Placing a SPST switch between RTC CLK and Microcontroller

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newbie_hs

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I am using an PCA85073A RTC in my design. On some boards w the supercapacitor is getting discharged very fast (5V to 1.1V in half an hour.) This happens when we turn off the microcontroller power supply.

Because we are feeding logic signals or voltages into a MCU that is unpowered.

To avoid this issue I decided to keep an switch between RTC CLK and Microcontroller. The switch will be powered from the same power supply used for powering the Microcontroller.



When the Microcontroller is off the Switch will be in off state.The RTC will be running and signal is coming upto S pin of the switch.

My question is will the super cap discharge via input protection diodes of Switch(ADG801) when it is off
 

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Hi,

You gain nothing.
In first case the current will be drawn via the microcontroller internal protection diodes.
in the other case ... the current will be drawn via the analog switch internal protection diodes.

Why don´t you power the analog switch from the super capacitor?

BTW: what´s the clock frequency?

Why not switch OFF the CLK with an I2C command?
Why not using an 74LVC1Gxx buffer (powered from microcontroller side)?
Why no passive solution? Like a schottky diode between RTC (C) and microcontroller (A) and a pull up resistor at the microcontroller side.


Klaus
 
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KlausST said:
in the other case ... the current will be drawn via the analog switch internal protection diodes.
Click to expand...
Okay can I place a relay

KlausST said:
Why don´t you power the analog switch from the super capacitor?
Click to expand...
I can do that. That is one option

KlausST said:
Why not switch OFF the CLK with an I2C command?
Click to expand...
I need to check it. I think we can do it
KlausST said:
Why not using an 74LVC1Gxx buffer (powered from microcontroller side)?
Click to expand...
In this also the current will be drawn via the analog switch internal protection diodes. Please correct me if I am wrong.

KlausST said:
Like a schottky diode between RTC (C) and microcontroller (A) and a pull up resistor at the microcontroller side.
Click to expand...
In this case also when the microcontroller is off the schottky will conduct and current will flow through the clamp diodes of uC
 

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newbie_hs said:
In this also the current will be drawn via the analog switch internal protection diodes. Please correct me if I am wrong.
Click to expand...
What analog switch are you talking about?
What internal protection diodes are you talking about?
Did you even read a datasheet?

newbie_hs said:
In this case also when the microcontroller is off the schottky will conduct and current will flow through the clamp diodes of uC
Click to expand...
It would be nice if you draw a sketch to prove your current flow.

I try to help .. but if you know better ... I don´t like to waste your time.

Klaus
 
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KlausST said:
I try to help .. but if you know better ... I don´t like to waste your time.
Click to expand...
Yes you helped me a lot and I will be always thankful for that. Till my death
KlausST said:
It would be nice if you draw a sketch to prove your current flow.
Click to expand...

I will draw the waveform
--- Updated ---

KlausST said:
It would be nice if you draw a sketch to prove your current flow.
Click to expand...
Please see the waveform below.

 
Last edited:

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Hi ,

I re-read your post. I did not take care of the diode orientation.My mistake.

Please see the waveforms .During off condition I can see some spikes

When uC is on

When Controller is off
I can see a spike in the output waveform
 

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newbie_hs said:
During off condition I can see some spikes
Click to expand...
This happens because you are using an 1A schottky rectifier diode instead of a lower capacitance signal diode.

In the real circuit, any capacitance between RTC Clkout pin and other circuit nodes (diode/switch/buffer input whatsoever capacitance) causes a certain current spike, creating an additional amount of RTC supply current proportional to clock frequency. Even the unloaded Clkout pin does when activated. The point is to reduce it to an acceptable value.
 
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KlausST said:
What is R2 good for?
Click to expand...
Sorry for the delay in my response.
I considered R2 as the input impedance of Microcontroller GPIO.
Even without R2 ,I am getting the same spike
--- Updated ---

May I know you mean this spikes are unavoidable.
I tried to run the simulation using this diode. But the spikes are still present.
 
Last edited:

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There isn't much to say about the spikes, voltage dv/dt * C = current spike. Why do you worry about it?

Real waveforms will be different, lower dv/dt, uC input capacitance, voltage above Vdd is clamped by CMOS input diodes.
 
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Hi,

R2: Makes no sense. If you want to simulate the microcontroller, then don´t use (only) an R, because C and D may have more influence on the spike behaviour. Especially when the microcontroller is powered OFF.
Did you tell what kind of microcontroller you use? (so we

Spike: Again, don´t post riddles. What´s your concern with the spike? Do you think it´s good or bad? And FvM already told you where it comes from.
So: Talk to us.

Klaus
 

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