What are phase lock circuits used for?

[danny davis]

What are phase lock circuits used for?

I know the circuit locks phase IN phase together , but why would you need or want that?

Without a phase lock circuit, it's called free running

Are Phase Lock Circuits used just to lock the input and output together?

To test if a phase lock circuit is working, you put oscilloscope channel#1 on the input and oscilloscope channel#2 on the output on a unit

The problem here for me is that I can only use the Probes Ground on Channel#1 only

Oscilloscope channel#2 probe doesn't have a ground, so the sine waveform is noisy and moves back and forth very fast some techs call it oscillation or noise

I set the time base on the oscilloscope so that I can see the phase shift so they are so very close together, that they are in phase

The problem is that the channel#2 waveform is so noise and moving I can't take a measurement

I can't use the probe ground on channel#2 because my manager says that it will fry the oscilloscope because there is a DC offset or a potential difference voltage from earth ground, and it will cause a SHORT CIRCUIT

If a circuit has DC offset on the ground or a potential difference from earth ground, the oscilloscopes ground will SHORT CIRCUIT the DC offset or potential difference

My manager told me that I can use a DVM meter and measure the input AC HOT line Black wire with the outputs AC HOT line Black wire, It measured 7 volts AC going from input HOT to output HOT. He said this is the phase angles voltage of the inputs sinewave and the outputs sinewave. Next I measured using the DVM meter to the Neutrals White wire from the input to the Neutrals White wire at the output of the unit, it measured 1.4 volts AC. He said this is a phase angle voltage, the phase angle is the inputs waveform and the outputs waveform, the phase shift between the input and output

I have seen techs use an Ez jumper wire to extend the oscilloscopes probe at the probes tip rather than using it on the ground clip. When you use an Ez Hook jumper wire on the ground clip, it picks up to much noise. But when you use an Ez hook jumper wire on the probes tip and use the Ez hook to measure test points, it doesn't have noise on the signal.

The Probes ground clip is so short it's so hard to make measurements, So you use an Ez hook on the probes lead instead of extending the probes ground clip

I'm just confused as why they just don't make a oscilloscope probe that has a longer probe tip

 

[Darktrax]

Hmm - OK then, taking from the top..

What are phase lock circuits used for?
I know the circuit locks phase IN phase together , but why would you need or want that?
Without a phase lock circuit, it's called free running

From your question - Guess we are starting with oscilliscopes, where the waveform being measured deflects the trace vertically, and the timebase oscillator deflects horizontally, and the only way you get it to display stationary need two
things.
1. The timebase frequency must be related to the display frequency by some whole number. To display (say) 4 cycles, the timebase has to run at one fourth of the display frequency.
2. The timebase needed to run IN PHASE , so that the start of a Wave displayed begins at the left

Back in the middle of the last century, when oscilliscopes looked like these --> HERE
it was found that provided the timebase oscillator was already free running at nearly the right frequency, then coupling a small amount of display waveform into it would cause it to "lock", to display a stationary wave.
Later, better circuits were used to force phase lock, using triggering, where the oscillator was forced to wait between sweeps until the next correct wave "triggered" it.

BUT - phase lock circuits are used for much more than that. For decades, every TV scan used phase lock. The way we make most oscillators run at a correct frequency is to phase lock it to a stable (crystal oscillator) reference. In EVERY mobile phone, every TV, every radio transmitter, every computer.

To test if a phase lock circuit is working, you put oscilloscope channel#1 on the input and oscilloscope channel#2 on the output on a unit

Maybe not - it depends on a lot of things. It requires a very full understanding of the circuit, and the expected frequencies. I would not expect a local oscillator running at Ghz to be directly probed by any scope, though I might have a go at a divided down frequency. There are parts of phase lock loops that cannot be probed without the probe itself seriously affecting or even stopping the oscillation.

The problem here for me is that I can only use the Probes Ground on Channel#1 only
Oscilloscope channel#2 probe doesn't have a ground, so the sine waveform is noisy and moves back and forth very fast some techs call it oscillation or noise
I set the time base on the oscilloscope so that I can see the phase shift so they are so very close together, that they are in phase
The problem is that the channel#2 waveform is so noise and moving I can't take a measurement

Honestly? - there is no good reason for that! I also lose the ground lead on my probe, but it does not stop me contriving some bit of wire, or braid, or copper foil. Search keywords "oscilloscope technique tutorial". There is lots of information on how to do this. One of my best probes is a 20:1 hookup made from a 50-ohm coaxial cable, with just the right value series resistors at the probe end to form a potential divider with the cable impedance itself. It easily displays picoseconds risetimes without overshooting and ringing, and its made up of junk!

If a circuit has DC offset on the ground or a potential difference from earth ground, the oscilloscopes ground will SHORT CIRCUIT the DC offset or potential difference

OK - if you have a significant voltage between "safety" ground and the place you clip the ground of the probe, then you have a serious safety problem. The only damaging shock I ever received was from trying to plug a grounded BNC cable into a scope that had a mains leak.
BUT.. there is often huge amounts of noise on the frame (safety) ground. The mains ground is not the same as the probe ground!

To measure a waveform in a circuit where the local ground may be at a high DC offset, or even at very high AC voltage.. CAN'T EVER put the probe ground clip on it!!, there is a way available with nearly all multi-channel scopes.

You use BOTH probes, set up to have the same gain, and their ground clips connected together, but that is all. The ground clips do not connect anywhere else. Channel #2 is set to "INVERTED" and the scope is set to "ADD" Channel #1 and channel #2. both probes are put on the calibration square wave together, and the fie gain is adjusted if necessary to null out the waveform. to nothing.

Then, the probe of Channel #2 is place on the "ground" that has the DC offset, and the probe of Channel#1 is placed on the point you want to measure. The common-mode noise just disappears from cancellation. In this sense, the DC offset voltage is seen as part of the unwanted common mode "noise"

DO NOT just stick the ground lead of a probe in just anywhere where you want to measure relative to. Even if it is a grounded point, it may well have a lot of noise you do not want to end up on your trace. Again, there is lots of information on how to properly use a scope to get around these problems.

My manager told me that I can use a DVM meter and measure the input AC HOT line Black wire with the outputs AC HOT line Black wire, It measured 7 volts AC going from input HOT to output HOT. He said this is the phase angles voltage of the inputs sinewave and the outputs sinewave. Next I measured using the DVM meter to the Neutrals White wire from the input to the Neutrals White wire at the output of the unit, it measured 1.4 volts AC. He said this is a phase angle voltage, the phase angle is the inputs waveform and the outputs waveform, the phase shift between the input and output

I am sorry I cannot help you much here, because I don't fully understand what you are measuring. I guess there is some phase shift between the input and output of the kit you have. It sounds correct to me. We do not have black wires for AC in my country.

I have seen techs use an Ez jumper wire to extend the oscilloscopes probe at the probes tip rather than using it on the ground clip. When you use an Ez Hook jumper wire on the ground clip, it picks up to much noise. But when you use an Ez hook jumper wire on the probes tip and use the Ez hook to measure test points, it doesn't have noise on the signal.

Again, I have some difficulty imagining the scenario.

I can tell you that if the frequency is low, and 50 or 60 Hz is definitely very low, then there is often no problem in extending the probe. The probe tip is at very high impedance. 1MOhm at x10. It will easily pick up noise. The noise pickup is proportional to the area of the loop formed by the lead from the probe tip, and the lead from the probe ground. You can easily entend a probe and its ground at low frequency, by using a twisted pair of the same gauge hookup wire. The only important thing is to have the ground connection close enough to the measure point, or at least run near metal to minimize the noise pickup loop area. If the voltage being measured is of such power that loading it with a few kilo-ohms would not affect the measure accuracy at all, then putting some kilo-ohms across the probe input (at the probe) will shunt most stray noise. Use a value high enough not to get hot!

I'm just confused as why they just don't make a oscilloscope probe that has a longer probe tip

Probes need to deliver measurements where putting the probe onto the point being measured does not modify the measure much by its presence. There is no way to measure without extracting some energy. Being high impedance makes it very prone to error from size capacitance alone. Not a problem at 50 Hz, but definitely hard to do at 50mhz! Special frequency compensated attenuators need to be used, both in the probe and in the scope. The cable from the probe to the scope is often a very fine wire in a tube with ferrite magnetic loading around it. Longer is worse!
The smaller the probe bits, the better to measure without overshoots. My probe has a special springy metal tab to use instread of a ground lead. When it gets this tough to measure, I build in instrumentation routes if I have to, to get the signal out at 50 ohms monitoring point. They are made that way to deal with high frequency signals of feeble power.

My apologies for it being such a long post - the nature of your questions seemed to need it.

 

[danny davis]

It's a UPS AC input 120 , AC output 120

120 AC out of the outlet from the wall goes into the UPS unit and comes out AC 120 on the output of the UPS unit

When you measure from HOT input and HOT output its 7 VAC , when you measure Neutral input and Neutral output it's 1.4 volts AC . these voltages are phase angle voltage my manager said

DVM black probe goes to input HOT, Red probe goes to output HOT = 7 VAC
DVM black probe goes to input Neutral, Red probe goes to output Neutral = 1.4 VAC

These are phase angle voltages , the 7 VAC is a phase angle

The 1.4 VAC on the neutral is a DC offset I think because it should be zero

It's a UPS AC input 120 , AC output 120

120 AC out of the outlet from the wall goes into the UPS unit and comes out AC 120 on the output of the UPS unit

When you measure the inputs HOT on oscilloscope channel#1, outputs HOT is on oscilloscope channel#2, Trigger channel is on channel#1

Channel#2 has noise because I can't use the oscilloscopes probe ground on the units outputs neutral

It will short out the DC offset voltage or potential difference DC voltage my manager said, it will cause a short circuit

The 120 AC outlet from the wall has a earth ground, the Unit has a ground , but these is a potential difference between grounds so the oscilloscope will short it out

 

[Darktrax]

OK - I understand.
UPS use a switched-mode power supply to convert the battery DC to 120V AC. Designs vary as to charge management, and energy hold-up during power fail, but the main point is that both the AC at the output is normally isolated from the input by the higher frequency switcher circuit. The safety frame ground is carried through.

Most that I have encountered can be treated nearly as if they were an isolating transformer. In use, if you must, then it is OK to ground the neutral of the output upstream of a RCD (residual current circuit breaker). Measuring between the (live) of the input and the (hot?) of the output is very likely to show some reading from across the switch-mode circuits, and this would include the switcher high frequency ferrite transformer parts. There could well be power factor correction circuits also. I cannot see why such a measure is vey meaningful, but it does somewhat represent the difference between the UPS output and the AC mains.

Unless you are into using power inverter units capable of returning energy to the power grid, there is no reason that the phase of the UPS output should be synchronised with the AC input. The output of the UPS is never going to be connected in parallel to a live AC mains, or at least, not where I want to be close to. Its OK to have a phase difference here.

The input impedance of a DVM meter is usually about 11MegOhm for DC, and maybe still very high for AC. If you connect one probe to ground, and so much as touch a finger onto the other, the pickup from cables and fields around you will show a reading, much like you get when you touch the end of a scope probe. It is very likely to show some volts across the hot ends of a UPS, and they could be the leftover from the fact the UPS output may be phase-delayed, and also not precisely equal to the AC input.

Connecting a DVM meter across the input and output live leads of a UPS may be easy enough, because the DVM is nicely isulated and isolated. To do this measurement with a scope MUST NOT be done in any manner other than using the two-probe-INVERT-ADD method I described earlier. This measure would show the difference between input and output. To try and display the UPS output at the same time as the AC mains input, to see the phase difference, requires more than two channels. Two probes get used up on the AC input, and another two on the UPS output.

 

[danny davis]

UPS use a switched-mode power supply to convert the battery DC to 120V AC.

Yes, you're right, do you know how this concept works? the UPS uses IGBT at a very high frequency which I can't see on my oscilloscope

the main point is that both the AC at the output is normally isolated from the input by the higher frequency switcher circuit

Yes true, it is isolated so there is a potential difference between AC input and the UPS output so if I use the oscilloscope channel#2 Probe ground , it will cause a SHORT circuit and fry the oscilloscope

Do you know how the high frequency switch works?

the difference between the UPS output and the AC mains.

Yes there is a phase difference, using a DVM meter to measure the AC mains Line/HOT to the UPS outputs Line/Hot wire gives a voltage of 7VAC

This 7VAC is the phase angle voltage from AC mains input to UPS output

Do you know anything about this phase angle measurement done using an DVM meter like this?

7 VAC means a phase angle degree of what?

To do this measurement with a scope MUST NOT be done in any manner other than using the two-probe-INVERT-ADD method I described earlier. This measure would show the difference between input and output. To try and display the UPS output at the same time as the AC mains input, to see the phase difference, requires more than two channels. Two probes get used up on the AC input, and another two on the UPS output

That oscilloscope would have to have 4 isolated channels to do 2 different ADD functions right?

You're saying I need to ADD channels#1 and #2 for the AC input
ADD channels#3 and #4 for the UPS output

- - - Updated - - -

When you put the AC off the UPS , the UPS is running on batteries

The Phase lock circuit is OFF and the UPS output is Free Running

The Batteries have a battery charger circuit is only on during when the AC is ON. The battery charger circuit uses a PWM power supply so when you turn on the batteries you can hear the low frequency of the PWM of the battery charger

I'm not sure why it's a low frequency PWM signal when the batteries are apply to the battery charger circuit, do you know why?






- - - Updated - - -

What is circulating current? and Reactive loads?

With a current meter it measure amps , but since it's circulating current it's different why?

When it's circulation current it's a reactive load?

"A non linear load means the current and voltage are not in phase"
"A linear load means the current and voltage are in phase"
" Reactive Load means? circulating current

A light Bulb is a linear load

Using a Transformer and a light bulb will be a Non-Linear load because of the transformer? the transformer puts the current and voltage not in phase? because its a inductor + resistor , phases are not in phase ?

What power supplies are High Impedance? is it tube, solid state, switching power supplies, Linear power supplies, how do you know or tell?

What power supplies are Low Impedance? is it tube, solid state, switching power supplies, Linear power supplies, how do you know or tell?

A High Impedance power supply creates a voltage drop, but what makes a power supply high impedance? or low impedance?

What is DC component?

At my work , they say DC component is the "TIME GAP" or " Cross over" section between the positive and negative cycles at the zero crossing point when you have a push pull network

Is this Time gap or Cross over called DC component?

When You have a PWM push pull output stage , the DC component is an adjustment to center the Positive and Negative cycles AC waveform to zero volts. If you don't then there will be a time gap which is called DC component

I'm not sure the difference between DC component VS DC offset

 

[danny davis]

The unit is a UPS system
1.) The UPS unit , takes the AC 120 outlet and converts it to DC and then converts it back to 120 VAC
2.) The inputs and outputs are isolated because of the higher frequency switcher circuit

No sure how the high frequency switcher circuit works? do you? I can't even see the high frequency on my oscilloscope because the frequency are so high, it just looks very messy on my oscilloscope

3.) There is a phase shift between the UPS outputs and the AC Mains input
4.) The Phase Lock circuit , locks the UPS output AC 120 with the AC mains input 120 VAC

why does the input and output of an UPS system need to be in phase?

To try and display the UPS output at the same time as the AC mains input, to see the phase difference, requires more than two channels

How do you set up the oscilloscope to measure the UPS output at the same time as the AC mains input?

5.) When you turn off the AC mains breaker on the UPS unit, it's in DC mode which is called " inverter mode"
The inverter circuit converts DC into AC

I'm not sure how the Invert circuit converts DC back to AC 120 Volts , but it uses a High frequency switcher circuit which is a PWM signal which controls IGBT driver chips & IGBT transistors

When you turn off the AC mains breaker on the unit, It's in DC mode which is in the Inverter mode

**The UPS input and UPS output is not "Phased locked" anymore, It's only Phase locked when the AC breaker is turn on apply AC mains 120 VAC from the wall outlet into the UPS unit.

I'm not sure why the UPS unit's output needs to be In phase and Phased locked with the UPS input?

In Inverter Mode, It's just running on DC batteries

So It's called Free Running because on the oscilloscope the output waveform is Free Running and the phase shift gets longer as the output waveform keeps running away from the input waveform on a 2 channel oscilloscope

 

[danny davis]

The UPS unit's input and output is Isolated, that's why there is a potential difference

Since there is a potential difference that the reason why they use a phase lock circuit to keep the phases of the input and output of the UPS IN phase so there is no voltage or potential difference

When there is a phase difference between input and the output , there is a potential difference which can cause a short circuit when the UPS goes into "static bypass switching"

The reason why they have a phase lock circuit is because of a static bypass switching circuit

I'm not sure what is does or how it work as of yet

Another reason why the have a phase lock circuit or to keep the input and output of the UPS in phase, is because if a load is a motor it needs to be in sync with the AC Main's input

If the output of the UPS is free running and not phase locked, the motor ( a load ) will start to shutter and cause problems because it's not in sync with the AC main's

When the output of the UPS is free running, there is a potential difference between the input and output because of the phase difference which will cause a short circuit when it goes into a static bypass switching

 

[danny davis]

3 Reasons why the phase lock circuit is used to lock the UPS input and output together

1.) Synchronization (Phase Lock or Sync.): Generally refers to synchronizing the output of
the Inverter to the Utility mains, in order to allow smooth transfer of load from mains to Inverter
and vice versa.

2.) Frequency stability: Deviation of output frequency from nominal value when the UPS is not
synchronized to mains power (at mains outage). The stability is generally +/- 0.5%, with RC
oscillator and +/- 0.05% with crystal oscillator

3.) Parallel Operation: Operation of two or more systems with outputs connected to a common
Load Buss for Redundancy or power enhancement purpose. To enable parallel operation the UPS
systems should have equal output voltages, operate synchronously (same frequency and same
phase), and have load-sharing capabilities.


Linear Load: load comprised of linear (non switching) components, such as, resistors,
capacitors, inductors, motors, lamps, transformers etc.

Non-Linear Load: Load comprised of switching components, such as diodes, rectifiers,
Thyristors, Switching or Pulse modulating systems or circuits. Non-linear loads generate
current and voltage harmonics with integral multiple frequencies of fundamental source
frequency.

 

[ArNvC]

Not sure what are you trying to do?

 

[BradtheRad]

It is good to want to learn. You are asking the right questions. However only a few of them can be answered by a brief reply. Many of the questions are suitable to start individual threads, rather than one after the other in a single post.

Using a Transformer and a light bulb will be a Non-Linear load because of the transformer? the transformer puts the current and voltage not in phase? because its a inductor + resistor , phases are not in phase ?

With an ohmic load, the primary and secondary will be in phase over a wide range of frequencies.

However as the frequency is reduced, response starts to be affected by the inductive time constant (L/R). The flux field can become a driver of the action (for a while after a change in frequency, until things settle). The result is that the secondary does not immediately respond in exact phase to the primary.

As for the light bulb, it is not altogether linear. Its resistance increases with the voltage/current applied to it.

 

[danny davis]

As for the light bulb, it is not altogether linear. Its resistance increases with the voltage/current applied to it.

The voltage and current are IN phase right? even if the resistance increases with the temp

What u mean by its not altogether linear?

 

[BradtheRad]

The voltage and current are IN phase right? even if the resistance increases with the temp

Yes, V and A are in phase. (Unless the filament has inductive properties due to its coiled shape, however I don't remember seeing anyone state that the filament is an inductor.)

What u mean by its not altogether linear?

An incandescent bulb might draw 1A at 12V.
However it does not draw .5A at 6V.
But more like .5A at 3V.

Response is non-linear because the temperature of the filament changes its Ohm value.

 

[ArNvC]

Hi danny,

You need to know the Ohm's Law. The Incandascent Lamp dosent follow the Ohm' law. This means the V-I curve as it should be linear to be a Ohmic load, is actually not linear in case of Lamp. The reason as given by BradtheRad in Post #12. A linear resistance, a load more so always follows Ohm's Law.

Thanks

 

[danny davis]

The Incandascent Lamp dosent follow the Ohm' law.

What law doesn't it follow? since ohm's law doesn't work

Response is non-linear because the temperature of the filament changes its Ohm value.

So the filament doesn't follow ohms law you're saying? than what "law" is it following?

 

[BradtheRad]

So the filament doesn't follow ohms law you're saying? than what "law" is it following?

The filament certainly obeys ohm's law...
However it is not what we call an ohmic resistance in the classic sense. It's R value changes, whereas a resistor's R value is constant.

The filament is a non-linear component.

Screenshot showing response of a bulb and a resistor, with voltage-vs-ampere curves.

 

[Darktrax]

The filament certainly obeys ohm's law...
However it is not what we call an ohmic resistance in the classic sense. It's R value changes, whereas a resistor's R value is constant.

BradtheRad is right in that at any instant during the heating up of the filament, Ohm's law is obeyed in that the current will behave according to the resistance it finds. The resistance goes up rapidly to find the stable point where the self-heating is balanced by the loss in radiation from the filament.

Similar to filaments, the cold resistance of heaters , like ceramic furnace elements, is so low when cold they can overload the ability of power circuits to supply the start-up. They have to be "soft-started" to avoid circuit breaker tripping. Yet once hot, the current is at a satifactory level. (They have other issues like resistance increase with age, but that is a separate thing).

Filaments, like in lamps vary in construction, but deliver all the energy as heat, mostly infra-red radiation, and light, I think about 2% to 4% light depending on whether the filament is "coiled coil" construction or not. Given the energy into the filament, it has a mass and specific heat. It will get to a temperature. Where it limits is when the losses by radiation (see Stefan's Law) gets to equal the energy input.

What makes the behaviour non-linear is because the radiation law involves the fourth power of temperature.

Sometimes the stable point cannot happen. eg. A glass rod with mains power wires connected around a section, if heated by flame until it starts to conduct, will then be able to sustain the conducting until, beyond a critical point, the self-heating will run away into sudden avalanche breakdown, and the whole mess melts. I saw this myself. Unlike conducting metal where the resistance increases with temperature, glass does the opposite.

How things behave as conductors depends on the material properties, and the physical structure. If you want gross non-linearity, try a diode!

PS. Yes - the folk who point out that this thread has drifted well off-topic are right. Maybe we have wrung out all there is on this.

 

[danny davis]

What makes the behaviour non-linear is because the radiation law involves the fourth power of temperature.

This is good to know , that it went from ohms law to radiation law , temperature made the light bulb non-linear

So Temperature is non linear or it makes a component non linear

 

[ArNvC]

I = V / R still holds, but R is not a constant and will depend on temperature.If we were to increase the supply voltage in a real lamp circuit, the resulting increase in current would cause the filament to increase temperature, which would in turn increase its resistance, thus preventing further increases in current without further increases in battery voltage.