I've known the diode drop the voltage by 0.6V.

But why is the dropped voltage different depend on the resistor at the below pictures?

 

I've known the diode drop the voltage by 0.6V.

But why is the dropped voltage different depend on the resistor at the below pictures?

Because the remaining voltage drops across the resistor.

The diode and resistor will be the 2 components of the voltage divider and 0.7V drop across diode and the remaining (5 - 0.7)V if source is 5V drop across the resistor.

 

hi JD,
The resistance response of the diodes forward voltage versus current is not linear, ref this image.

E

 

Ohm's law: V=I*R

Plus the voltage drop over the diode is "0.5v" to "0.7v", we take it " 0.6v" middle, on some diodes its "0.2v", and even at " 0.3v-0.4v", in real devices the diode will conduct a little.

 

See

 

The diode that you are using has a voltage drop of "0.9v, 1v", check the data sheet.

 

I've known the diode drop the voltage by 0.6V.

But why is the dropped voltage different depend on the resistor at the below pictures?

Hi,

The most basic answer is that two things in series drop voltages, and the one with the higher resistance drops the most voltage. So if you increase one resistance that resistance will drop more voltage and thus the other device in the series string will have less voltage across it now.

The basic rule is:
The sum of the voltage drops around a series circuit is equal to zero.

 

Hi,

The most basic answer is that two things in series drop voltages, and the one with the higher resistance drops the most voltage. So if you increase one resistance that resistance will drop more voltage and thus the other device in the series string will have less voltage across it now.

The basic rule is:
The sum of the voltage drops around a series circuit is equal to zero.

Resistor is in series with the diode and resistance limits the current flow. Higher the resistance more voltage is dropped across the resistor and less current flows through it.

I = V/R

If R is increased then I through the series circuit reduces.

V = IR

If R is increased then V (drop across resistor) across resistor increases.

V = Vd + Vr

Vr = (V - Vd) / I

Vd = (V - Vr) / I

I = (V - Vr - Vd) / (Rr + Rd)

 

In other words "Kirhov's law" for the voltage and "Ohm's law for the voltage over the resistor. Normally you should know the drop over the diode (if we take 0.6V) and than you should calcualte the remaining voltage over the resistor and the current that you will receive. It must be enough to drive the diode.

Like mentioned in the previous posts.

 

Resistor is in series with the diode and resistance limits the current flow. Higher the resistance more voltage is dropped across the resistor and less current flows through it.

I = V/R

If R is increased then I through the series circuit reduces.

V = IR

If R is increased then V (drop across resistor) across resistor increases.

V = Vd + Vr

Vr = (V - Vd) / I

Vd = (V - Vr) / I

I = (V - Vr - Vd) / (Rr + Rd)

Hi,

I was quoting Kirchhoff's second law, which is the sum of voltage drops in a conservative series circuit (or loop if you will) is equal to zero, which implies that if more voltage is dropped across one element then less is dropped across the other if the source voltage remains unchanged. So it does not matter if we have a diode or resistor as the second element because if we make one resistance higher, the other resistance will see less voltage drop. The diode of course follows the diode law, but that's a secondary issue. If it is a second resistor that resistor follows Ohm's Law of course, but that is also a secondary issue from this point of view.

In other words, if we have two elements A and R where R is a resistor and A is an unknown device and they are in series and powered by a 10v constant DC voltage source and the drop across R is 5 volts, then the drop across A is 5 volts. But if then we change R such that it's drop is 6 volts, that means the drop across A must now be 4 volts. So the drop across A changed as a simple consequence of the increase in the other element's voltage drop. That's Kirchhoff's second law as it applies to a conservative system.
Note that we did not have to solve for the current because we already knew the voltage drops of 2 out of 3 elements. To actually solve the circuit we may have to consider the current, but once given the resistor's voltage drop we then right away know the drop across the second passive element without having to solve for the current. In the practical case this would mean measuring the voltage drop across the resistor R with a suitable voltmeter.

 

I've known the diode drop the voltage by 0.6V.

But why is the dropped voltage different depend on the resistor at the below pictures?

The two different values of resistor (10Ω and 10kΩ) result in different amounts of current through the diode. Diode forward voltage is not constant, and 0.6V (or 0.65V or 0.7V, or whatever your preference) is merely an approximation. In reality, diode forward voltage varies as a roughly logarithmically function of forward current.

 

The two different values of resistor (10Ω and 10kΩ) result in different amounts of current through the diode. Diode forward voltage is not constant, and 0.6V (or 0.65V or 0.7V, or whatever your preference) is merely an approximation. In reality, diode forward voltage varies as a roughly logarithmically function of forward current.

Hi,

Yes but as i explained in my previous post, it may not matter what kind of device is present in place of the diode because once we know the voltage drop across the 10k resistor we know the drop across the second element,which can be a diode or another resistor or a lump of coal

The question was kind of short though, so it could be that he could not understand how the diode voltage changed when he took it to be a *constant* 0.6v in the past. In that case we *do* need the diode law to explain this.

 

hi JD,
The resistance response of the diodes forward voltage versus current is not linear, ref this image.

E
View attachment 133367

Thank you.

 

See
View attachment 133368 View attachment 133369

Thank you very much.