as example, the amount of current will be there is 20mA.
then, customer will be able to make a decision how much cap they can use and can't use by having the step as above.
could you give us such information?
I am not sure where you are getting 20mA? 3.3V/6.3Ohms = 528mA.
This current is unacceptable and will not change from your capacitor size. You will have to add a series resistor to the output.
The real math you need to check is whether the transition rate will be fast enough for the frequency you are operating at.
Lets say you add a 100 ohm series resistor to the output with a 50pF capacitor load then you can calculate transition rate as 2.2*R*C = 2.2*106.3*50e-12 = 11ns rise time. So If you are trying to operate the part with a 20ns period (10ns high and 10ns low) or at 50Mhz then your capacitor is too large to let that happen.
Thats the equation you want to follow.
On the behalf of Takahashi-san, I want to re-start from the following E2E.
My customer wants to use SN74LVC1G08 as load switch.He is thinking the following condition/circuit.
The reason of this device selection is small size and use this device for other section.
For implement this function, could you please advise us? Also, please let me know what issue does he comes up.
Thanks and best regards,M.HATTORI.
In reply to Motoyasu Hattori:
* .48uF capacitor requires at least a 100 Ohm output series resistor or device will be damaged.
* .48uF and 100Ohm series resistor will give 250uS from 0V to 3.3V.
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