Simple Diodes Q&A - Constant Voltage Drop & Output Regulation

[Bassalisk]

I have a fairly stupid question about diodes.

Constant voltage drop means that, no matter how much i put a voltage on a diode, there will be always 0,6V drop on it?

Can I use this a simple output regulator?

http://pokit.etf.ba/get/3935756d6354860fdc6a3b72f60c66db.jpg

 

[berkeman]

I have a fairly stupid question about diodes.

Constant voltage drop means that, no matter how much i put a voltage on a diode, there will be always 0,6V drop on it?

Can I use this a simple output regulator?

http://pokit.etf.ba/get/3935756d6354860fdc6a3b72f60c66db.jpg

No.

Have a look at the I-V characteristic graph part-way down this wikipedia page:

http://en.wikipedia.org/wiki/Diode

The forward voltage drop is fairly constant for a range of forward currents, but does grow with current (until you burn out the diode with too much current).

 

[Jiggy-Ninja]

I have a fairly stupid question about diodes.

Constant voltage drop means that, no matter how much i put a voltage on a diode, there will be always 0,6V drop on it?

Can I use this a simple output regulator?

http://pokit.etf.ba/get/3935756d6354860fdc6a3b72f60c66db.jpg

Kirchoff's voltage law. The diode and source are the only elements in the circuit, so they ust have the same voltage.

It's not constant. The current rises exponentially with relation to the voltage.

Alternately, the voltage rises logarithmically with relation to the current.

The breaking point where voltage starts to rise very slowly as the amps climb is usually between 0.6-0.7V (my textbook uses 0.7V). For small currents (mA range), that's a good approximation. For much larger currents though, the voltage will be noticeably higher. A few dozen amps might give you a 1V drop across the diode.

If you put more than that across a diode, say 10V, the current through it would be extremely high. Far more than enough to completely destroy it.

You need a resistor in series with the diode to dissipate the remaining voltage, though your setup still isn't very good. 0.6V is pretty low. Google "Zener diode voltage regulator" for something a little better.

 

[Bassalisk]

Kirchoff's voltage law. The diode and source are the only elements in the circuit, so they ust have the same voltage.

It's not constant. The current rises exponentially with relation to the voltage.

Alternately, the voltage rises logarithmically with relation to the current.

The breaking point where voltage starts to rise very slowly as the amps climb is usually between 0.6-0.7V (my textbook uses 0.7V). For small currents (mA range), that's a good approximation. For much larger currents though, the voltage will be noticeably higher. A few dozen amps might give you a 1V drop across the diode.

If you put more than that across a diode, say 10V, the current through it would be extremely high. Far more than enough to completely destroy it.

You need a resistor in series with the diode to dissipate the remaining voltage, though your setup still isn't very good. 0.6V is pretty low. Google "Zener diode voltage regulator" for something a little better.

Thats what has been confusing me, Zener diode. It basically works the same way, just inverted and on greater voltage.And if i put a resistor in parallel with a Zener diode, it would be on the same voltage as the inverted Zener diode right?
Why wouldn't this work with a normal diode, just 0,6 v?Lets make Ideal assumptions for CVD models. Graphs are linear.

 

[Studiot]

Firstly, yes you could use the diode as the basis for a simple regulator.

This type is known as a shunt regulator.

It is called this because it is in parallel with the load. It works in conjunction with the load to draw a constant total current from the source. If the load varies the shunt regulator varies its own draw to compensate.

As JN noted you have to draw this current through a resistor from the source.
This is a series resistor, not a parallel one.

The diode may be used in forward voltage mode like this and indeed is used this way for this purpose within some integrated circuits.

However you can only get 0.6 volts this way.

If you reverse the diode and use it in zener mode then a wider range of stabilised voltages are available and, of course, the zener diode is designed and optimised for this purpose.
(All diodes have a reverse breakdown voltage that 'could' be used this way, just as you 'could' use a zener in the forward mode. But both such operations would be sub optimal).

There are, however simple better ways to implement shunt regulators using transistors.

go well

 

[Bassalisk]

Firstly, yes you could use the diode as the basis for a simple regulator.

This type is known as a shunt regulator.

It is called this because it is in parallel with the load. It works in conjunction with the load to draw a constant total current from the source. If the load varies the shunt regulator varies its own draw to compensate.

As JN noted you have to draw this current through a resistor from the source.
This is a series resistor, not a parallel one.

The diode may be used in forward voltage mode like this and indeed is used this way for this purpose within some integrated circuits.

However you can only get 0.6 volts this way.

If you reverse the diode and use it in zener mode then a wider range of stabilised voltages are available and, of course, the zener diode is designed and optimised for this purpose.
(All diodes have a reverse breakdown voltage that 'could' be used this way, just as you 'could' use a zener in the forward mode. But both such operations would be sub optimal).

There are, however simple better ways to implement shunt regulators using transistors.

go well

Yea i get it, that's what i wanted to know. The possibility. I know Zener is mainly designed for this. But can you explain why you need to put a series resistor along with a parallel one?

And wouldn't my resistor on the parallel side be on 0,6 volts? Thats the main point, to maintain constant voltage drop, only current varies?

 

[Studiot]

It's all in the attachment.

I said and repeat you do not want a parallel resistor.

(You can have a parallel resistor as a load, but it then has nothing to do wit the regulation)

 

[Jiggy-Ninja]

Yea i get it, that's what i wanted to know. The possibility. I know Zener is mainly designed for this. But can you explain why you need to put a series resistor along with a parallel one?

And wouldn't my resistor on the parallel side be on 0,6 volts? Thats the main point, to maintain constant voltage drop, only current varies?

A resistor in series with the diode.

Remember Kirchoff's voltage law: The algebraic sum of all voltages in a circuit must be 0V.

If your source is something like 5V, and you want to use a diode to regulate it down to 0.6V, whee does the 4.4V left go? It can't just disappear, it has to be accounted for somewhere.

That's where the series resistor comes in. The current through the resistor dissipates that 4.4V as heat, and leaves 0.6V for the diode.

 

[Bassalisk]

I think i get it, thanks

I was going after this.

But just with regular diode.

 

[Studiot]

The circuit diagram is fine.

The breakdown (approx zener) voltage of a 'regular' diode will not be less than 50 volts, may be a few hundred.

What voltage do you actually require?

And what are you actually trying to do? The regulated current available from a zener is usually quite low, unless you have a very high power one.

 

[Bassalisk]

I am just trying to understand. I am asking if you COULD do it, not actually need it. They don't teach you this stuff in class. And my questions pop out from everywhere. Trying to get some practical thinking. Why, when, how, why not.

 

[Studiot]

So have you got it now?

Zener diodes are made in breakdown (zener) voltages of 3.3 to about 300 volts in small steps.

Regular diodes are made not to breakdown in reverse mode so they have a parameter called peak inverse voltage (PIV) which is usually from about 50 - 5000 volts, but the steps are

50, 75, 100, 150, 200, 400...

 

[Bassalisk]

Yes I did. In practice, people needed this feature of diodes, this CVD of 0,6 Volts so they designed a special one just to do that am I right? Hence the Zener diode.

Thank you very much

 

[Studiot]

In practice, people needed this feature of diodes, this CVD of 0,6 Volts so they designed a special one just to do that am I right? Hence the Zener diode

Where did I say that? Zeners are used in reverse mode, just as you have drawn.

The lowest possible zener voltage is 3.3 volts.

Zener diodes are made in breakdown (zener) voltages of 3.3 to about 300 volts in small steps.

I did say that the forward voltage of 0.6 volts is sometimes used to provide a lower reference. Obviously you can stack these up to 0.6, 1.2, 1.8 etc.

However these days the modern reference at that level is the bandgap device which has a reference voltage of about 1.2 volts.

 

[Bassalisk]

Where did I say that? Zeners are used in reverse mode, just as you have drawn.

The lowest possible zener voltage is 3.3 volts.



I did say that the forward voltage of 0.6 volts is sometimes used to provide a lower reference. Obviously you can stack these up to 0.6, 1.2, 1.8 etc.

However these days the modern reference at that level is the bandgap devioce which has a reference voltage of about 1.2 volts.

Yea correction, that's what i meant. I know that Zener diodes work in reverse bias. Sorry.

 

[Bassalisk]

So just to make things clear, I studied this rectifier with Zener diode.

[PLAIN]http://pokit.etf.ba/get/34b33911e875befc948d82819dddf3f4.jpg

Here is the graph of this circuit.
http://pokit.etf.ba/get/0951ccb3d7c640faba2986bed3a892fe.jpg

Now, this gives a constant voltage of 5 Volts on the R8 resistor with red hooks.

But I just want to make things clear what is going on when sinusoid goes through its period. Corrections are welcome.

When without Zener diode, we would have a small fluctuation of voltage, that ripple voltage, but the Zener diode keeps the voltage constant, on the resistor and diode.

But something must to compensate for that constant voltage, current must change somewhere. So in addition to keep a CVD on resistor R8 of 5 volts, through Zener diode goes weaker or stronger current depending what time sinusoid is at. That resistor in series R7 is there to dissipate that current into Joule energy.

Probably a lot wrong, but as a i said corrections are welcome.

 

[Studiot]

No the zener does not affect the ripple.

Regulation by zener can only affect variations in the load. It does not affect variations in the supply - which is what ripple is.

C4 is there to remove as much of the ripple as practicable. It charges to the peak value of the AC supply from the bridge and supplies current to the load when the instantaneous value of the AC supply is below this peak. I see it is a reasonable value 1mF for this purpose.

The purpose of R8 is to provide a (light) load for the power supply when there is no other load connected. Normally we would expect to see a much lower value of load than 10k.

 

[Bassalisk]

In a nutshell, will I have constant current AND voltage through load?

 

[Studiot]

Of course not, unless the load is constant.

The earlier stuff I gave you said that the total current taken by both the zener and the load is constant.

As the load changes its resistance so the zener changes its current draw to maintain this total.

Oh and you have a voltage across the load, not through it.

 

[Bassalisk]

Sry not native English speaker. Ok i get it. Thank you very much. You have been very helpful

 

[Jiggy-Ninja]

No the zener does not affect the ripple.

Regulation by zener can only affect variations in the load. It does not affect variations in the supply - which is what ripple is.

C4 is there to remove as much of the ripple as practicable. It charges to the peak value of the AC supply from the bridge and supplies current to the load when the instantaneous value of the AC supply is below this peak. I see it is a reasonable value 1mF for this purpose.

The purpose of R8 is to provide a (light) load for the power supply when there is no other load connected. Normally we would expect to see a much lower value of load than 10k.

Zener would suppress the ripple, as long as the ripple from the supply never went low enough to take the Zener out of regulation. The voltage across the resistor will vary, while the Zener's voltage stays relatively constant.