# why there is rectifier after the HF transformer in the AC-DC isolated SMPS?

Discussion in 'General Electronics Chat' started by xchcui, May 17, 2014.

1. ### xchcui Thread Starter Member

May 12, 2014
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Hi.

My question is refer to the isolated SMPS which convert 120vac-220vac to ___ (whatever)DC voltage.

I read that after the DC inverted by the oscillator to high frequency AC voltage,it pass through high frequency transformer and in the output of the transformer,it pass through a rectifier.
Why there is a rectifier in that stage?
I know that there is variation in voltage,but isn't the HF voltage flowing in the same direction?(pulses above the zero)?
So if the voltage is not changing its direction,for what do we need recifier after the HF transformer(after the converted AC HF voltage)?

2. ### Alec_t AAC Fanatic!

Sep 17, 2013
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The output of the transformer is high frequency pulses going both above and below the desired DC level; therefore it needs a rectifier to convert it to the required DC.

3. ### crutschow Expert

Mar 14, 2008
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A transformer cannot carry DC so any output from a transformer is AC with no DC component, even if the input is a pulsed DC. In effect the DC point is shifted to half the average voltage of the pulse at the output (the same as if were coupled through a series capacitor).

4. ### studiot AAC Fanatic!

Nov 9, 2007
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There is a very simple reason that has so far been missed.

It does not matter if the feed is alternating or unidirectional, sine waves, triangular waves or pulses, the reason is the same.

The voltage output from the transformer rises and falls.
This output is fed into a reservoir capacitor which charges up to the desired DC voltage output.

Or does it?

In order to charge the reservoir capacitor the input voltage from the transformer must be greater than the existing voltage on the capacitor.

If it is not, the capacitor voltage will drive the current in the opposite (unwanted) direction.

So we place a diode in the feed to prevent this happening.

5. ### ErnieM AAC Fanatic!

Apr 24, 2011
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+1

(filler letters here)

6. ### vk6zgo Active Member

Jul 21, 2012
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Indeed!,but take away the capacitor,& put a 10X 'scope probe in its place.

Without the rectifier,you will see a waveform evenly distributed around zero volts.
With the rectifier,you will see a waveform which is no longer evenly distributed around zero volts.
Instead,the negative or positive peak of the waveform will be at zero volts.

If the capacitor is now replaced:-

Without the rectifier,the capacitor will attempt to charge one way on the +ve half cycle,& the other way on the -ve half cycle.
Not at all good for an Electrolytic Capacitor!

With the rectifier,the pulses are unidirectional,so the capacitor can only attempt to charge in one direction.

7. ### xchcui Thread Starter Member

May 12, 2014
90
0

I attached two photo in order to understand how does the voltage wave look like,when it get into the transformer and out of it.
Does the voltage wave on the input and the output of the HF transformar looks like photo 1 or photo 2?
If it looks like photo 1(pulsed a bove the zero and below the zero)i can understand that the rectifier rectified the negative pulses,but if it looks like photo 2(all the voltage pulses are a bove the zero/unidirectional)i don't see what there is to rectified(it is already unidirectional).
Maybe,the answer lies between the lines in yours answers,but i think if i will understand how does the voltage wave looks like,on the input and the output of the HF transformer,it will help to understand the issue.

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8. ### studiot AAC Fanatic!

Nov 9, 2007
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You missed the essential point of my post and what you say is not strictly true.

ErnieM caught it OK.

Apart from the fact that a capacitor can only charge in one direction (it discharges in the other)

Whether the waveform reverses or just goes only down to(wards) zero there will be a time when the instantaneous waveform voltage is less than that of the capacitor.
For this part of the waveform the diode is essential to prevent the capacitor discharging the other way.

9. ### studiot AAC Fanatic!

Nov 9, 2007
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It may look like either! or something else again!

You should understand that a transformer is an isolating device.

So it depends entirely what your reference is as to where the zero line falls.

You can connect either end or somewhere in the middle of the winding to 'zero'.

10. ### vk6zgo Active Member

Jul 21, 2012
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In the real world,a diode is nice with unidirectional signals but not necessary.

Let us posit a squarewave generator with internal resistance equivalent to that of the transformer secondary.
This can be drawn as an ideal generator in series with its internal resistance R.

If a capacitor C is now connected across the output of this combination,it appears as an ideal squarewave generator driving a basic Integrating Network of time constant τ=CR.

For convenience,our generator produces a squarewave of period 2τ,of Vmin zero volts & Vmax,10volts..

C is initially uncharged,so when the first Vmax half cycle is applied to it,will try to charge to that voltage,through R.

In one time constant τ=CR,C can only charge to approx. 63% of Vmax,or in this case,6.3v.

During the second half cycle,where the gen o/p is Vmin=0volts,C attempts to discharge,but can only discharge to 37 % of the 6.3v,or 2.33v.

During the next half cycle,C charges to 2.33v plus 63% of the difference between 2.33v & 10 volts.for a total of 7.67volts.

Back to a zero volts half cycle,so C discharges to 37% of 7.67 volts,or 2.89 volts approx.

After each cycle,the charge on C increases,eventually coming close to,but never quite reaching Vmax.

C requires approx. to discharge fully,so it cannot do this during the time when the generator voltage is less than Vmax.

If a capacitor is chosen such that is shorter than one half cycle of the input signal,what you suggested will happen,otherwise the above will be the case.

"Apart from the fact that a capacitor can only charge in one direction (it discharges in the other)"

Can you run that by me again?

Take an uncharged capacitor,place it across an ohmmeter.

Initially,the cap will look like a low resistance,but as it charges up,the resistance reading increases until it pretty much "plateaus".

Now reverse the meter.

Initially it will look like a high resistance,then as the cap discharges,it drops down to a lower value,then increases again as the cap charges in the opposite polarity.

Last edited: May 19, 2014
11. ### studiot AAC Fanatic!

Nov 9, 2007
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Here is the circuit you describe.

I have indicated the conventional current flow during time t1, the first half cycle when the waveform is at +10 volts.

During this time the capacitor is defined as charging, since current is flowing in.

What is the direction of current flow in period t2?

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12. ### vk6zgo Active Member

Jul 21, 2012
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In period t2,the capacitor is discharging through R & the ideal generator.
(As you state,in the opposite direction to the charging current.)

The capacitor doesn't know where its charge came from,----as far as it is concerned,it always had that charge,& is starting with a "clean slate".

The only way it can completely discharge is in 5 timesτ,where τ=CR,but we only give it the time interval t2, (which,for convenience, I have stated is the same as τ).

In that time interval,it can only discharge to 37% of the charge present at the completion of t2.

This is the simplest case,but all other combinations of C & R,will cause an increase in charge on C over time,except for those cases where the duration of one half cycle is equal to,or greater than 5τ.

Simple integrating networks are used in practice to convert unidirectional pulse trains into DC levels.

My comment re charging a capacitor in either direction was not for the unidirectional case,but in reference to my earlier posting:-

"Without the rectifier,the capacitor will attempt to charge one way on the +ve half cycle,& the other way on the -ve half cycle.
Not at all good for an Electrolytic Capacitor!

With the rectifier,the pulses are unidirectional,so the capacitor can only attempt to charge in one direction"

And what I assumed to be your response to it;-

"Apart from the fact that a capacitor can only charge in one direction (it discharges in the other)"

A capacitor will charge on ,say, the +ve half cycle, of an ac signal as it has no way of foreseeing that a-ve half cycle is coming,only to have to discharge as the incoming voltage changes through zero,& increases in the other direction,discharging it & charging it in the other direction.

13. ### studiot AAC Fanatic!

Nov 9, 2007
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Remember first, the purpose of this thread.

We are discussing a power supply, not a lightly loaded signal conditioning circuit.

So, in normal running, there will be a significant load in parallel with the generator output resistance during any discharge part of the cycle.

Nor do can we carefully select the shape of the input waveform. And we certainly can't apply your statement above for any general alternating or pulsed unidirectional waveform. I noted this in post#4.

Finally , whilst I have never disagreed that your configuration or similar will have a 'pumping' effect, note again the question "why is there a rectifier?", not "why is there not a rectifier?".

Even for your conditions above a rectifier will improve efficiency.

I was, and still am, trying to provide a compact explanation to a beginner, that does not contain any actual inaccuracies.

14. ### vk6zgo Active Member

Jul 21, 2012
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In your "compact" explanation,you made a blanket statement,& I disagreed with it---You seem to be now qualifying that statement.

"And we certainly can't apply your statement above for any general alternating or pulsed unidirectional waveform. I noted this in post#4."
My statement can be applied to rectangular pulsed unidirectional waveforms directly.
I chose a squarewave of period equal to CR to make the description easier.
Different pulse widths or repetition frequencies will modify the maths,but not the fact of the increasing charge on C.

Of course,a different unidirectional waveshape again modifies the result,but the statement is still true.

Obviously,a bi-directional alternating waveshape will not cause a net DC charge over time.----I never said it would!

Last edited: May 20, 2014
15. ### studiot AAC Fanatic!

Nov 9, 2007
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You have failed to address (one of) the most important players in a power supply, namely the load. Adding this will completely destroy your carefully arranged maths.

Any analysis that depends upon a subsequent connection that does not load the circuit presented (as yours does), will fail in a power supply.

Further you have avoided the issue of why there is a diode and how this diode prevents the partial discharge of the reservoir capacitor and also allows actual control of the voltage on the capacitor so that it can be maintained at the desired level, independent of the load.

So no, I am not qualifying my statement.

Last edited: May 20, 2014
16. ### vk6zgo Active Member

Jul 21, 2012
677
85
Prior to the advent of capacitor filtered power supplies,rectifiers were used to operate DC machines from AC.
They produced horrible lumpy "pulsating" DC,but it was perfectly OK for such uses.

The basic use of rectifiers was,& still is,to convert AC to DC.

The fact that in capacitor filtered power supplies they prevent capacitor discharge is an important secondary function,but in normal power supplies,without the rectifier,there would be no charge on the capacitor in the first place.

Yes,in my described circuit,a diode would allow the capacitor to charge quicker,& thus increase the efficiency,but ,as I showed,a diode is not essential in the unidirectional case.

17. ### xchcui Thread Starter Member

May 12, 2014
90
0
I understood that if the wave form that came out from the HF transformer is pulsed unidirectional,then only one diode is enough in the rectifier in order to prevent the current to flow at the other direction,at the moments that the voltage of the secondary winding is lower than the capacitor voltage.
But i don't understand why wouldn't the wave form of the output HF transformer be all the time unidirectional.
Lets say(in general) that the 120vac or 220vac get into the smps from the main wall socket,then there is a rectifier with 4 diodes that rectified it to 120DCV or 220DCV that flow to a large filter capacitor,then there is switch mode system that invert the DC to high frequency pulses that get into the transformers.
The wave form that get into the switch by the chip component commands is indirectional,it is flow in one direction.If you open the switch the voltage is zero,if you close the switch,the voltage rises instantly to a specific voltage,but at the same direction,doesn't it?(i refer to the creation of the HF for the transformer).
So,i don't understand why you are talking a bout wave form that the flow is bidirectional?

18. ### studiot AAC Fanatic!

Nov 9, 2007
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Let's wrap this up.

Firstly I explained in post#9 about the position of zero and the waveform, and others here have also done so.
It depends entirely how you connect the transformer.

Here is a pictorial view.

The transformer is shown with 3 possible grounding connections, along with the resultant waveforms.
1) If you ground one end of the transformer secondary you will get negative pulses.
2) If you ground the middle you will get a wave with positive and negative part cycles.
3) If you ground the other end you will get positive pulses.

Now you need to understand that this is a special transformer, called a pulse transformer.

It is special because it is designed to pass the pulses shown. Normal transformers do not work well with this type of waveform.

For convenience we will discuss connection 3 - Positive pulses.

The waveform is called a 'pulse train', for very good reasons.

The SMPS contol works like this:

The pulse train is fed through the single diode (yes you are correct only one is needed) to a capacitor.

You say you now understand that the diode is there to prevent the capacitor partly discharging throught the transformer secondary when there is no pulse present, so no more need be said of this.

The control circuitry monitors the current out to the load and/or the voltage on the capacitor and adjusts the pulses to counteract any change. That is it tries to keep the voltage on the capacitor constant.

The pulses are all of the same height (voltage) and repetition rate, since changing these makes for inconvenient circuitry.

But the width or duration of the pulse can is controlled by the controller.

So when the load is light the pulse duration is short and when the load is heavy the pulse duration is much longer.

This controls the quantity of charge each pulse puts onto the capacitor to keep the voltage steady.

This is a bit like filling a tank up with water by adding buckets, whilst at the same time drawing off a flow for some purpose.

When the draw off flow is low the bucket is only partly filled. When the draw off flow is high the bucket more nearly full.

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19. ### studiot AAC Fanatic!

Nov 9, 2007
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I think you have taken our discussion far enough off-topic.

Do you not intend to help the OP?

20. ### vk6zgo Active Member

Jul 21, 2012
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No,I'll leave that up to you-----_Bye!