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lotec

Apr 15, 2013
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Hi All,
I am looking at building an astable multivibrator using an ne 555. While looking for circuits I came accross one that allowed for independantly controlliing the on and off times, by providing a separate charge and discharge path to the capacitor by using blocking diodes.
Another website talked about that circuit and said that they had to be fast diodes if it is going to be used at high frequencies.

My question is this.
Take a diode like IN1007. Used as a blocking diode, would this or any other cheap common diode, block instantaneously or do they take some time to establish that blocking action. If so then the higher the frequency probably the less blocking occurs. Most time I here the word fast with diodes it is usually talking about the recovery time. Does this mean recovery from an avalanch type reverse breakover situation. Can someone please explain recovery time and do diodes that are not fast delay before blocking.

Thanks in advance
Wil.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Imagine a real diode is like a "perfect diode" with a capacitor in parallel.

As the voltage reverses, some current flows to charge the capacitor. This is one way of viewing recovery time.

Typically we're only interested in the recovery time between forward current flow and a reverse blocking, but I'm sure there's a similar effect in the other direction too.

Small signal diodes (such as 1N4148 and 1N914) are faster. Typically they have a much smaller junction area. This allows less current, but also causes less capacitance.
 

lotec

Apr 15, 2013
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Hi Steve,
Thanks for taking the time to help me here,

I hope I have understood you rightly. As current goes forward through the diode some break down occurs and the recovery time means the time taken after the current has stopped for the diode to repair its self so that current cant leak or flow back the way it came

.As for the blocking I took that to mean that a diode in its recovered conditiion may or may not instantly block but it may absorb some of the juice because of its capacitance.

The bloke who brought this to my attention told me about his circuit, and how the lights took a couple of seconds to go out when the power is turned off, due to capacitance in the diodes. I think in his own way he was talking about recovery time as you just explained it to me as the reason for why they may not block instantly. Taking another look at the circuit, at a certain frequency recovery time will be an issue. I think my confusion arose from wrongly understanding recovery time and miss understanding what he said.

Many thanks
wil.
 

Harald Kapp

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As current goes forward through the diode some break down occurs and the recovery time means the time taken after the current has stopped for the diode to repair its self so that current cant leak or flow back the way it came
That's not so. In the forward direction no break down occurs. Instead you have an electrical field between the p-doped region and the n-doped region of the diode. That's where the voltage drop occurs.
This field is sustained by opposite charges that collect on both sides of the p-n junction. The function can be thought of as a capacitor that needs to be discharged for the diode to stop conducting.

There's no break down and/or repair involved.
 

duke37

Jan 9, 2011
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If you are using these diodes in a 555 circuit, then almost anything will do since the 555 is not that fast.
stadard diodes 1000ns
fast diodes 150 to 200ns
ultrafast 25 to 75ns
schottky <10ns

The current in a 555 circuit wll be low so use the 1N4148.

Any long delay for the circuit to die when the power is turned off will be due to energy retention in electrolytic capacitors.
 
Last edited:

lotec

Apr 15, 2013
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Hi Harald Kapp thanks for pointing that out to me.

Hi Duke37 thanks for those specs. very handy.
 

BobK

Jan 5, 2010
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A 1N4148 actaully has a recovery time of 4ns.

Bob
 

lotec

Apr 15, 2013
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Thanks Bob,
I was looking at about 2khz, so I think I should be well covered and then some.
 
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