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Can we use zener diode as a rectifire"bridge circuit"

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i am a new member of this forum and i am a beginner so i have a question in my mind for many days
can we use zener as a rectifire in bridge circuit
plz answer me as soon as possible
 

Karkas

Member
If you use the Zener diode, it will allow the flow of current where it hasn't to flow, i mean if the current is exceeds the Izk when it is reverse biased it will allow it to flow, and then it won't be exactly rectifing.
 

kchriste

New Member
Forum Supporter
The long answer is "yes". As long as the zener voltage of the diodes is more than the peak voltage of the incoming AC. The short question is why would you do this when ordinary diodes are cheaper and more suitable?
 

BrownOut

Banned
The zener voltage would need to be 2X the peak AC voltage. Or I guess you could just do your regulation ahead of filtering. Doesn't strike me as a very good idea though.
 

colin55

Well-Known Member
Using a zener diode is a very good and very clever way to get a defined output voltage.
I have used it in the follwing circuit:

**broken link removed**

But you will notice the output voltage is generatd from a capacitor-input -arrangement.
You will need to include a current-limiting resistor before the bridge to prevent damage to the zener.

This type of circuit has a low output capability due to the rectifier being classified as a SHUNT REGULATOR.

The zener voltage would need to be 2X the peak AC voltage
This is entirely untrue.
The zener voltage is actually the PIV rating of the diode!!!!!
You can use any diode and the PIV rating of the diode becomes the zener voltage of the bridge above.


If you use the Zener diode, it will allow the flow of current where it hasn't to flow, i mean if the current is exceeds the Izk when it is reverse biased it will allow it to flow, and then it won't be exactly rectifing.
This is mudled thinking and is basically untrue.

This is obviously a new concept for most electronics personnel and so many get this aspect of a zener diode wrong.
 
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colin55

Well-Known Member
zener forward voltage + zener reverse voltage > peak AC voltage

What is the zener forward voltage?
What is the zener reverse voltage?

A zener diode is an ordinary diode such as 1N4001. A 1N4001 is a zener diode with a 100v rating.
This rating is called its peak inverse voltage or the voltage it can withstand in the reverse direction.
The forward voltage for all diodes is about 0.7v
The reverse voltage for a 1N4001 is 100v.
When a 1N4001 is used in a bridge, the output voltage will be 100v plus 0.7v FROM THE OTHER DIODE IN THE BRIDGE. The diagonally opposite diode can be a zener or ordinary diode. In either case it will produce 0.7v across it during this part of the cycle.

The zener voltage would need to be 2X the peak AC voltage.
The 2V you are talking about is the requirement of an ordinary diode in a power supply to prevent it "zenering," as some power supplies produce an AC voltage that is twice the "expected" voltage due to
the circuit having capacitors that charge during part of the cycle. This does not have to be taken into account when a zener is used in the power supply above as the energy from the excess voltage is absorbed by the zener.
 
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BrownOut

Banned
Quote:
The zener voltage would need to be 2X the peak AC voltage

This is entirely untrue.
The zener voltage is actually the PIV rating of the diode!!!!!
You can use any diode and the PIV rating of the diode becomes the zener voltage of the bridge above.

We were expecting the OP wanted a conventional rectifier.
 

Karkas

Member
I understand what Colin55 said, but now i have a question for you, what about the current that goes through the zener? how much would it be? i think it doesn't matter if it makes it to the load anyway and finally in the load you have the same current as you had a conventional bridge, but would it?
 

ericgibbs

Well-Known Member
Most Helpful Member
I understand what Colin55 said, but now i have a question for you, what about the current that goes through the zener? how much would it be? i think it doesn't matter if it makes it to the load anyway and finally in the load you have the same current as you had a conventional bridge, but would it?

hi,
Look at these 2 sims, one with 560R and the other with 1 meg as load.

Note the Izener currents.
 

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ericgibbs

Well-Known Member
Most Helpful Member
hi karkas,
#
To demonstrate the point 'brownout' was making about a transformer driven rectifier bridge, look at this image.:eek:

Its not suitable for a transformer source.:)
 

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Karkas

Member
Thanks for the answer Ericgibbs, but i see that the current is divided in the same amount for the two zener diodes, for the forward biased and the reverse biased in every half-cycle, but what i don't understand is the difference in the waveforms,
in the first image:
those waveforms shouldn't be for D3 and D4? i mean that's the usual waveform that you get when you make the measure for those diodes.

in the second image:
I can see that the current takes positive and negative values because the zener allows it to flow in both directions, but shouldn't the voltage be constant at 18 V, when the corrent is positive through it?

And about the transformer driven rectifier, i'm sorry i don't understand.
Could you explain?
 

colin55

Well-Known Member
When you include zener diodes in the bridge, the bridge operates completely differently to a normal bridge.
You must include a resistor between the zener diode and the transformer as the zener and resistor form a circuit known as a SHUNT REGULATOR.
Look up SHUNT REGULATOR to see how this type of circuit works.
 

Karkas

Member
Excuse me, as I see it, the output will be clipped at 18 V once it takes that value, well i think exactly 17.3 V because once the zener diode is at 18 V it will maintain that voltage in that mesh, am I right?
 

Karkas

Member
Why 18.6 V? i simulated it first without the capacitor and i actually obtained the clipped rectified wave, but it was clipped at 13.4V, i used a load of 576 Ohm and the limiter resistor was 10k, then i put the capacitor and it was oscillating and it remains contant, well oscillating between 13V and 13.4V then i changed the load to 10K and it remains constant at exactly 13.5V, and if i change the limiter to 1K, i have again the little oscillation, but it is not at 18.6V it is at 13.4 V,
why? i've already checked, by the way the zener is 1N5248B
 

dougy83

Well-Known Member
Why 18.6 V?
It's wrong. For an 18V zener in the configuration posted, the voltage across the bridge will be 18V + 0.6V = 18.6V. The voltage across the load capacitor will be less than this by 2 diode drops; so what you said (17.3V) is correct (assuming 0.7V diode forward voltage).

Why 18.6 V? i simulated it first without the capacitor and i actually obtained the clipped rectified wave, but it was clipped at 13.4V, i used a load of 576 Ohm and the limiter resistor was 10k, then i put the capacitor and it was oscillating and it remains contant, well oscillating between 13V and 13.4V then i changed the load to 10K and it remains constant at exactly 13.5V, and if i change the limiter to 1K, i have again the little oscillation, but it is not at 18.6V it is at 13.4 V,
why? i've already checked, by the way the zener is 1N5248B
Increasing the load (i.e. decreasing the load resistor) will cause the output to droop more. It will be charged throughout the input cycle which will give an oscillating voltage on the load.
 

Karkas

Member
Thanks dougy, but i still have those 13.5V, and can't see where are the other 3 or 3.3V
 
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