Diodes confuse me

  • Thread starter ness9660
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Usually I can pick up most concepts pretty quickly, but EE just makes me feel dumb.

Ive managed to grasp most concepts we have covered this semester, except one, diodes. I really dont understand how to determine forward or reverse bias. For example:

http://img276.imageshack.us/img276/7519/diode2jw.gif [Broken]



Now, I know D1 and D2 are in reverse bias, and D3 is forward biases, so that Vout=-5 + .6 = -4.4V.

But I dont understand how any of that was determined. I can manage, sometimes, to correctly complete simple circuits that have one Voltage source, one diode and one resistor. For those simple ones I use the method of assuming a diode state and solving for voltage, but this situation above baffles me.


Is there some simple trick that I am missing?
 
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  • #2
Astronuc
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ness9660 said:
Usually I can pick up most concepts pretty quickly, but EE just makes me feel dumb.

Ive managed to grasp most concepts we have covered this semester, except one, diodes. I really dont understand how to determine forward or reverse bias.


Now, I know D1 and D2 are in reverse bias, and D3 is forward biases, so that Vout=-5 + .6 = -4.4V.

But I dont understand how any of that was determined. I can manage, sometimes, to correctly complete simple circuits that have one Voltage source, one diode and one resistor. For those simple ones I use the method of assuming a diode state and solving for voltage, but this situation above baffles me.
Are you sure about that image? D2 and D3 have the same bias, i.e. both have -5 V and share a common node, according to the image.

See - http://hyperphysics.phy-astr.gsu.edu/Hbase/solids/diod.html

http://www.allaboutcircuits.com/vol_3/chpt_3/1.html
 
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  • #3
ranger
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Is there some simple trick that I am missing?

Gererally: to forward bias a diode, the anode must be more positive than the cathode or LESS NEGATIVE.
to reverse bias a diode, the anode must be less positive than the cathode or MORE NEGATIVE.
 
  • #5
Astronuc
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Are the v[itex]_\gamma[/itex]'s given for D2 and D3?
 
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Astronuc said:
Are the v[itex]_\gamma[/itex]'s given for D2 and D3?

Yeah, .7v for D2 and .6v for D3. .7V for D1 also
 
  • #7
berkeman
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The solution is mostly correct, and only D3 is forward biased enough to turn it on hard. However, D2 is on a little bit, so it's not correct to say that D2 is "reverse biased". A diode is only reverse biased if the voltage of the cathode is higher than the voltage of the anode.

The best way to think about circuits like this is to sketch the diode V-I plot from the diode equation. Different values of Is will give you different forward voltages versus current (and temperature). The diode V-I plot has Id = -Is for reverse bias (where Va-Vk < 0), it goes through the point 0,0, and rises up to the right. Va-Vk is the horizontal axis, and Id is the vertical axis. Depending on the diode and temperature, you will get numbers in the 0.5V to 0.75V range for the forward voltage drop at moderate forward currents.

So in the original question, do as the solution suggests and figure out which diode is on the hardest first (taking the most A-K current). Since both D2 and D3 are going to clamp Vout near -5V, D1 has its K near 0V and its anode near -4.4V, so it is reverse biased and only has Is flowing through it from K-A. Then look at D2 and D3 -- D3 is said to have the lower forward voltage drop, so it will be taking the majority of the current and clamping Vout near -4.4V. D2 is therefore at the spot on its diode V-I plot where V=0.6V, and it will have a smaller positive current flowing through it compared to D3, but D2 is definitely not off or reverse biased. You can work out the numbers from the diode equation, but D2 will probably have about 1/10 of the forward current that D3 does.

Makes more sense now?
 
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  • #8
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berkeman said:
The solution is mostly correct, and only D3 is forward biased enough to turn it on hard. However, D2 is on a little bit, so it's not correct to say that D2 is "reverse biased". A diode is only reverse biased if the voltage of the cathode is higher than the voltage of the anode.

The best way to think about circuits like this is to sketch the diode V-I plot from the diode equation. Different values of Is will give you different forward voltages versus current (and temperature). The diode V-I plot has Id = -Is for reverse bias (where Va-Vk < 0), it goes through the point 0,0, and rises up to the right. Va-Vk is the horizontal axis, and Id is the vertical axis. Depending on the diode and temperature, you will get numbers in the 0.5V to 0.75V range for the forward voltage drop at moderate forward currents.

So in the original question, do as the solution suggests and figure out which diode is on the hardest first (taking the most A-K current). Since both D2 and D3 are going to clamp Vout near -5V, D1 has its K near 0V and its anode near -4.4V, so it is reverse biased and only has Is flowing through it from K-A. Then look at D2 and D3 -- D3 is said to have the lower forward voltage drop, so it will be taking the majority of the current and clamping Vout near -4.4V. D2 is therefore at the spot on its diode V-I plot where V=0.6V, and it will have a smaller positive current flowing through it compared to D3, but D2 is definitely not off or reverse biased. You can work out the numbers from the diode equation, but D2 will probably have about 1/10 of the forward current that D3 does.

Makes more sense now?


Thanks, it makes much more sense now.


Just testing latex below here:


[itex]
theta=(s&pi;x)/&lambda;
[/itex]
 

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