# Tag Info

## Hot answers tagged voltage-divider

8

Yes, there are online calculators that will allow you to find resistor pairs that will closely approximate the desired ratio. For example, this one. If you wanted to have (say) 3.0V out and your reference is 0.8V and we're ignoring bias current, and we're using E96 series (1%) values we get the following results: Obviously you can scale the values by ...

6

The 220 ohm pull down sets an operating point for the system. Without it, the fan would never turn off. With it, the fan will only turn on if the NTC resistance is low enough to pull the transistor base above 0.7V base-emitter threshold. With a 220ohm pull-down, that transition will begin to happen when the NTC resistance gets below 2.6k ohms (about 55-57 ...

6

I while back I wrote a script that accomplishes the same thing as the Douxchamps site above, but not limited to dividers - it does an exhaustive search on any system you feed it. I haven't published it on GitHub have by request published it on GitHub. For posterity, here it is in its entirety, with your 3.0/0.8 voltages in an example at the bottom: "&...

4

Voltage dividers aren't especially useful for LEDs. LEDs are current driven devices, not voltage driven devices. Voltage dividers provide a voltage. Voltage dividers aren't good when there's a load attached to them. The load changes the resistance and therefore the divider ratio. To make a somewhat stable voltage divider, you need at least about 10 times ...

4

You do not need - and should not introduce - more reference voltage supplies than you have to. In this case you do not need any. Just have a non-inverting amp with gain of 10/3 and offset of -9. Any ratiometric multiple of this circuit will do: simulate this circuit – Schematic created using CircuitLab Standard resistor values that get you pretty ...

3

You need amplification since: $$\frac{5-1}{4.2-3} = 3,333$$ So you could use a differential amplifier: to get the desired result: V1 is the voltage you want to read. V2 shifts the output up and V3 is there to match the minimum input voltage. Both V2 and V3 can be provided by simple voltage references like a TL431 and you may need precision resistors to ...

3

If you had a buffer amplifier with gain of 1 between the stages then the first equation would be correct (assuming no load on the output). Since the second voltage divider loads the first one, the output signal is affected, and obviously will be less than in the non-interacting case. The loaded output of the first voltage divider becomes Rx/(Rx + R1) where ...

3

Well, let's make a mathematical closed solution. I know that this is maybe above the OP's knowledge, but I think it is important to show it in combination with the other answers given. The Shockley diode equation, gives the relation between the voltage across and the current trough a diode: \text{I}_\text{D}=\text{I}_\text{S}\left(\exp\left(\frac{\text{q}\...

3

There exactly one voltage that will develop at the junction of R1, R2, and the LED in steady state. The LED is a semiconductor (i.e. a non-linear device) which makes it complicated to reason about in analytically precise terms. I would practically think about this circuit by omitting R2 initially and analyzing the circuit that way. In that light (pun ...

2

Most audio circuits have a low-end cutoff frequency in the 20 Hz range. So, even at 50/60 Hz, they will have a small attenuation, and a noticeable phase shift (perhaps 10-20 degrees). Why do you care about the waveform so much ? The AC line is quite 'dirty' and most loads don't care. Motors and (incandescent) lamps don't care. Laptops, TVs, and other loads ...

1

My answer is simple: The transistor must be controlled by voltage, not by current, since it is used as a voltage comparator with 0.7 V threshold (the base-emitter voltage VBE). This voltage threshold can be thought of as an opposing reference voltage that is subtracted from the input voltage. If the difference is positive, a base current flows and the ...

1

I only get HIGH from life. ;) but the PIR goes high with IR activity and needs a pull-down resistor for say 50uA or 3.3V/50u = 67k nearest value to indicate absence. If you were powering from 5V your ESP might latchup and might get very hot. You better check that 5V is not being used.

1

I have directly connected 120VAC mains to a voltage divider and directly digitized that and it works fine, although you really really do have to be careful about having energized neturals/grounds and stuff like that, which you seem to be aware of. Note that it is probably a good idea to put transient protection on one or both sides of the bridge if you want ...

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