How to take the internal circuitry of an IC into consideration when running simulation?

Thread Starter

Yami

Joined Jan 18, 2016
282
Hi guys,
Random question, but it's been something that I have been wondering for sometime.
I been trying to simulate the control circuitry for the 'mode' control function of the TDA8924 amplifier chip. I have to assume a value for the 'load' of that particular pin. In this case I use a 100K resistor. This is the only way to figure out what the voltage at the pin would be due to the external components.
So my question is that is there a certain value I should be using for this pin 'load'? Different values for different kind of pins?Maybe a hint given on the datasheet? I can think of other instances where I might encounter the same issue, with different ICs I mean.
Thanks in advance for the help
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ronsimpson

Joined Oct 7, 2019
684
I think there is an error in the data sheet. 'Max voltage on mode pin is 5.5V' One diagram shows a 5.6V Zener I think it should be a 5.1V Zener.
It is clear that when the switch is at "off" the voltage is 0V.
When the switch is in "on" the voltage is the Zener voltage. I think 5.1V but maybe 5.6V which is over spec.
In "mute" mode the voltage should be about 2.6V +/-0.4V.
Assume Zener=5.0V for the sake of round numbers. R4 makes the top half of a voltage divider. Likely in side the IC there is a second 39K to ground. The two 39K resistor will cause of voltage of 1/2 Zener voltage. 5.0V/2=2.5V.
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I think the resistor inside the IC is 39k not 100k.
 

Thread Starter

Yami

Joined Jan 18, 2016
282
I think there is an error in the data sheet. 'Max voltage on mode pin is 5.5V' One diagram shows a 5.6V Zener I think it should be a 5.1V Zener.
It is clear that when the switch is at "off" the voltage is 0V.
When the switch is in "on" the voltage is the Zener voltage. I think 5.1V but maybe 5.6V which is over spec.
In "mute" mode the voltage should be about 2.6V +/-0.4V.
Assume Zener=5.0V for the sake of round numbers. R4 makes the top half of a voltage divider. Likely in side the IC there is a second 39K to ground. The two 39K resistor will cause of voltage of 1/2 Zener voltage. 5.0V/2=2.5V.
View attachment 207453
I think the resistor inside the IC is 39k not 100k.
Thanks @ronsimpson, I didn't mean only this instance. But in general. This was what I was messing around today. Can I ask though how you deduced that its a 39K resistor inside the IC.
 

ronsimpson

Joined Oct 7, 2019
684
I tried to explain that "mute" is defined by 2.6V+/-0.4V.
R4=39k. All this we know.
What we need to find is what resistor, along with a R4=39k, will take 5.1V and reduce it to 2.6V. (voltage divider formula)
 
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