# Diode bridge rectifier, single phase, full wave, rippled output

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#### KhaledOsmani

##### Full Member level 6
Hello

Im using a single phase full wave bridge rectifier made of four diode connected as shown in Fig(a).
The input is an AC voltage source of 120Volts.
When Im branching the voltmeter at the terminals of the 100uF electrolytic capacitor, and switch the voltmeter to AC mode, it keeps outputting AC voltage!

the AC voltage should be rectified and no AC components must be there, how come I still read an AC voltage on its output? Could it be that there are harmonics on the output, and the voltmeter reads them, contrarily to a DC motor that takes the average value of the rectified output?

Should I add any type of filter on the output terminals, so that I can get a pure DC signal?
Im intending to input the rectified voltage, after making voltage division, to a digital system. aren't these AC voltages would harm the digital system?

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the diodes used are 1N4007 and the capacitor in parallel is of value of 100uF

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#### rahdirs

the AC voltage should be rectified and no AC components must be there, how come I still read an AC voltage on its output? Could it be that there are harmonics on the output, and the voltmeter reads them, contrarily to a DC motor that takes the average value of the rectified output?
How much voltage are you reading ? Large value or a small ripple ?

#### FvM

##### Super Moderator
Staff member
That's a meaningless question without telling the load resistor value.

#### ark5230

If you take 1.5 V dry cell and attempt measuring voltage using multimeter on AC range, it will show more than 1.5 V AC.
If it is so then:
What you are measuring is not real pure AC.
To be able to measure the AC component in the DC output you will have to include a capacitor in series with the multimeter with multimeter in the AC range.
Even then the AC component at the filtered output will depend on the load connected as pointed out by FvM.
The value of capacitor will also matter if the reactance / impedance becomes comparable to the internal resistance of the multimeter used.

points: 2

### d123

points: 2

#### KhaledOsmani

##### Full Member level 6
Hi

This is the first time I see that a dry cell could show an AC value on the multimeter. Thanks.
Assuming that I branch the multimeter directly to the output, regardless of the load, (the load later on would be a PIC18F microcontroller), after adding a series capacitor through a branch of the multimeter, should I keep seeing an AC value?

This all means that the bridge is working fine, but there exists some sort of harmonics?

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80V AC do you consider it large?

#### Audioguru

You have ripple because the value of your 100uF filter capacitor is too low for the amount of current you are drawing in the load.
Increase the value of the filter capacitor to 1000uF, then 4700uF then 10000uF to see the ripple almost vanish.

I think your multimeter is garbage. Mine does not measure AC from a battery.

KhaledOsmani

### KhaledOsmani

points: 2

#### betwixt

##### Super Moderator
Staff member
This all means that the bridge is working fine, but there exists some sort of harmonics?
Nothing to do with harmonics at all. The capacitor will charge with current fom the diodes and discharge into the load. As the charge current can only flow when the voltage from the bridge is higher than across the capacitor, it will only 'top up' on the peaks of the rectified waveform, leaving an unstable signal across the capacitor. If you really are measuring AC, what you are seeing is the rise of voltage during the charge period and it's decline as it discharges into the load. This is the normal 'ripple' voltage refered to earlier. You can reduce (but not eliminate) it by increasing the capacitor value as Audioguru stated although you should consider that it will also increase the inrush current as you first apply the AC.

If your meter measures a voltage across a battery on it's AC range, it just means it has nothing inside it to block DC. In other words it will give an inaccurate reading if measuring AC and there is also some DC present as well. You can eliminate the DC part of it's measurement by adding a blocking capacitor external to the meter, in series with one of the probes. Use a sensible value, maybe 1uF or more and non-polarized. This will still let you measure the AC part of the signal (the ripple) without also measuring the DC part.

Brian.

points: 2

### d123

points: 2

#### KhaledOsmani

##### Full Member level 6
Hello

I tried what Audioguru said, and branched 1mF at the output of the bridge rectifier ( by parallelly connecting five 2200uF together), the result was the same and i was seeing ac voltage from the output terminals of the bridge.

As for betwixt, i tried to connect non polarized capacitor in series with one of the two probes of the multimeter, from 10nF to 1uF the result was that i wasnt able to neither see DC nor AC voltage for a silver battery cell that has 7,5V.

Please note that the source voltage im taking is the landline phone branchements, that has 50v in steady DC 12 V AC when ringing

#### hobbyckts

If you take 1.5 V dry cell and attempt measuring voltage using multimeter on AC range, it will show more than 1.5 V AC.
If it is so then:
What you are measuring is not real pure AC.
To be able to measure the AC component in the DC output you will have to include a capacitor in series with the multimeter with multimeter in the AC range.
Even then the AC component at the filtered output will depend on the load connected as pointed out by FvM.
The value of capacitor will also matter if the reactance / impedance becomes comparable to the internal resistance of the multimeter used.

I don't think empty battery will read AC and never got that. It's impossible

#### SunnySkyguy

Choose C based on Ipeak currents and ripple current % and thus %ripple voltage. 10% ripple voltage means Ipk=Iavg/10% similar for 1% is 100x Iavg for peak current during 1% of time. Thus often equates to huge inrush, at low ripple V and if Iavg step size is full the Vdc swings are huge. So PWM at high rates are preferred with buck coil. Or Triac pre-regulator or LC filter or RC filter.

Primitive bridge and Cap for high power does not work with small cap.

Consider C= 40kuF per Amp Avg with ICL inrush current limiter(NTC) and PTC protection, both cheap metal oxide parts for critical current levels.

40mF/A=dt/dV = 25 ms/V ripple then if using 100Hz or 10ms pulses , you can estimate ripple vs desire such as this guideline of 10ms/25ms/V = 0.4 V ripple.

Imax is also limited by ESR of 2 diodes and ESR of cap, so must be very low ESR to avoid thermal damage or use high Rs in series to reduce Ipk and expect worse load regulation , ESR.total/Rload min-max

there are lots if free RCL diode calculators on web for bridge design basic values.

KhaledOsmani

### KhaledOsmani

points: 2

#### betwixt

##### Super Moderator
Staff member
As for betwixt, i tried to connect non polarized capacitor in series with one of the two probes of the multimeter, from 10nF to 1uF the result was that i wasnt able to neither see DC nor AC voltage for a silver battery cell that has 7,5V.
Not what I meant! The battery will NEVER produce AC, what we are tying to establish is whether your meter measures a DC voltage when set to an AC range.

Do this:
1. place your probes directly across the battery and measure the voltage on the AC range, you should see 0V.

If you still measure a voltage it means your meter has no DC isolation capacitor when using the AC ranges so -

2. Connect the 1uF non-polarized capacitor in series with one of the meter probes and measure the voltage at your bridge rectifier to see what the real AC part of the voltage is.

Brian.

KhaledOsmani

### KhaledOsmani

points: 2

#### ark5230

Does it mean that it is ripple free and you are done?
In no load condition ripple would be negligible.

#### SunnySkyguy

Some old meters read Vac as 2x Vdc + Vac-p because they use rectifier without ac coupling cap.

If so, use series cap to meter on Vac.

Dry cell Vac is simply Irms* ESR of battery.
Same with ESR of Caps. If ripple exists and C is huge, ripple is determined by ratio of ESR of Cap/load R * I load.

Batteries have huge Capacitance (Farads) but finite ESR.

KhaledOsmani

### KhaledOsmani

points: 2

#### KhaledOsmani

##### Full Member level 6
Hi

@betwixt My multimeter is crap (sinometer) and does not have the insulation you spoke about. I did place in series a non polarized capacitor with one of the two probes and tested the DC output of the bridge: it keeps outputting an AC voltage with almost a double magnitude than the initial DC.

What I need is a stable DC output voltage, since I will be using a 100k - 10k voltage divider and input it to a PIC18F.

Please not that the initial input voltage source is not a regular single phase home based supply, it is the two terminals of a landline phone. I don't have a deep idea about its mixed signals, but in theory it must have a 50V DC when steady, 12V AC when placing a ringing call.

I've been reading articles about filtering the DC output of a bridge rectifier, and seen something called LC pi filter wich consists of two capacitors with an inductor in between.

If I branch the output of the voltage divider with 10mF electrolytic capacitor, then add a 1mH inductor and finally branch another 10mF electrolytic capacitor and get the final output DC voltage from it, wouldn't I get a stable DC voltage, with no AC components??

From another part, I have delt before with LM317, so if I branch the output DC voltage of the bridge, connect to 10mF ele. Cap. Then input it to the LM317, also wouldn't be a way to have a pure neat DC voltage output on the LM317 output terminal??

Since this whole system will later be injected to a digital system (PIC) i shouldn't have any noise, ripples, AC harmonics on its input. I will try tomorrow with a more efficient multimeter, that shows 0V on a dry cell when switched to AC and see what I get after the bridge and 10mF (thanks to Audioguru) capacitor parallelled to it. If still AC exist, i will continue with the 10mF capacitor and continue it to form an LC pi filter (thanks to SunnySkyGuy).

Thanks for all of your interest, and looking forward to achieve my goal.

#### SunnySkyguy

Remember to use 10 to 40mF per amp on 100 Hz for low ripple.

But for high power LDO's become very inefficient unless very low drop like <500mV at rated current. Or superlow , <100mV drop.

Cost , efficiency and regulation tolerance are all tradeoffs which make SMPS more attractive, as LC filters are much more expensive and bigger at 100 Hz than 100 kHz, and have poor PF and conducted line noise is a pulse wave before additional filtering.

KhaledOsmani

### KhaledOsmani

points: 2

#### KhaledOsmani

##### Full Member level 6
Hello

I've branched the output of the bridge with a 10mF electrolytic capacitor, and tested the value of the voltage after making a voltage divider of 100k - 10 k in series, so I had (using the multimeter: Greenlee dm-110), 2,38V DC, which is correct, since it must be 4.5V DC but there were voltage drops due to the diodes.

When switching to AC mode, in the multimeter I see this:
It starts with a high value (not constant, sometimes 200, sometimes 60 mV AC) then this whatever value starts decreasing to stabilize between 0.3mV and 0.1mV. If I wait much more time looking at the multimeter, it tends to 0V AC after almost 15 seconds of connecting the system to the input terminals. It fluctuates between 0 and 0.1mV AC.

What we see here is that the ripple almost vanish, but still exist, and I think it is not yet safe to enter it to a PIC micro-controller, so I will use instead of the 10mF capacitor and LC filter, such that I will double the value of the capacitance of the two shunt capacitors, to become 20mF, and choose L according to the following:

Since the frequency of the signal is <100Hz:

Wrf = 2*pi*100
wc = wrf * (1/1-0.1333)
L = 35.6/wc [mH]

Wouldn't this be sufficient to delete all ripples? and the system would be safe enough to enter it to the PIC?

Attached there is the Design.

Thanks

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in the design2 it shows the resistors value to make the DC rectified input to the PIC no more than 4.5V

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You have ripple because the value of your 100uF filter capacitor is too low for the amount of current you are drawing in the load.
Increase the value of the filter capacitor to 1000uF, then 4700uF then 10000uF to see the ripple almost vanish.

I think your multimeter is garbage. Mine does not measure AC from a battery.

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#### SunnySkyguy

Where is your 3 terminal regulator?

#### betwixt

##### Super Moderator
Staff member
I would urge extreme caution! The landline is not ground on one side and voltage on the other, both wires are 'floating' and you can only draw current from between them. If you draw current, the line will be seen as 'off hook' which will prevent incoming calls and prevent a telephone sharing the line from making outgoing calls. The 50V is not a stable voltage, that is it's unloaded voltage, it is derived from a constant current source which means it will vary according to the load you place across it. Your notion of 12V AC for ringing may be true in your country but in most places it is 75V and ADDED to the 50V on the line, taking it's peaks up to around 125V!

In European countries the maximum current you are allowed to draw in the 'on hook' condition is 100uA so if your line is similar, you will not be able to power your circuit without risk of the telco detecting a fault. You are not allowed to draw ANY current from either wire to ground so you will have to isolate your circuit and not have any other connections to it. Drawing power from phone lines for anything other than approved telephones is a bad idea!

Incidentaly, your bridge rectifier is wired wrongly and if you are tying to measure an analog voltage at the tap between the resistors you should take note of the limitations of ADC source impedance.

Brian.

#### Audioguru

We do not know the country that rings a telephone line with only 12V and we do not know the ringing frequency and cadence.
In Canada and the US a telephone line rings with 90VAC at 20Hz, superimposed on the 50VDC. When a small current is drawn by a phone going off-hook then the call is answered and the ringing is stopped.

I think the surge of your filter capacitor charging will answer a ringing telephone line. Actually the capacitor will charge to the 50VDC of the on-hook telephone line, causing the line to go off-hook until the charging current stops. Then if a telephone connected to the same telephone line goes off-hook, the telephone will probably be destroyed by the unlimited current of the capacitor discharging into the telephone.

##### Super Moderator
Staff member
L, C1, C2 together resonate at 48 Hz. This could lead to oscillations which can get out of hand (more so if incoming mains is 50 Hz). Destruction of components is possible.

There are two-stage capacitor filters, which have a resistor where you have L. You will not get resonance with such a filter. It would require experimentation to find a satisfactory combination of values.

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