significance of high resistances

hello guys

I want to ask one very basic thing which I have encountered many times during circuit studies and that is the significance of the high resistance in the circuit.

Actually I want to know that how a voltage is sensed across a circuit if there is no flow of current , for example in the figure below:



In this figure it is written that The value of the output pull-up resistor must be high enough to avoid excessive power dissipation, and yet low enough to allow adequate supply to switch the load circuitry being driven by the output. Now what I want to know is that if the value of the resistor is in the megaohms range than will it allow the current to flow through it and if not than how come the voltage is sensed across the output without the flow of current for example in the case of the voltage divider.

Also if the current flow is zero in the above diagram than it is beneficial for the above comparator because it will not be loaded by the incoming circuit as the load and the bias condition and the voltage supply to the comparatoir will not be effected.

and if there is current flow than is there any value of resistance for which the current is zero and the voltage is equal to the no load or terminal voltage across the output.

also I want to know that how is it possible to realize the voltage with no current flow across the circuit with the use of high value resistances , and can it be realized without the use of the high value resistances.


Thank you

piyushpandey said:

Actually I want to know that how a voltage is sensed across a circuit if there is no flow of current ,

Click to expand...

In practical circuits, there will always be some current flowing... Usually opamps / comparator inputs work with extremely small input currents, but there's still some. Also that resistor divider shown in above circuit will have current flowing through it, and current will flow through Rpull-up whenever Vout is lower than +Vcc.

In this figure it is written that The value of the output pull-up resistor must be high enough to avoid excessive power dissipation, and yet low enough to allow adequate supply to switch the load circuitry being driven by the output.

Click to expand...

I'm assuming that in this case the comparator is better at driving its output to the negative supply, than to the +Vcc supply, and the pull-up resistor is there to 'help' the comparator drive its output more stronger / faster towards the +Vcc voltage. With the output low, the pull-up resistor forms an additional load which, strictly speaking, you'd want to not have (and which increases current consumption of the circuit). The datasheet for this IC should have more on this.

With a value too high, it wouldn't 'help', since it wouldn't have any significant effect. With a value too low, the additional load (with output low) would be too much. So there is an optimal value, which would depend on circuit that's driven by the output & the frequencies / kind of signals being operated on.

In any circuit like this, there's always a potential that signal noise increases when you use higher-value resistors in input circuitry. But that also depends on things like a clean power supply, circuit board layout etc.

and if there is current flow than is there any value of resistance for which the current is zero and the voltage is equal to the no load or terminal voltage across the output.

Click to expand...

:?: Current through the pull-up resistor is simply (+Vcc - Vout) / Rpull-up, that's 0 when Vout = +Vcc (and non-zero when Vout is lower than +Vcc).

also I want to know that how is it possible to realize the voltage with no current flow across the circuit with the use of high value resistances , and can it be realized without the use of the high value resistances.

Click to expand...

See above, and check datasheet of this comparator. To be honest, your post doesn't make much sense... :?::roll:

And note to mods: why is this in "Embedded Systems and Real-Time OS" ? It's more an "Electronic Elementary" or "Analog Circuit Design" question...