I mean I want to know something such as thevenin and nortan ? what if I have such a resistor with resistance R, and a diode? diode will have two directions > but please answer with not carring to its direction
DEPENDS ON THE RESISTOR VALUE, REMEMBER CURRENT TAKES PATH OF LEAST RESISTANCE. IT IS POSSIBLE THE DIODE WILL NOT ENOUGH LIGHT UP DUE TO LACK OF SUFFICIENT CURRENT. AS A POSSIBLE SOLUTION, PLOT THE IV CURVE OF THE LED, THEN SUPERIMPOSED THE RESISTOR CHARACTERISTICS. WHEN THE DIODE CURVE CROSSES THE RESISTOR LINE THEN THE DIODE CURRENT WILL INCREASE FASTER THAN THE RESISTOR CURRENT. IF THE RESISTOR LINE IS BELOW THE DIODE CURVE THEN THE CIRCUIT WILL LOOK LIKE THE RESISTOR DOES NOT EXIST.
Okay - you got me - my eyes saw one thing and my mind said another I do have an excuse CRS and it gets worst with age.
And resistor not existing is an asymptote but in practical terms we can neglect its effects. If you want to be a Phd about it then you must always include it.
Also, my point was to make it clear that it doesn't just suddenly "look like the resistor does not exist", it progresses towards that, albeit, rather quickly
HI There; Thank you verymuch my friends for your answers. So you mean that we will have just a diode. so what if we have these materials in series with eachother? what we will have for the answer ? (I mean a resistor and a diode in series) Thanks
Not true. Some current will flow through the resistor, which will skew that characteristic (more or less, depending on the value of the resistor).
The following thought-experiment will exemplify this:
Imagine you have a variable resistor across that diode. Now, imagine what happens as you adjust that resistor closer and closer to zero ohms. Them, image what happens when you reach zero ohms. Does the system still have the characteristics of a diode?
in my openion, it depends on our circuit, you know if the current flow in the same direction with diode, (I mean if diode passes currents) so there will be no current flowing in the resistor, but because of my less practical knowledge I dont know what will happen if we have no resistance . I'll be happy that if possible help me.
By "no resistance" I'm going to assume you mean "zero resistance" and say ALL of the current will flow through the zero resistance path. Now, in the real world, there will likely be some resistance (unless the shunt is a superconductor), in which case the current will divide between the two.
So, there are several modes to look at: 1. diode forward biased. 2. diode reverse biased but not in avalanche or breakdown. 3. diode in avalanche or breakdown.
For case 2, the if any current flows through the diode it will be the reverse leakage current. But, if the "resistor" has 0 ohms, then, still, all of the current will flow through that 0 ohm path. No current will flow through the diode. Otherwise, the current will divide between the two. In most cases the current through the diode will be negligible.
For case 3, the diode will have a fairly constant [reverse] voltage across it and the current will be hard to predict (in the case of a voltage source connected directly across the resistor diode pair).
For case 1, there will also be a fairly constant [forward] voltage across the diode--not as constant as in case 3, though. Again, the current will be hard to predict.
What will happen if a voltage source is placed directly across the resistor diode pair in case 1 and case3? The current through the diode will rise to a limit determined by the internal resistance of the source -- or it will generate lovely curls of acrid smoke (in the case that the voltage source has a low enough internal resistance to allow enough current to flow to fry the diode ;)
Because the diode is a non-linear device with characteristics that vary with temperature (much more so than most resistive materials) and that vary between parts (no two are exactly alike -- of course that can be said for some resistive devices, too) -- it's a little difficult to come up with precise currents, especially since the math is a bit hairy:
Basically the circuit becomes a diode with a resistor across it and another resistor (the internal resistance of the voltage source) in series with the resistor/diode pair.
For case 1, if you approximate the diode voltage at somewhere between .6 to .7 for small currents and upwards of 1 to 1.5 volts for large currents, then knowing that the voltage across all parallel circuit components is the same, the resistor will have the same voltage across it as the diode. So, you can simply use ohm's law to determine the current in the resistor [Ir = Vd/R], and the current in the diode will be the total current minus the current in the resistor. The total current can be determined with the following formula:
It = (Vs - Vd)/Rs
where *It* is the total current, *Vs* is the magnitude of the source voltage (measured with no load attached), *Vd* is the forward voltage of the diode, and *Rs* is the internal resistance of the voltage source.
If the voltage source is perfect (i.e. has an internal resistance of 0) and if the diode is capable of handling any amount of current, then what will happen when you put the resistor/diode pair across the voltage source is this: the forward voltage of the diode will rise until it is equal with that of the voltage source. If the magnitude of the voltage source is much more than a volt or two, there will be a massive amount of current flowing through that diode -- image how much current would have to be flowing in a diode for the forward voltage to be at, say, 5 volts -- remember, it's an exponential relationship!
Case 3 is similar to case 1. The diode will have it's avalanche or breakdown voltage across it, and the current will rise until limited by the source resistance. The parallel resistor will have the same voltage across it and ohms law applies.
Yes, but that would mean the circuit would behave like a diode without a resistor. I figured Saijad was smart enough to know that, so I interpreted his words as a grammatical error.
✨ The discussion addresses the electrical behavior of a resistor and diode connected in parallel. The outcome depends on several factors including the type of input voltage (AC or DC), polarity, resistor value, diode type, and diode peak inverse voltage (PIV). Generally, current favors the path of least resistance, so if the resistor has very low resistance, most current will bypass the diode, potentially preventing it from conducting or lighting up if it is an LED. The diode's I-V characteristic curve compared to the resistor's linear characteristic determines current distribution; when the diode's curve surpasses the resistor's, diode current increases significantly. In practical terms, the resistor's effect can often be neglected if its resistance is high relative to the diode's conduction path, but it never truly "disappears." Different diode biasing conditions (forward, reverse, avalanche) affect current flow, with reverse leakage current being minimal unless breakdown occurs. The discussion also clarifies that zero resistance in parallel means all current flows through the resistor path, effectively bypassing the diode. The question of series connection of resistor and diode was briefly raised but not deeply explored.
TL;DR: A silicon diode in parallel with a resistor sets about 0.6–0.7 V across both; “current will divide between the two.” Source resistance ultimately limits current in forward or breakdown modes. [Elektroda, Steve Lawson, post #21664404]
Why it matters: This helps you predict current sharing, avoid diode damage, and design safe clamps or shunts in real circuits.
What happens when a resistor and diode are connected in parallel on a DC source?
Both parts share the same voltage. In forward bias, the diode fixes about 0.6–0.7 V, so the resistor current is Ir ≈ Vd/R. Total current is set by source resistance; the diode takes the balance. [Elektroda, Steve Lawson, post #21664404]
Does all current go through the diode if it’s forward biased?
No. Current splits between the diode and the resistor based on their I–V curves and the source’s internal resistance. “Otherwise, the current will divide between the two.” [Elektroda, Steve Lawson, post #21664404]
What if I short the diode with a 0 Ω resistor in parallel?
A true 0 Ω shunt takes 100% of the current, and the diode current is 0 A. Any nonzero shunt will split current with the diode. [Elektroda, Steve Lawson, post #21664404]
How does the circuit behave in reverse bias?
Below breakdown, only tiny reverse leakage flows through the diode; the resistor carries almost everything. In breakdown, diode voltage is roughly constant while source resistance limits current. [Elektroda, Steve Lawson, post #21664404]
Will the resistor ever ‘not exist’ when paralleled with a diode?
It asymptotically looks negligible as the diode clamps voltage, but the resistor still draws Vd/R current. It never truly disappears. [Elektroda, Steve Lawson, post #21664402]
Is this the same as having just a diode in parallel?
Not exactly. The resistor skews the composite I–V curve and increases total current at the diode’s clamped voltage. Behavior depends on R and source resistance. [Elektroda, Steve Lawson, post #21664402]
How do I estimate currents quickly without solving exponentials?
Use a diode drop approximation. Assume Vd ≈ 0.6–0.7 V (small I) or ≈1–1.5 V (large I), then compute Ir = Vd/R and It = (Vs − Vd)/Rs. [Elektroda, Steve Lawson, post #21664404]
What’s the failure or edge case I should watch for?
With a low-R source, diode current can skyrocket in forward or breakdown mode and overheat the junction—visible smoke is possible. Add series resistance. [Elektroda, Steve Lawson, post #21664404]
How does AC input change the behavior?
During positive half cycles (forward bias), the diode clamps near its forward drop and shares current with the resistor. During negative cycles, leakage flows unless breakdown is reached. [Elektroda, Steve Lawson, post #21664404]
What did the original thread mean by ‘no resistance’?
One post clarifies that “no resistance” means infinite resistance (open circuit), not zero. Precision avoids confusion in analysis. [Elektroda, David Adams, post #21664405]
What if the resistor and diode are in series instead of parallel?
Series connection limits diode current. You pick R so Vd is reached without exceeding current limits; it’s a safer, predictable configuration. [Elektroda, Steve Lawson, post #21664402]
How can I visualize current sharing between a resistor and a diode?
Plot the diode’s I–V curve and overlay the resistor’s load line at the applied voltage. Their intersection gives operating current and voltage. [Elektroda, David Adams, post #21664396]
Can a reverse-biased diode with a parallel resistor be used as a clamp?
Yes. In breakdown, the diode holds near its breakdown voltage while the resistor and source resistance set current. Ensure current stays within limits. [Elektroda, Steve Lawson, post #21664404]
Do LEDs behave differently in this parallel setup?
Principles match, but LED forward voltage varies with color and current. The resistor will still take Vd/R at the LED’s drop. [Elektroda, Steve Lawson, post #21664402]
Quick how-to: analyze a diode‖resistor at a given supply voltage.
Approximate the diode drop (e.g., 0.7 V) or its breakdown value.
Compute resistor current Ir = Vd/R and total It = (Vs − Vd)/Rs.
What does “PIV” or reverse voltage rating mean here?
It’s the maximum reverse voltage before breakdown risk. In this parallel circuit, ensure supply never exceeds the diode’s reverse rating. [Elektroda, Steve Lawson, post #21664404]
