Could you explain the terms forward bias and reverse bias in relation to diodes and what the voltage drop means? I understand that diodes have a very high resistance in the ‘opposite’ direction, so what does breakdown voltage mean / relate to the shape of your graph, and why is the value at which the current approaches infinity, 0.6 V?
Here’s another comment with a link that should cover all this material https://oxfordpat.wordpress.com/comment-page-1/#comment-378
thank you very much for your reply.
Can you please explain this point you’ve made? “Then you need to think about the abnormal case — a big discharge. This is another keyword: charge needs to flow to ground, so a big current needs to flow. From the graph we can see that the protection circuit allows any amount of current to flow but the voltage never exceeds the forward voltage drop, so the voltage across the amplifier inputs is limited and this protects the amplifier.”
Sorry that wasn’t very clear, and it was maybe stretching a point. I meant that the word ‘discharge’ is a keyword, but that is maybe putting it a bit strongly. In any case a discharge (like a lightning discharge, for example) involves a sudden big current flow from a high potential to a low potential.
If the discharge current goes through the amplifier it could destroy it. But in this case the diode protection circuit works like a lightning conductor would, providing a path of near zero resistance as long as the voltage across it is equal to the forward voltage drop of the diode: in this case a huge current can flow through the protection circuit, but the voltage across the amplifier inputs never exceeds the forward voltage drop of the diodes.
Wouldn’t the resistance also be near zero for voltages that exceed the voltage drop across the diode?
Yes, that’s right, but because the resistance is zero the voltage can’t get any higher.
but a discharge voltage, for example a lightning strike, would succeed the voltage drop across the diodes, so how can it be that the voltage doesn’t exceed the voltage drop?
When the lightning bolt strikes, consider the path that current takes from the charged cloud to the ground. It goes through the ionized air (with some quite big resistance), then through the diode circuit (with zero resistance), and then to ground. The total voltage from cloud to ground is big, but it is all across the parts of the circuit that have a resistance. It is an iron law that a resistor of zero ohms has zero volts across it. Of course, this is how the circuit actually protects the amplifier.
What was your thought process when you try to tackle this kind of questions (when knowledge is applied in unfamiliar real life situation)? I have come across diodes before in AS, and I know their characteristics. However, when I need to apply my knowledge to these type of questions, I usually get really stuck. Do you have any technique that you use, and what should I do to improve on these type of questions?
Thank you very much!
Hi Claire, thanks for a very interesting question.
In this case you are asked to sketch a graph and so that is the place to focus your energies, especially since the second part of the question sounds so weird, the reasoning being that the graph might shed some light on the second part. When sketching a graph of a physical situation there are a few techniques to remember.
So in this case, thinking about the slope, plus a bit of knowledge about diodes, gets us a long way.
If we now think about the second part of the question there are some more techniques to consider.
Look for keywords, especially adjectives
Don’t forget the adjectives — they are always there to communicate some information. The amplifier is a ‘sensitive’ one for measuring ‘small’ signals. So we are considering small changes in voltage around the origin, and in this case we can assume that the signals are less than the forward voltage drop. In this normal case the protection circuit just looks like an extremely big resistor. Then you need to think about the abnormal case — a big discharge. This is another keyword: charge needs to flow to ground, so a big current needs to flow. From the graph we can see that the protection circuit allows any amount of current to flow but the voltage never exceeds the forward voltage drop, so the voltage across the amplifier inputs is limited and this protects the amplifier.
And when you have solved a problem, look back over it and try to think about techniques and arguments that you used, and try to refine them so that you can apply them in similar circumstances.
I hope this is some help. Finally, I know this is easy to say and harder to do, but think positively, stay calm, and concentrate on what you do know, not what you don’t.
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