Thank you for these videos! I have a question for this video: how do the equations used determine *reverse* saturation current density? Nothing about the set-up of the problem seems to indicate to me that the diode being analyzed is reverse-biased, but I must be missing something. Is it because we are just using Vbi for the potential, instead of (Vbi - Vd), thus implying that there is no forward bias and therefore we are looking at the reverse bias case?
Ah, good question. We're not assuming it's forward or reverse biased, we just want to see what current would flow at *any* bias, and then from that, we *infer* what the reverse saturation current density is (it's the coefficinet out front that doesn't include the exponential).
I have attempted a simple and intuitive explanation for the diode current dependence on saturation current. The diode current I includes the saturation current represented by the symbol Io. The current for p-n junction diode current for an applied voltage V is I = Io {exp(V/VT) - 1} where Io = {AqD(p)p_no}/L(p) + {AqD(n)n_po}/L(n) In the expression for Io, 'A' is the area of cross-section of the junction, D(p){D(n)} is the diffusion constant for holes{electrons}, L(p){L(n)} is the diffusion length for holes{electrons} and p_no{n_po} is the thermal equilibrium concentration of holes{electrons} in the n-type{p-type} material. Here, Io represents a term which incorporates the thermal equilibrium concentrations of holes and electrons. While diode current I takes into account the applied voltage V, Io does not. Io is multiplied by a factor {e (exp[V/VT]) -1} dependent on the amount of applied voltage and it is this amount of charge carriers which is injected across a forward biased or reverse biased junction. When forward biased, the factor {e (exp[V/VT]) -1} increases with more and more biasing voltage and the current increases sharply with bias. Note: The current in a diode under forward bias and in constant thermal equilibrium is due to the movement of the holes and the electrons moving and recombining on and on and on. When reverse biased, the factor {e (exp[V/VT]) -1} decreases sharply and settles at Io when the bias is large. Here, what it means is that only the thermal equilibrium concentration of carriers participate in conduction. So, the term saturation current Io. Space does not permit me to explain more here but you may refer to the textbook 4 mentioned below. The conduction processes of p-n junctions can be easily understood if Current is understood properly by taking a unified approach to electrostatics and circuits. Electrostatics and circuits belong to one science, not two. These are discussed usually separately in textbooks and science and engineering courses. Watch the two videos listed below to learn about current and the conduction process and surface charges (using a unified approach to electrostatics and circuits) which set up the electric field whose line integral is the potential difference. The battery produces the emf. The last frame of video #1 lists textbooks which discuss all these topics in more detail. 1. th-cam.com/video/REsWdd76qxc/w-d-xo.html 2. th-cam.com/video/8BQM_xw2Rfo/w-d-xo.html I have discussed the exact idea on the relation between diode current and saturation current in detail in Chapter 9 and Appendix E of textbook 4 listed in the video #1.
Would'nt it have been simpler to use ni^2=n*p to get n_p0 ni^2=2.25e20 p=N_a (as we are in p)=10e16 n=n_p0 so n_p0=2.25e4 ? which close to your results did i miss something, isn't it at equilibrium ?
Pls someone should help me solve this...solve, A germanium diode carries a current of 1.56mA at room temperature when a forward bias 0.19V is applied. Estimate the reverse saturation current at room temperature?
Appreciate it, sir.When I was trying to calculate n_p0, I find myself somehow trying to use ni squared divided by p_p0(which is NA). But I realized soon, this result is the concentration of electrons on the p side when no voltage is added to the diode.When a voltage is added to the diode, the concentration of the minority carrier changes dramatically, which means we should use the equation that connects the concentration of the minority carrier on the one side to the majority carrier concentration on the other side. When we have calculated the value of n_po, maybe we can plug that value into the equation n_p0 x p_p0 = ni² to calculate the acurate p_p0 and then compare it with NA to see if our assumption of NA≈p_p0 is reasonable.
I thnik that you actually can use np0=ni squared divided by pp0 even though there is a applied forward voltage. And this is because np0 indicates the concentration in equilibrium state which will not be affected by the applied voltage.
Yup. Although you have to be cared about the sign. If you apply the voltage so you are making a larger electric field, then yes it’s Vbi+V (this is called reverse bias).
Why is Jsat called reverse saturation current though it is calculated in forward bias condition?
Thank you for these videos! I have a question for this video: how do the equations used determine *reverse* saturation current density? Nothing about the set-up of the problem seems to indicate to me that the diode being analyzed is reverse-biased, but I must be missing something. Is it because we are just using Vbi for the potential, instead of (Vbi - Vd), thus implying that there is no forward bias and therefore we are looking at the reverse bias case?
Ah, good question. We're not assuming it's forward or reverse biased, we just want to see what current would flow at *any* bias, and then from that, we *infer* what the reverse saturation current density is (it's the coefficinet out front that doesn't include the exponential).
I have attempted a simple and intuitive explanation for the diode current dependence on saturation current.
The diode current I includes the saturation current represented by the symbol Io.
The current for p-n junction diode current for an applied voltage V is
I = Io {exp(V/VT) - 1} where
Io = {AqD(p)p_no}/L(p) + {AqD(n)n_po}/L(n)
In the expression for Io, 'A' is the area of cross-section of the junction, D(p){D(n)} is the diffusion constant for holes{electrons}, L(p){L(n)} is the diffusion length for holes{electrons} and p_no{n_po} is the thermal equilibrium concentration of holes{electrons} in the n-type{p-type} material.
Here, Io represents a term which incorporates the thermal equilibrium concentrations of holes and electrons. While diode current I takes into account the applied voltage V, Io does not.
Io is multiplied by a factor {e (exp[V/VT]) -1} dependent on the amount of applied voltage and it is this amount of charge carriers which is injected across a forward biased or reverse biased junction.
When forward biased, the factor {e (exp[V/VT]) -1} increases with more and more biasing voltage and the current increases sharply with bias.
Note: The current in a diode under forward bias and in constant thermal equilibrium is due to the movement of the holes and the electrons moving and recombining on and on and on.
When reverse biased, the factor {e (exp[V/VT]) -1} decreases sharply and settles at Io when the bias is large. Here, what it means is that only the thermal equilibrium concentration of carriers participate in conduction. So, the term saturation current Io.
Space does not permit me to explain more here but you may refer to the textbook 4 mentioned below.
The conduction processes of p-n junctions can be easily understood if Current is understood properly by taking a unified approach to electrostatics and circuits.
Electrostatics and circuits belong to one science, not two.
These are discussed usually separately in textbooks and science and engineering courses.
Watch the two videos listed below to learn about current and the conduction process and surface charges (using a unified approach to electrostatics and circuits) which set up the electric field whose line integral is the potential difference. The battery produces the emf.
The last frame of video #1 lists textbooks which discuss all these topics in more detail.
1. th-cam.com/video/REsWdd76qxc/w-d-xo.html
2. th-cam.com/video/8BQM_xw2Rfo/w-d-xo.html
I have discussed the exact idea on the relation between diode current and saturation current in detail in Chapter 9 and Appendix E of textbook 4 listed in the video #1.
n(p0)= ni^2/NA
Indeed it would have been easier to calculate it that way
Would'nt it have been simpler to use ni^2=n*p to get n_p0
ni^2=2.25e20
p=N_a (as we are in p)=10e16
n=n_p0
so n_p0=2.25e4 ? which close to your results
did i miss something, isn't it at equilibrium ?
Pls someone should help me solve this...solve, A germanium diode carries a current of 1.56mA at room temperature when a forward bias 0.19V is applied. Estimate the reverse saturation current at room temperature?
Hint: figure out what the exponential factor contains (e^V_bias/phi_t).
Hello, I'd like to know why in the beginning you assume Na=Nd.
Great question! Absolutely no physical reason, just to keep things simple. In general you will have different doping on each side.
Appreciate it, sir.When I was trying to calculate n_p0, I find myself somehow trying to use ni squared divided by p_p0(which is NA). But I realized soon, this result is the concentration of electrons on the p side when no voltage is added to the diode.When a voltage is added to the diode, the concentration of the minority carrier changes dramatically, which means we should use the equation that connects the concentration of the minority carrier on the one side to the majority carrier concentration on the other side. When we have calculated the value of n_po, maybe we can plug that value into the equation n_p0 x p_p0 = ni² to calculate the acurate p_p0 and then compare it with NA to see if our assumption of NA≈p_p0 is reasonable.
这里面有一个前提，这是一个无限长的PN结
I thnik that you actually can use np0=ni squared divided by pp0 even though there is a applied forward voltage. And this is because np0 indicates the concentration in equilibrium state which will not be affected by the applied voltage.
By saying that I mean this equation always holds in equilibrium state whenever there is an applied voltage.
In fact, you just need to pulg the equation of Vbi into the former equation to get:np0=ni^2/NA
Can I obtain reverse sat current without Area being given?
If you have the physical device you can measure it. Otherwise, no, you can only have the reverse saturation current density.
What if you applied a voltage, would vbi become vbi + applied voltage?
Yup. Although you have to be cared about the sign. If you apply the voltage so you are making a larger electric field, then yes it’s Vbi+V (this is called reverse bias).
Sir, is there any full forms for the terms Ln & Lp?
diffusion lengths for n-type and p-type
Ln=(Dn x τn)^1/2 ,Lp = (Dp x τp)^1/2 ,where Dn,Dp are diffusion coefficients and т time (s)
Thanks a lot!!
Thankyou sir ..
0:56 Your G looks like 6...
You need to be careful when write G...