How does diode dissipate energy?

[long]

I just realized that I don't really understand how diode consumes energy.

Say, we have a diode in series with a resistor, there is a constant 0.7V voltage drop on it. So we all know that this diode dissipates energy of 0.7 * Current Watt thru this diode.

However, we know that 0.7V is mostly used to overcome the potential barrier present in the in P-N interface.

So this 0.7V is NOT used to overcome ohmic resistance, but to overcome a potential resistance.

So, where does the energy go away? textbook says it becomes heat, but by my above line of thinking, I am not fully convinced.

What is your take? thank you for discussing it with me.

 

[KrisBlueNZ]

The lost energy definitely is converted to heat... But I would also like to know how this happens, if anyone can explain it in a simple and understandable way.

 

[(*steve*)]

The electrons have to climb a 0.7V hill. They get all hot and sweaty doing that.

OK, electrons don't sweat! The diode remains dry, but it does get hot.

Note that the voltage drop is dependant on current. For rectifier diodes a voltage drop of 1V or even higher is not unexpected at the limits of their rated current.

 

[duke37]

The electrons lose energy when they jump down an energy level, some give out light when they do this. I never did understand it and gave up when the concept of reciprocal space was introduced.

 

[Harald Kapp]

Between the diode's N-doped side and the P-doped side an internal potential barrier develops. By applying an external voltage electrons are pushed across this potential barrier leading to the flow of current. The electrons from the N-doped side recombine with holes in the P-doped side, losing their energy in the form of heat.
This energy is supplied by P=V*I from the external source.

More reading here and follow the link at the end of the tutorial.

 

[Ratch]

I just realized that I don't really understand how diode consumes energy.

Say, we have a diode in series with a resistor, there is a constant 0.7V voltage drop on it. So we all know that this diode dissipates energy of 0.7 * Current Watt thru this diode.

However, we know that 0.7V is mostly used to overcome the potential barrier present in the in P-N interface.

So this 0.7V is NOT used to overcome ohmic resistance, but to overcome a potential resistance.

So, where does the energy go away? textbook says it becomes heat, but by my above line of thinking, I am not fully convinced.

What is your take? thank you for discussing it with me.

I have explained how a junction diode works before, but it appears that a lot of folks don't still don't understand it. First and foremost, a junction diode is a diffusion device. It works on the same principle that a drop of ink diffuses throughout a glass of water. When the diode is first manufactured, its P and N type material are bonded together. The electrons from the N-side and the holes from the P-side diffuse into each other's territory and annihilate each other to form a boundary region free of of charge carriers called a "depletion region". When the immobile N-type material on the boundary lose their electrons and become positive ions, and the immobile P-type material on the boundary lose their holes and become negative ions, an equilibrium is reached because the positive ions on the N-side oppose further diffusion of the holes and the negative ions on the P-side oppose further diffusion of the electrons. This gathering of opposite charges cause a barrier voltage. The diode is packaged and shipped in this equilibrium condition.

When inserted into a circuit, a forward bias voltage lowers the barrier voltage and allows a current to exist through the diode. This bias voltage itself does not propel the charge carriers through the depletion region, diffusion does. But, the bias lowers the barrier voltage so that more charge carriers can diffuse and allow a current to exist. Because the bias voltage controls the charge carriers by diffusion instead of directly by an electric field, the current is proportional to the exponent of the voltage. This exponential relationship is true of any diffusion system, not just in semiconductors. A reverse bias does the opposite and can completely cut off the current.

Now, let me debunk a few myths. A current will exist in a diode anytime it is forward biased above zero. The current becomes significant at 0.6-0.7 volts, but it is still there at lower voltages. The current varies with the voltage, which usually changes only a few millivolts. The current can be calculated by Shockley's diode equation, http://en.wikipedia.org/wiki/Diode_modelling . Voltage across the diode controls the current not the other way around. Although you can invert Shockley's equation and get voltage in terms of current, that is only what the voltage should be if that current were present. The physics of the diode determine what controls what, and you cannot explain how you can control a diode with current like you can do so with voltage. Energy is dissipated in a diode by heat and light if it is a LED. The rate of heat energy dissipation is the current times the voltage across the diode.

Ratch

 

[long]

Ratch:

Very appreciate you explain this in detail and it is always good to solidify the foundation of discussion. Thank you!

I do apologize that I adopt a much simpler version of the diode working mechanism, and might be misleading at the first place. True, a diode is a diffusion device plus if there is any forward-bias, it will have some current. 0.6-07V bias voltage is for so called "fully-on" condition.

I just want to understand:

In a resister, when current becomes larger, the classic material textbook says more electron is moving and causing more collisions happen, hence it converts to heat.

But as you have explained, the bias voltage's effect is to lower the barrier. So how is this voltage related to energy consumed by a diode?

I have explained how a junction diode works before, but it appears that a lot of folks don't still don't understand it. First and foremost, a junction diode is a diffusion device. It works on the same principle that a drop of ink diffuses throughout a glass of water. When the diode is first manufactured, its P and N type material are bonded together. The electrons from the N-side and the holes from the P-side diffuse into each other's territory and annihilate each other to form a boundary region free of of charge carriers called a "depletion region". When the immobile N-type material on the boundary lose their electrons and become positive ions, and the immobile P-type material on the boundary lose their holes and become negative ions, an equilibrium is reached because the positive ions on the N-side oppose further diffusion of the holes and the negative ions on the P-side oppose further diffusion of the electrons. This gathering of opposite charges cause a barrier voltage. The diode is packaged and shipped in this equilibrium condition.

When inserted into a circuit, a forward bias voltage lowers the barrier voltage and allows a current to exist through the diode. This bias voltage itself does not propel the charge carriers through the depletion region, diffusion does. But, the bias lowers the barrier voltage so that more charge carriers can diffuse and allow a current to exist. Because the bias voltage controls the charge carriers by diffusion instead of directly by an electric field, the current is proportional to the exponent of the voltage. This exponential relationship is true of any diffusion system, not just in semiconductors. A reverse bias does the opposite and can completely cut off the current.

Now, let me debunk a few myths. A current will exist in a diode anytime it is forward biased above zero. The current becomes significant at 0.6-0.7 volts, but it is still there at lower voltages. The current varies with the voltage, which usually changes only a few millivolts. The current can be calculated by Shockley's diode equation, http://en.wikipedia.org/wiki/Diode_modelling . Voltage across the diode controls the current not the other way around. Although you can invert Shockley's equation and get voltage in terms of current, that is only what the voltage should be if that current were present. The physics of the diode determine what controls what, and you cannot explain how you can control a diode with current like you can do so with voltage. Energy is dissipated in a diode by heat and light if it is a LED. The rate of heat energy dissipation is the current times the voltage across the diode.

Ratch

 

[(*steve*)]

The energy dissipated in all devices is simply due to the work being done in order to move some amount of charge through the device.

If you ignore the storage of energy in the device, then that work is measured as V * I * t and is measured in Joules. Typically we're interested in the rate at which work is being done, and that's (V * I * t) / t, or simply V * I, and is measured in Joules per second, or Watts.

In this respect, the only difference from a practical perspective between a diode and a resistor is that a resistor has a far more linear relationship between V and I than a diode.

Regardless of the reason that a device limits the current for a particular voltage (or conversely drops a particular voltage at a given current), work is being done to move charge through the device. This work may be useful (e.g. in a LED) or produce a useful by-product (e.g. a zener diode), or may simply be wasted (or indeed, in a heater, it may be desired).

It's difficult to know exactly what you're after. Ratch has given you a description of what the work is achieving (or what is achieved by the work, I'm not sure which one causes the other) and maybe you're interested in that. Or maybe you're wondering if there's something different between V*I for a diode compared with V*I for a metal film resistor.

As an example, if you have a diode which has a particular current passing through it, and this results in a particular voltage drop across it (ratch would say the opposite) *AND* you're not concerned with changes due to temperature etc., then you could replace that diode with an appropriate value resistor and you could not tell the difference as long as the operating conditions remain the same.

 

[Ratch]

Ratch:
I just want to understand:

In a resister, when current becomes larger, the classic material textbook says more electron is moving and causing more collisions happen, hence it converts to heat.

But as you have explained, the bias voltage's effect is to lower the barrier. So how is this voltage related to energy consumed by a diode?

Voltage is the energy density of the charge (joules/coulomb). Charge is going to go from a higher energy density (voltage) to a lower density if a conduction path exists. In a conductive resistor, charges are going to move from the higher voltage (energy density) to the lower one. During their travel, they are going to encounter quantum obstacles like ionic cores, electric fields, repulsion from other particles, etc. Again I emphasize, this is not a mechanical friction phenomenon but a quantum one. It is going to take energy to overcome those obstacles, and the particles are going to arrive at the opposite end of the resistor with less energy then they started with. The the same number of electrons with less energy means a lower voltage or voltage drop across the resistor. This loss of energy is dissipated as heat.

Shockley's diode equation gives the current vs the voltage. Multiplying current and voltage together gives the power dissipated by the diode. With a plot, you can find the current at a particular voltage on the curve. That proves that a diode has a nonlinear resistance and a particular power dissipation at any point in the curve.

Ratch

 

[long]

....Or maybe you're wondering if there's something different between V*I for a diode compared with V*I for a metal film resistor......

yeah, that was my question... thank you for your explanation.

 

[long]

thank you Ratch.

Guess I should find a physics book to study how heat is generated from this process...

Voltage is the energy density of the charge (joules/coulomb). Charge is going to go from a higher energy density (voltage) to a lower density if a conduction path exists. In a conductive resistor, charges are going to move from the higher voltage (energy density) to the lower one. During their travel, they are going to encounter quantum obstacles like ionic cores, electric fields, repulsion from other particles, etc. Again I emphasize, this is not a mechanical friction phenomenon but a quantum one. It is going to take energy to overcome those obstacles, and the particles are going to arrive at the opposite end of the resistor with less energy then they started with. The the same number of electrons with less energy means a lower voltage or voltage drop across the resistor. This loss of energy is dissipated as heat.

Shockley's diode equation gives the current vs the voltage. Multiplying current and voltage together gives the power dissipated by the diode. With a plot, you can find the current at a particular voltage on the curve. That proves that a diode has a nonlinear resistance and a particular power dissipation at any point in the curve.

Ratch

 

[Arouse1973]

I know this might be a simple explanation and Ratch covered all of this, but what about this? When a charged particle as in this case an electron is accelerated in a conductor due to the application of an electric field it will collide with ions (atoms that have lost an electron) in the conductor. Some of the kinetic energy the charged particle has is transferred to the ion and released as heat.
Adam