High resistance, no current. What's the point of a potential then?

Question

I am referring to this example:

1. Why does it matter that the IC has a high resistance? What does this do for/to the IC?

2. If there is a potential but no current, what is the point of using the IC when in the end no current would run through? Would the device even work?

Answer 1

We can apply voltage divider rule only when same current is running through these resistors. This means, if the IC is supposed to draw current, sum of two currents: one towards the IC and another towards the R2 passes through the resistor R1. So there is difference in the currents flowing throught R1 and R2. Look at the schematic I have added. I have just tried to show how the current flows when there is low resistance in parallel circuit.

So by considering high resistance of IC, we can neglect the current flowing through the IC and apply the voltage divider rule.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

Not every gate needs current for it to operate. MOSFETs are an example of voltage controlled devices whereas BJTs are current controlled devices.

A high input impedance means that no significant current will be drawn and as such components can be chosen accordingly. It also means that the voltage between the divider won't change according to current draw because no current wil be drawn. As such it is safe to use a voltage divider to get a wanted voltage.

Answer 3

The formula for the voltage divider's output

\$V_{out} = V_{in}\cdot \frac{R2}{R1+R2}\$

only applies if you're not drawing any current of the output node. This because the formula above presumes that the same current flows through both the resistors R1 and R2.

In other words, to have a stable output voltage Vout you either have to take the required output current into your calculation or it has to be small compared to the current through R2. For that reason you cannot use a simple voltage divider to power e.g. a microcontroller.

A potential without a current is a signal, either an analogue one or a digital one. In this case you'd route it on a high impedance input pin of a device and therefore the application with the voltage divider works (as no current is drawn). You'd do this e.g. to lower the logic level of an input signal from 5V to 3.3V.

Answer 4

1. Try doing the problem again with a much lower restance for the IC input, you'll find you get a very different answer. That is why it matters. Most (modern) IC inputs (not power supply pins) can be modeled as a large resistor.
2. Again, this problem is talking about an input pin, not a power supply pin. The input pin could be one of the sense terminals of an op-amp or it could be an input to a digital circuit IC. So whatever the IC is, it will sense the voltage at that input and respond accordingly.

This problem would be very different for a power supply input because power supply inputs can rarely be modeled as a simple resistor. Hence, you would not design a bias circuit using a simple voltage divider for a power supply.