What is ripple voltage and current

Ripple Voltage

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Space Interference

4.6.2 INPUT RIPPLE NOISE

A low input ripple voltage must be maintained in a converter; otherwise, the regulator will not function properly. A regulator can only tolerate a limited input ripple voltage. There is always a limit on the ripple in the DC input voltage of the converter. In order to maintain a low ripple voltage there is a need for a large input filtering capacitance C, as shown in Figure 4.22 . A common practice is to allow the ripple to be 30% of the peak value of the minimum allowed DC voltage on the S/C power bus. For example, if the average power bus voltage is 28 V and the minimum allowed voltage (or undervoltage) is 22 V, then the maximum ripple voltage allowed into the power converter should not be more than 0.30 (22 V) = 6.6 V, which translates to about 35 V maximum allowed voltage.

Figure 4.22 . Typical input filter in most power converters.

Figure 4.22 shows the input circuit to a typical switching power supply. The minimum capacitance C required is given by

where P is the output power of filtering or input power to the switching-mode power supply (SMPS), f the DC input ripple frequency, and V_min the minimum bus-allowed DC voltage.

Case Study – The Variable Speed Drive

6.4.3 Capacitance calculation (rectifier)

We have established that the voltage ripple (at 300 Hz for a 50 Hz network) at the filter output point needs to be reduced by 17 dB. It is arguably safer, and simpler, to work with a reduction of 20 dB (factor 10). To do this, we note that our LC filter presents a 2 nd order behavior, with an eigenfrequency:

The reduction slope is –40 dB/decade for this type of filter, so a half-decade will be sufficient to produce a factor 10 ripple reduction, on the assumption that this is purely sinusoidal at 300 Hz (this simplifying hypothesis was also used in Chapter 3 ). The resonant frequency of the filter should therefore be placed a half-decade lower than the frequency of the ripple; a half-decade corresponds to a factor of 10 1 / 2 = 10 = 316 . We therefore take:

As we have already calculated the value of L required to satisfy the ripple criterion for the current supplied by the rectifier, we simply need to determine the required capacitance:

Surge Reduction

JAMES W. CLARK , in AC Power Conditioners , 1990

Design Example

It is desired to use a single silicon suppressor, placed across the output of a bridge rectifier operating directly from the power line, to suppress spikes arriving from the power line. The following selection process is offered as a typical process to be followed, resulting in direct selection of a suppressor from the manufacturer’s data book.

Determine the maximum DC plus ripple voltage on the output of the bridge, plus maximum tolerance.

Select a suppressor to have a reverse standoff voltage equal to or greater than the maximum circuit voltage, as defined in step 1 above. This selection will allow for operation over the temperature range of −65 to + 175°C.

Define the waveshape or duration of the transient.

Determine the peak pulse current the suppressor will be required to pass. Check this against the maximum peak pulse current shown on the device data sheet. If the predicted current exceeds the device rating, all is not lost. Something can be done, as will be discussed later.

If the current to be handled is within the device rating, proceed to determine the maximum peak pulse power the device must handle. This value is obtained by multiplying the transient peak pulse current by the maximum clamping voltage of the selected device.

If peak pulse power is within the maximum rating of the device, use the device selected. To evaluate this, the follow criteria should be used: a.

For a transient pulse width of an exponential delay, define “pulse time” as the duration of the pulse from the 10% amplitude point on the leading edge to the 50% amplitude point of the trailing edge. From the device data sheet determine the peak pulse power rating of the device for the computed pulse time of the incident pulse.

For transients that approximate a sine wave in pulse waveshape, use the total pulse width as the pulse time and multiply the corresponding peak pulse power value by 0.75 to obtain the effective device rating.

For square-wave-shaped transients, multiply the peak pulse power value for the corresponding pulse time by 0.66 to obtain the effective device rating.

If the transient is a rapidly damped sine wave, or rapidly damped square wave, with one time constant within eight cyclic, rate the same as if the device were subjected to just one cycle of the highest amplitude in the wave train.

For any other transient conditions, consideration must be given to the average power–handling capability of the device, and/or the total transient energy must be summed and evaluated in respect to thermal derating.

If the peak pulse power of the incident pulse is greater than the rating of the device, then several devices may be stacked in series to increase the power rating. When stacking the same power-rated devices, the current rating increases in proportion to the number of devices stacked. If it is impossible to achieve the necessary power rating by stacking the devices in series, parallel stacking can be done effectively for devices rated below 100 V. Close matching, about 200 mV between each device, is necessary to assure even loading of the transient between all the devices. Series–parallel combination stacking can be done for higher voltages and power.

Modeling of the Photovoltaic Irrigation Plant Components

Capacitor selection

The design criterion for the capacitor is that the ripple voltage across it should be less than 5%. The average voltage across the capacitor C is given by Eq. (2.68) ( Oi, 2005b ):

The value of the capacity C is calculated with the following equation ( Oi, 2005b ):

where R is the equivalent load resistance which is given by ( Oi, 2005b ):

Resonant Converters and Soft Switching

3.1.2 Resonance

Resonance is an interesting way of introducing current and voltage ripples so that they pass through zero at precise instants, while avoiding energy dissipation. Based on this principle, “inductance/capacitor”-type assemblies are particularly useful for soft switching in power electronic converters. 2 Consequently, LC circuits are always found in operating converters using soft switching; these circuits create resonance. For this reason, we often speak about resonant (or quasi-resonant) converters rather than soft switching converters.

This subject is complex and would fill a whole book on its own (e.g. [ CHE 99 ] is devoted to this subject). In this chapter, we will focus on a single example, which is used wholly or partially in a wide range of applications: a DC–DC resonant converter, using two cascaded converters (as shown in Figure 3.1 ):

Figure 3.1 . DC-DC resonant converter

a single-phase inverter connected to a resonant load;

a diode rectifier (which is part of the resonant load).

Signal Processing Circuits

David L. Terrell , in Op Amps (Second Edition) , 1996

Lower Frequency Limit.

The lower frequency limit is the frequency that causes the ripple voltage to exceed the maximum allowable level (determined by the design requirements). It can be estimated by applying the basic discharge equation for capacitors, which is

where V₀ is the initial charge of the capacitor (V_pk), V is the voltage to which the capacitor will discharge (assumed to be 0), and v_C is the minimum allowable voltage on the capacitor. For this discussion, the lower frequency limit will be considered to be the frequency that causes the ripple voltage across C₁ to be 1 percent of the DC voltage. Having made this definition, we can apply a simplified equation to determine the lower frequency limit:

In the case of the circuit in Figure 7.18 , we estimate the lower frequency limit as

Power Supplies

Choosing the Reservoir Capacitor and Transformer

If we have designed our supply to have a ripple voltage of 5% of supply voltage, then for 90% of the time the transformer is disconnected, and the output resistance of the power supply is determined purely by the capacitor ESR and associated output wiring resistance. This is why changing reservoir capacitors from general purpose types to high ripple current types produces a noticeable effect on the sound of an amplifier; they have a lower ESR (but a higher price).

The transformer/rectifier/capacitor combination is a non-linear system. This makes its behaviour considerably more complex than the ideal Thévenin source, so we need to investigate it over different periods of time.

In the short term (less than one charging cycle), the output resistance of the supply is equal to capacitor ESR plus wiring resistances. This will be true even for very high current transient demands, which may appear in each and every charging cycle, provided that they do not significantly deplete the charge on the capacitor. All that is required is that the capacitor should be able to source these transient currents. To be able to do this, the capacitor needs a low ESR, not just at mains frequencies, but also up to at least 40 kHz, because a Class B power output stage causes a rectified (and therefore frequency doubled) version of the audio signal to appear on the power supply rails. (See Chapter 6 for explanation of Class B.) We can cope with this requirement by using an electrolytic capacitor designed for use in switched mode power supplies as the main reservoir.

A power amplifier may significantly deplete the charge in the reservoir capacitor, causing output voltage to fall either by drawing a sustained high current, due to a continuous full power sine-wave test, or by reproducing a short, but loud, sound – such as a bass drum.

Supplying a constant load is relatively easy, because we know exactly how much current will be drawn, and we simply design for that current. If the ripple voltage for a sensible ripple current is higher than we would like, then we simply add a regulator to remove it.

The difficulties start when we want to supply a changing load. It might seem that if the power amplifier is rated at 50 W continuous into 8 Ω, then all we have to do is to calculate what load current that implies, and design for that current. The drawbacks of this approach are more easily demonstrated using a transistor amplifier, where the load is directly coupled to the output stage and the power supply is very simple (see Figure 5.14 ).

Figure 5.14 . Typical power supply for transistor amplifier.

Considering our 50 W 8 Ω example:

Therefore, for a sine wave:

But we have to supply the peak current, which is √2 greater, at 3.5 A:

But we have to supply the peak voltage, which is √2 greater, at 28.3 V. Transistor amplifiers can typically swing to within about a volt of rail, so we might just tolerate ±29 V rails, and a power supply capable of delivering ±29 V at 3.5 A is implied. We therefore need 203 W per channel and 406 W for a 50 W stereo amplifier! This is a very large and expensive power supply, and we would need some astonishingly good reasons for using it.

The key to the problem lies in the class of the output stage. If the output stage operates in Class A, then the quiescent current equals the peak current required at maximum power output, in this case, 3.5 A. If each channel genuinely draws a constant 3.5 A from the ±29 V power supply, then we really do need the 406 W power supply. The classic Krell KSA50 50 W stereo amplifier drew 300 W from the mains at idle [5] , suggesting that it wasn’t quite true Class A, but it was certainly far closer than most Class A pretenders.

The reservoir capacitor value was easy to determine using our earlier formula and 5% ripple voltage criterion, but the transformer is quite a different matter. It is possible to determine the requirements of the transformer exactly, using the graphs originally devised by Schade [6] . In practice, the required transformer information may not be available, so a practical rule of thumb is to make the VA rating of the transformer at least equal to the required output power.

If our example stereo 50 W amplifier output stage becomes Class B, then each channel still supplies 3.5 A to the load on the crests of the sine wave, but at other points in the cycle the required current from the power supply is much lower. The effect of the reservoir capacitor is to average the fluctuating current demand, and for a full-wave rectified sine wave:

The average supply current is 2.2 A, so a 250 VA transformer would be chosen.

We could further argue that the amplifier does not operate at full power all the time and that the short term musical peaks requiring maximum output power do not last long. A smaller transformer could therefore be used, since the reservoir capacitor could supply the peak currents. This is a very seductive argument, and many commercial amplifier manufacturers have been persuaded by it, since £1 extra on component cost generally adds £4–5 to the retail price.

We do not have to work to such tight commercial considerations and, within reason, the bigger the mains transformer, the better it is.

Источник

Ripple Voltage

Related terms:

Download as PDF

About this page

Space Interference

4.6.2 INPUT RIPPLE NOISE

A low input ripple voltage must be maintained in a converter; otherwise, the regulator will not function properly. A regulator can only tolerate a limited input ripple voltage. There is always a limit on the ripple in the DC input voltage of the converter. In order to maintain a low ripple voltage there is a need for a large input filtering capacitance C, as shown in Figure 4.22 . A common practice is to allow the ripple to be 30% of the peak value of the minimum allowed DC voltage on the S/C power bus. For example, if the average power bus voltage is 28 V and the minimum allowed voltage (or undervoltage) is 22 V, then the maximum ripple voltage allowed into the power converter should not be more than 0.30 (22 V) = 6.6 V, which translates to about 35 V maximum allowed voltage.

Figure 4.22 . Typical input filter in most power converters.

Figure 4.22 shows the input circuit to a typical switching power supply. The minimum capacitance C required is given by

where P is the output power of filtering or input power to the switching-mode power supply (SMPS), f the DC input ripple frequency, and V_min the minimum bus-allowed DC voltage.

Case Study – The Variable Speed Drive

6.4.3 Capacitance calculation (rectifier)

We have established that the voltage ripple (at 300 Hz for a 50 Hz network) at the filter output point needs to be reduced by 17 dB. It is arguably safer, and simpler, to work with a reduction of 20 dB (factor 10). To do this, we note that our LC filter presents a 2 nd order behavior, with an eigenfrequency:

The reduction slope is –40 dB/decade for this type of filter, so a half-decade will be sufficient to produce a factor 10 ripple reduction, on the assumption that this is purely sinusoidal at 300 Hz (this simplifying hypothesis was also used in Chapter 3 ). The resonant frequency of the filter should therefore be placed a half-decade lower than the frequency of the ripple; a half-decade corresponds to a factor of 10 1 / 2 = 10 = 316 . We therefore take:

As we have already calculated the value of L required to satisfy the ripple criterion for the current supplied by the rectifier, we simply need to determine the required capacitance:

Surge Reduction

JAMES W. CLARK , in AC Power Conditioners , 1990

Design Example

It is desired to use a single silicon suppressor, placed across the output of a bridge rectifier operating directly from the power line, to suppress spikes arriving from the power line. The following selection process is offered as a typical process to be followed, resulting in direct selection of a suppressor from the manufacturer’s data book.

Determine the maximum DC plus ripple voltage on the output of the bridge, plus maximum tolerance.

Select a suppressor to have a reverse standoff voltage equal to or greater than the maximum circuit voltage, as defined in step 1 above. This selection will allow for operation over the temperature range of −65 to + 175°C.

Define the waveshape or duration of the transient.

Determine the peak pulse current the suppressor will be required to pass. Check this against the maximum peak pulse current shown on the device data sheet. If the predicted current exceeds the device rating, all is not lost. Something can be done, as will be discussed later.

If the current to be handled is within the device rating, proceed to determine the maximum peak pulse power the device must handle. This value is obtained by multiplying the transient peak pulse current by the maximum clamping voltage of the selected device.

If peak pulse power is within the maximum rating of the device, use the device selected. To evaluate this, the follow criteria should be used: a.

For a transient pulse width of an exponential delay, define “pulse time” as the duration of the pulse from the 10% amplitude point on the leading edge to the 50% amplitude point of the trailing edge. From the device data sheet determine the peak pulse power rating of the device for the computed pulse time of the incident pulse.

For transients that approximate a sine wave in pulse waveshape, use the total pulse width as the pulse time and multiply the corresponding peak pulse power value by 0.75 to obtain the effective device rating.

For square-wave-shaped transients, multiply the peak pulse power value for the corresponding pulse time by 0.66 to obtain the effective device rating.

If the transient is a rapidly damped sine wave, or rapidly damped square wave, with one time constant within eight cyclic, rate the same as if the device were subjected to just one cycle of the highest amplitude in the wave train.

For any other transient conditions, consideration must be given to the average power–handling capability of the device, and/or the total transient energy must be summed and evaluated in respect to thermal derating.

If the peak pulse power of the incident pulse is greater than the rating of the device, then several devices may be stacked in series to increase the power rating. When stacking the same power-rated devices, the current rating increases in proportion to the number of devices stacked. If it is impossible to achieve the necessary power rating by stacking the devices in series, parallel stacking can be done effectively for devices rated below 100 V. Close matching, about 200 mV between each device, is necessary to assure even loading of the transient between all the devices. Series–parallel combination stacking can be done for higher voltages and power.

Modeling of the Photovoltaic Irrigation Plant Components

Capacitor selection

The design criterion for the capacitor is that the ripple voltage across it should be less than 5%. The average voltage across the capacitor C is given by Eq. (2.68) ( Oi, 2005b ):

The value of the capacity C is calculated with the following equation ( Oi, 2005b ):

where R is the equivalent load resistance which is given by ( Oi, 2005b ):

Resonant Converters and Soft Switching

3.1.2 Resonance

Resonance is an interesting way of introducing current and voltage ripples so that they pass through zero at precise instants, while avoiding energy dissipation. Based on this principle, “inductance/capacitor”-type assemblies are particularly useful for soft switching in power electronic converters. 2 Consequently, LC circuits are always found in operating converters using soft switching; these circuits create resonance. For this reason, we often speak about resonant (or quasi-resonant) converters rather than soft switching converters.

This subject is complex and would fill a whole book on its own (e.g. [ CHE 99 ] is devoted to this subject). In this chapter, we will focus on a single example, which is used wholly or partially in a wide range of applications: a DC–DC resonant converter, using two cascaded converters (as shown in Figure 3.1 ):

Figure 3.1 . DC-DC resonant converter

a single-phase inverter connected to a resonant load;

a diode rectifier (which is part of the resonant load).

Signal Processing Circuits

David L. Terrell , in Op Amps (Second Edition) , 1996

Lower Frequency Limit.

The lower frequency limit is the frequency that causes the ripple voltage to exceed the maximum allowable level (determined by the design requirements). It can be estimated by applying the basic discharge equation for capacitors, which is

where V₀ is the initial charge of the capacitor (V_pk), V is the voltage to which the capacitor will discharge (assumed to be 0), and v_C is the minimum allowable voltage on the capacitor. For this discussion, the lower frequency limit will be considered to be the frequency that causes the ripple voltage across C₁ to be 1 percent of the DC voltage. Having made this definition, we can apply a simplified equation to determine the lower frequency limit:

In the case of the circuit in Figure 7.18 , we estimate the lower frequency limit as

Power Supplies

Choosing the Reservoir Capacitor and Transformer

If we have designed our supply to have a ripple voltage of 5% of supply voltage, then for 90% of the time the transformer is disconnected, and the output resistance of the power supply is determined purely by the capacitor ESR and associated output wiring resistance. This is why changing reservoir capacitors from general purpose types to high ripple current types produces a noticeable effect on the sound of an amplifier; they have a lower ESR (but a higher price).

The transformer/rectifier/capacitor combination is a non-linear system. This makes its behaviour considerably more complex than the ideal Thévenin source, so we need to investigate it over different periods of time.

In the short term (less than one charging cycle), the output resistance of the supply is equal to capacitor ESR plus wiring resistances. This will be true even for very high current transient demands, which may appear in each and every charging cycle, provided that they do not significantly deplete the charge on the capacitor. All that is required is that the capacitor should be able to source these transient currents. To be able to do this, the capacitor needs a low ESR, not just at mains frequencies, but also up to at least 40 kHz, because a Class B power output stage causes a rectified (and therefore frequency doubled) version of the audio signal to appear on the power supply rails. (See Chapter 6 for explanation of Class B.) We can cope with this requirement by using an electrolytic capacitor designed for use in switched mode power supplies as the main reservoir.

A power amplifier may significantly deplete the charge in the reservoir capacitor, causing output voltage to fall either by drawing a sustained high current, due to a continuous full power sine-wave test, or by reproducing a short, but loud, sound – such as a bass drum.

Supplying a constant load is relatively easy, because we know exactly how much current will be drawn, and we simply design for that current. If the ripple voltage for a sensible ripple current is higher than we would like, then we simply add a regulator to remove it.

The difficulties start when we want to supply a changing load. It might seem that if the power amplifier is rated at 50 W continuous into 8 Ω, then all we have to do is to calculate what load current that implies, and design for that current. The drawbacks of this approach are more easily demonstrated using a transistor amplifier, where the load is directly coupled to the output stage and the power supply is very simple (see Figure 5.14 ).

Figure 5.14 . Typical power supply for transistor amplifier.

Considering our 50 W 8 Ω example:

Therefore, for a sine wave:

But we have to supply the peak current, which is √2 greater, at 3.5 A:

But we have to supply the peak voltage, which is √2 greater, at 28.3 V. Transistor amplifiers can typically swing to within about a volt of rail, so we might just tolerate ±29 V rails, and a power supply capable of delivering ±29 V at 3.5 A is implied. We therefore need 203 W per channel and 406 W for a 50 W stereo amplifier! This is a very large and expensive power supply, and we would need some astonishingly good reasons for using it.

The key to the problem lies in the class of the output stage. If the output stage operates in Class A, then the quiescent current equals the peak current required at maximum power output, in this case, 3.5 A. If each channel genuinely draws a constant 3.5 A from the ±29 V power supply, then we really do need the 406 W power supply. The classic Krell KSA50 50 W stereo amplifier drew 300 W from the mains at idle [5] , suggesting that it wasn’t quite true Class A, but it was certainly far closer than most Class A pretenders.

The reservoir capacitor value was easy to determine using our earlier formula and 5% ripple voltage criterion, but the transformer is quite a different matter. It is possible to determine the requirements of the transformer exactly, using the graphs originally devised by Schade [6] . In practice, the required transformer information may not be available, so a practical rule of thumb is to make the VA rating of the transformer at least equal to the required output power.

If our example stereo 50 W amplifier output stage becomes Class B, then each channel still supplies 3.5 A to the load on the crests of the sine wave, but at other points in the cycle the required current from the power supply is much lower. The effect of the reservoir capacitor is to average the fluctuating current demand, and for a full-wave rectified sine wave:

The average supply current is 2.2 A, so a 250 VA transformer would be chosen.

We could further argue that the amplifier does not operate at full power all the time and that the short term musical peaks requiring maximum output power do not last long. A smaller transformer could therefore be used, since the reservoir capacitor could supply the peak currents. This is a very seductive argument, and many commercial amplifier manufacturers have been persuaded by it, since £1 extra on component cost generally adds £4–5 to the retail price.

We do not have to work to such tight commercial considerations and, within reason, the bigger the mains transformer, the better it is.

Источник