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| Trying to understand voltage divider circuit |
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| Youkai:
I'm trying to understand some of the nuance of Voltage divider circuits. I drew a simple circuit in LTSpice (attached) that I'll use for my references. I'll divide this post into assumptions and questions. Assumptions are things I think I know/understand based on the little bit that I know of electronics and some inferences. Please review and correct any misunderstandings. I understand that there may be a lot of corrections if one of my early assumptions is incorrect. Assumptions: * Both R2 and D1 drop 5v because voltage is the same in a parallel circuit * Whatever D1 is (diode, LED, Arduino, etc) it probably has lower resistance than R2. So D2 would be paired with a resister (R3) for current limiting. * If there is no resister in series with D1 and it's internal resistance is negligible compared to R2 then effectively all of the current goes through D1 * The combination of D1/R3 drops 5v because they are in series. Questions: * Why is R2 even necessary? If D1/R3 are dropping 5v what's the point of R2? * If D1 needs 5v but we throw R3 in there too it's not getting the full 5v. Am I imagining a need for R3? If not how do we compensate for that? * Maybe I'm misunderstanding how things rated for 5v work. If a component is rated for 5v does that mean it's rated for 5v with 0 resistance? i.e. it can handle all of the current that 5v produces? Thank you for your patience with my super newb-ness. |
| IanB:
You think R1 and R2 are set up to divide the voltage into 7 V and 5 V. But they will only do that as long as D1 is not present. If you include D1 in the circuit then the R1/R2 voltage divider will no longer do what you think and everything changes. In the real world voltage dividers are not used in this way, for this reason. They don't work effectively of the goal is to power something that will draw significant current. If you want to power something like an Arduino then you need to use a voltage regulator. If you want to power an LED then you need a current limiting circuit, not a voltage regulating circuit. So your circuit basically is hopeless and cannot do what you want. Trying to analyze it will not be fruitful. |
| hamster_nz:
How I would analyse this.... D1 is the 'right' way around, so will start conducting when (if?) about 0.7V is over it. It will take a lot of current before the voltage drop to rise over 1V - over 1A. So assume that the node where R1,R2 and D1 will be 0.8V or so. It will be relatively 'stiff' - it will require a big change in conditions for the voltage at that point to change much - when conducting diodes let "a lot" of current flow through them with ease - many amps for power diodes.. If the supply voltage is 12V, and there is 0.8V over D1 (and also R2) then there must be 11.2V over R1. If R1 has a high resistance it might be closer to 11.3 V. If R1 has a low resistance (say 10 Ohms) it might be 11V. R1 will never have as little as 7V over it - unless the doide has failed, or is installed backwards. The only exception would require very high currents - tens or hundreds of amps. Enough that all the magic smoke will escape from all the components. The other (less intuitive) way of looking at is would be that the R1+R2 voltage divider is a voltage source of 12V * (R2/(R1+R2)) volts, with a source impedance of (R1*R2)/(R1+R2) ohms. When you connect this to the diode, enough current flows through the diode to drop the voltage down to 0.8V. |
| Youkai:
--- Quote from: IanB on April 25, 2018, 04:13:47 am ---So your circuit basically is hopeless and cannot do what you want. Trying to analyze it will not be fruitful. --- End quote --- Hmm good to know. If this type of circuit is more or less useless why do I find so many damn tutorial videos about it then :( |
| hamster_nz:
Oh, I see what you are getting at. Using two resistors to drop 12V to 5V works, but it is very dependent on the ratio of current flowing through the load (in your picture D1) vs the current being 'wasted' in R2. So if R1 is 7 ohms, and R2 is 5 ohms, 1A will be flowing through the divider and with no load connected you will have 5V. If you draw of any appreciable amount of current to power an LED or Arduino or something it will cause the voltage at that point to drop. As a rough rule of thumb for this case, drawing off 10% of the current (100mA in this case) will cause the voltage to drop by 10%. If the load current jumps around (e.g. an Arduino blinking an LED) then the voltage will also jump around. If you want a stable 5V, and don't want to waste a lot of current, then you need a voltage regulator. |
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