I am referring to this example:

enter image description here

  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?

  • \$\begingroup\$ The IC's pin is not a power supply pin but a signal input pin intended to sense the voltage applied, where only parasitic resistances will contribute to the input impedance. \$\endgroup\$ – JimmyB May 24 '19 at 13:38
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    \$\begingroup\$ I like how the text states 10 MOhm and the schematic says 1 MOhm. \$\endgroup\$ – Arsenal May 24 '19 at 14:10
  • \$\begingroup\$ One analogy is a REAL constant voltage power supply with internal output resistance powering a resistor. What is the voltage across the resistor if it's very zero or 1 ohm? What is it if it's 1 mega ohm or infinite? Which scenario uses less power and more accurately has the desired voltage appear across the resistor? Remember, you just want to get the signal/information across so minimum power us best. Or just replace Rin with 0 or 1 ohm in your example and look at what the voltage across it will be. It's like trying to fill a leaky bucket. Takes more water (power) and the level is not accurat \$\endgroup\$ – DKNguyen May 24 '19 at 14:25
  • \$\begingroup\$ It may be worth mentioning also that this high resistance is not something intentionally put there, in most cases. Generally they make input pins as high-resistance as feasible, and the megohm resistor there represents leakage currents through nonideal devices. Ideally the resistance would be infinite, but of course that's not possible. \$\endgroup\$ – Hearth May 24 '19 at 14:43
  • \$\begingroup\$ 1. What are these input and output pins you guys are referring to? 2. Toor, what equation do I have to use if I want to see the effects of changing Rin to a smaller value? The equations above have nothing to do with Rin. \$\endgroup\$ – mikanim May 25 '19 at 9:50

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

  • \$\begingroup\$ Wonderful. Thank you. Your comment and the comment from Humpawumpa complement each other. \$\endgroup\$ – mikanim May 25 '19 at 9:46

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.


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.

  • \$\begingroup\$ I thought Vout and V_2 are the same: imgur.com/a/6Yo7zX7 \$\endgroup\$ – mikanim May 24 '19 at 12:38
  • \$\begingroup\$ Yes that is correct, does this somehow interfere with my answer? \$\endgroup\$ – po.pe May 24 '19 at 13:00
  • \$\begingroup\$ Not sure what I meant there. Yea your comment makes a lot of sense and thank you for the further explanation of its application. To your second paragraph, could you point me to a wiki page or a website that delves into that further? I'd like to look at some examples of these cases. \$\endgroup\$ – mikanim May 25 '19 at 9:42
  • \$\begingroup\$ Another question, what is the purpose of a signal in this case? What could it do for us? \$\endgroup\$ – mikanim May 25 '19 at 9:46
  • \$\begingroup\$ Cases meaning: cases where one cannot use a simple voltage divider. \$\endgroup\$ – mikanim May 25 '19 at 9:53
  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.


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