The tolerances in Resistors and Capaictors have a major impact on Analog circuits but not on Digital ones!

This is a statement that was made to me and I was asked to explain:

1. Is it true or not?
2. Explain your answer

I have no idea how to approach this, can anyone help?

 

Digital is usally all or nothing, so if a component is "close enough", it works.

Analog, on the other hand, is a bit like machining parts. Tolerances add up, if they add up the wrong way it doesn't fit.

 

In digital circuits, logic signals can be recognized as high or low bertween some voltage range. Thus, a small variation due to components tolerance is acceptable.

In analog circuits, tolerance of the components may affect the operation of the circuit at a high degree. Imagine that you design a band pass filter with a very high Q and the values of the components are not exactly as calculated due to tolerance. Because of the high Q of the filter a small variation in the components value can cause a significant variation on the bandwidth of the filter. This is an example of an analog circuit.

 

for analog it works something like
1.1 + 1.2 + 3.1 + 1.3 + 2.3 = 9

for digital
floor(1.1) + floor (1.2) + floor (3.1) + floor (1.3) + floor (2.3) =
1 + 1 + 3 + 1 + 2 = 8

9 ≠ 8

Now imagine if that were kiloVolts, 9kV is way bigger than 8kV...

 

So if I understand this properly.

Even if I use a 10% tolerance Capacitor instead of a 1% the quality of the signal will not be effected nor will the quality of the product?

 

Depends, a lot of digital circuits have analog components. If this is the case it can matter.

In other words, only if we have the circuit schematic we can say for sure.

 

Electrolytics have tolerances as much as -50% to +100%

To make an affordable product, the use of 10% resistors is common.

The "Hard Part" is identifying which components need to be 1%, no matter what type of component. In an oscillator or amplifier, some capacitors can be off by double with no adverse effect, others can be off by 10% and it will function poorly.

 

Thanks Guys, that helps clarify the answer for me