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Circuit Surgery Regular clinic by Ian Bell Capacitor Dielectric Absorption – + + – – – + – + – + – + + – + – – – + C harging vo ltage + + + + + + + + + + + + + + + + + + + + + + + + + R ecove ry vo ltage – – + + + – + – + + – – – + + – – + – + + + – – – + + – + – 52 + + Most, but not every month, LTSpice is used to support descriptions and analysis in Circuit Surgery. The examples and files are available for download from the PE website. + Simulation files – As the name suggests, dielectric absorption is due to the properties of the capacitor’s dielectric – the insulating material which is placed between the two conducting plates to form the capacitor. The dielectric is an insulator, so unless the breakdown voltage is exceeded, very little electric charge flows through it when a voltage is applied across the capacitor. However, significant internal charge redistribution does occur in the form of polarisation of dipoles. Dipoles are parts of a material which can have a distribution of charge (in simple terms, positive at one end and negative at the other). This can occur at the atomic, – + – Dielectrics C apacitor vo ltage U npolarised + about capacitors on the EEWeb forum: ‘I have a question, but have not found an answer on the Net. An electrolytic capacitor, if it is in short circuit for a long time, discharges and its voltage goes to zero. But if you remove the short circuit, the voltage on the terminals slowly rises again. Why? Thank you very much.’ As was quickly pointed out by forum expert Peter Traneus Anderson, this effect is called ‘dielectric absorption’. Other names are also used, such as ‘voltage recovery’ and ‘dielectric soakage’. As John indicates, dielectric absorption is a non-deal phenomenon which occurs in capacitors – it is not exclusive to electrolytic capacitors but tends to be larger in capacitors of this type. The effect is most obvious if a capacitor is charged for a long time, then discharged to zero volts very quickly and immediately left open circuit. After this, both real and ideal capacitors will start at zero volts. An ideal capacitor will stay at zero volts, but a real capacitor will develop a non-zero ‘recovery’ voltage across it after a time delay due to dielectric absorption. – J ohn Curtin asked a question – – – – – – – – – – – – – – – – – – – – – – – – P olarised by an applied electric f ield Fig.1. Polarisation of dipoles in a capacitor dielectric. ionic, molecular and higher structural levels (such as cells in biological tissues). An electric field applied to a dielectric (in our case, a voltage applied across a capacitor) causes the dipoles to rotate to be more aligned with the applied field (with no field applied they are randomly aligned) – see Fig.1. The field created by the aligned dipoles interacts with the applied field and threby allows the capacitor to store more energy (hold more charge) than if there was just a vacuum between the plates. The more polarisation, the more additional storage (so more capacitance for a given physical size of device). When the field is removed (capacitor discharged) the dipoles return to their random orientation – referred to as ‘relaxation’. The polarisability of the material is characterised by its permittivity. The permittivity of a material can be stated as an absolute value (symbol – Greek epsilon), or as relative permittivity ( r), which is relative to the permittivity of free space (vacuum) ( 0); that is, = r 0. The symbol (Greek kappa) is also used for relative permittivity. Materials used for capacitor dielectrics tend to have a large relative permittivity (sometimes referred to as ‘high K’) as this allows for smaller capacitors for a given capacitance value and voltage rating. The polarisation of dielectrics is a complex process because there are typically multiple types of dipole within materials and the polarisation and depolarisation C harging D ischarging O pen circuit T ime Fig.2. Behaviour of a capacitor exhibiting dielectric absorption. (relaxation) is not instantaneous. In simple terms, dielectric absorption is due to the fact that not all of the dipoles depolarise in the time taken to rapidly discharge a capacitor to zero voltage. They continue to hold charge (‘absorbed’ by the dielectric). Initially, if the capacitor is open circuited after discharge, this does not result in a voltage at the plates. However, the charge will get transferred to the capacitor plates over time as the dipoles depolarise. This will cause a charge build up on the plates leading to the recovery voltage that John asks about. Capacitor behaviour and measurement The behaviour of a capacitor exhibiting dielectric absorption is illustrated in Fig.2, which is a graph of the voltage across a capacitor against time, undergoing the charge/discharge/open-circuit sequence just mentioned. Subjecting capacitors to this process can be used to make measurements to observe or characterise their dielectric absorption. A circuit concept for doing this is shown in Fig.3. The specifics of implementation may vary; for example, in R 1 1 S1 2 3 V c R 2 C H igh- impedance vo ltmeter Fig.3. Circuit concept for observing or measuring capacitor dielectric absorption. Practical Electronics | November | 2021 the simplest cases manual switching could be employed, but automated switching using relays and various improvements to the basic idea can be incorporated. In the first part of the graph the capacitor has been charged for a long time, so the voltage across it is effectively equal to the charging voltage. With the circuit in Fig.3 this is achieved by applying VC through R1 with S1 in position 1. A ‘long time’ could be from minutes to hours. Typically, real capacitors would be charged to a suitable voltage (maybe their rated working voltage) when making dielectric absorption measurements, taking care to safely limit the initial current surge when charging is started. Applying the maximum working voltage to a discharged capacitor through too small a resistance (R1 in Fig.3) may result in a damaging current surge (eg, by connecting the capacitor directly to a power supply at that voltage). Once fully charged, the capacitor is discharged quickly relative to the charge time, this is achieved by connecting a suitable resistor across the capacitor (R2 in Fig.3, with S1 in position 2). The switchover from charge to discharge mode should occur quickly with the capacitor briefly open-circuit (S1 break-before-make). The capacitor will hold its charge during this time. Capacitors tend to discharge slowly if left charged and open circuit, so if the aim is to observe dielectric absorption, the time between charge and discharge should be brief. To allow the dielectric absorption phenomenon to be observed, the discharge time should be just sufficient to get very close to zero volts as quickly as possible. This means discharge through a low resistance (R2) – but, for real tests, the current must be limited to a value that will not cause any problems. Standard measurements may use fixed times and resistance values for discharge to ensure the same conditions are used when comparing devices. After the quick discharge the capacitor is disconnected from the discharge resistor so that it is open circuit (S1 in position 3 in Fig.3). The capacitor voltage will rise, as shown in the final section of the graph in Fig.2. Initially, the increase will be rapid, but the rate of change will slow over time. The final voltage reached is the recovery voltage. The voltage across it over time can be measured using a voltmeter, which should have a very high input impedance if accurate results are required. Dielectric absorption (DA) figures for capacitors are defined as the percentage of the recovery voltage (VR) relative to the charging voltage (VC) under specified conditions (charge and discharge time, discharge resistance, open-circuit time after which voltage is measured, temperature and so on): DA(%)= (VR/VC) × 100 Practical Electronics | November | 2021 As well as describing the behaviour of a capacitor exhibiting dielectric absorption, the preceding discussion has sketched a basic approach to measuring the DA value. You will find references to MIL-C-19978D and EN 60384-1 as standard test produces, but their relevance in the context of modern electronic systems may be debatable. For further discussion on this, and a more comprehensive exploration of measuring dielectric absorption, readers might be interested in a paper published online by Leslie Green: Practical Exploration of Dielectric Absorption in Capacitors for the 21st Century (see: http://lesliegreen. byethost3.com/articles/Dielectric.pdf). Effects of dielectric absorption The recovery voltage due to dielectric absorption varies significantly for different types of capacitors. Capacitors with polymer dielectrics, such as polyphenylene sulphide, polypropylene and polystyrene tend to have the lowest DA values, which can be well below 0.1%. Dielectric absorption is higher for ceramic capacitors (around 0.5 to 2.5%) but depends on the type of ceramic, with Class 1 C0G (NP0) having lower DA values than class 2 X7R. The highest DA values are for electrolytic capacitors, which can reach 10% or 15%. Dielectric absorption makes itself felt (literally in some cases) in different ways depending on the type of capacitor and the application. For high-value, high-voltage capacitors there is a possibility for the recovery voltage to deliver a significant, even lethal electric shock to someone handling the capacitor. Such capacitors are usually shipped and stored shorted out, but can still recover after the short is removed, so in situations where this is a potential problem people need to follow appropriate safety procedures. Dielectric absorption can also cause problems in some signal processing applications, with one of the most important examples being sample and hold circuits (see Fig.4) – these are commonly used to provide steady input voltage samples to analogue-to-digital converters (ADCs). The input signal is sampled by rapidly charging a capacitor to the value of the input voltage at the instant that a conversion is required (by closing the sample switch briefly). The sample switch is then opened, and the capacitor holds the value steady while the conversion is performed, buffered by a high-impedance unity-gain amplifier to prevent the capacitor from discharging. When used with a single input signal, in many cases the voltage will not change much between successive samples (with a smoothly changing signal) and dielectric absorption will not be a problem. However, some signals will have abrupt changes, and it is common for ADCs to be used with multiplexers so that a single ADC Multiplexe r I n1 I n2 I n3 – Sample switch + O utput ( to AD C ) C I nN Fig.4. Sample and hold with input multiplexer. The switches are electronic (typically MOS transistors). can convert signals from multiple input channels (see Fig.4). Here it is quite likely that the capacitor will be charged to near the maximum input voltage for one sample and close to zero on the next – if different channels are at opposite ends of the input voltage range. This creates a situation very similar to that shown in Fig.2. Even if the conversion takes place relatively quickly compared with the time taken to reach the full recovery voltage, the effect of dielectric absorption will result in an unwanted change in voltage on the sample capacitor – remember, the rate of change is relatively fast initially. This can result in errors in the converted value. For example, for a 12-bit ADC a change of only 0.025% of the full-scale voltage is sufficient for a change of the converted value (by 1 leastsignificant bit). Another circuit where dielectric absorption may have a significant effect is the op amp integrator (see Fig.5), one of the ‘standard’ op amp applications. The output of this circuit is proportional to the integral of the input voltage over time. In some situations, it is necessary to reset the integrator (or more precisely, set the initial conditions of the integration). This is shown in Fig.5, implemented as a switch discharging the capacitor, although setting a specific voltage other than zero may be required. The situation for the capacitor in Fig.5 is again very similar to the scenario in Fig.2. The capacitor may be at a relatively large value when the reset (discharge) occurs, and the resulting recovery voltage will be added as an error to the output from the circuit after the switch is opened. Model circuit It is possible to model the dielectric absorption of a capacitor using an equivalent circuit of the form shown in Fig.6, which R eset switch C V in R – V out + Fig.5. Basic op amp integrator circuit. 53 R A1 R A2 R A3 R AN C 1 C A1 C A2 C A3 C AN Fig.6. Equivalent circuit model for capacitor dielectric absorption. Fig.7. Equivalent circuit model with five RC pairs used for dielectric absorption and other nonideal capacitor characteristics. R S L S R L was published by Paul Dow in 1958 in a paper in the IRE Transactions on Electronic Computers. The model helps us understand Ω the behaviour of capacitors with respect to dielectric absorption and simulate these I nitially effects at circuit level without needing a uncharged 1 V 0 .1 µ F (0 V ) deep knowledge of the physics of dielectrics at the atomic and molecular level. The equivalent circuit comprises a set of N Fig.8. Example circuit for RC time RC circuits (RA1 and CA1 to RAN and CAN) constant – charging. in parallel with an ideal capacitor, C1. C1 has the rated value of the capacitor. The capacitors modelling the absorption (CA) will have a smaller value than C1 (the I nitially charged Ω 0 .1 µ F total absorption capacitance relative to to 1 V C1 determines the final recovery voltage). The resistor values will typically be large (megohms to giga-ohms) to account for Fig.9. Example circuit for RC time the slow change of the recovery voltage. constant – discharging. Intuitively, the model in Fig.6 explains the 1 .0 V behaviour shown 0 .9 V in Fig.2. The 0 . 8V long charge time 0 .7 V ensures that all the 6 3% charged 0 .6 V af ter R C = 0 . 1 ms capacitors in the 0 . 5V circuit are charged 0 .4 V to the applied D ischarged to 37 % af ter R C = 0 . 1 ms voltage, despite the 0 . 3V large resistor values 0 .2 V in the absorption 0 .1 V part. The quick 0 V discharge through 0 . 0 ms 0 . 2 ms 0 . 4 ms 0 . 6 ms 0 . 8ms 1 . 0 ms 1 . 2 ms 1 . 4 ms 1 . 6 ms 1 . 8ms 2 . 0 ms a relatively small resistor removes Fig.10. Example RC charge and discharge waveforms based on the charge from C1, Fig.6 (0 to 1.0ms) and Fig.7 (1.0 to 2.0ms). R A1 R A2 R A3 R A4 R A5 C A1 C A2 C A3 C A4 C A5 C 1 but the large resistors in series with the other capacitors mean that little charge is removed from them. During the open circuit period the capacitors in the equivalent circuit will share charge – specifically charge will move off CA1 to CAN, and onto C1 until the voltages on the capacitors equalise (if the capacitors are at different voltages this difference will appear across the resistors, causing current to flow, moving the charge between the capacitors). We observe the voltage on the terminals (across C1) rise – this is the recovery voltage, or the capacitor appearing to recharge itself. The large resistor values mean that this will take a relatively long time compared with the discharge of C1 just performed. The circuit in Fig.6 only models the effect of dielectric absorption, but other components can be added to cover other non-deal characteristics of capacitors if required. An example is shown in Fig.7, which uses five RC pairs to model absorption and also includes equivalent series resistance (ESR) (RS), equivalent series inductance (ESL) (LS) and leakage resistance (RL). Curves and time constants The concept of RC time constants is important in the dielectric absorption equivalent circuit. This can be illustrated by the basic charge and discharge circuits shown in Fig.8 and 9. The well-known exponential charge and discharge curves for these ideal RC circuits are shown in Fig.10. The time constant for an RC circuit is the value obtained by multiplying the resistor and capacitor values together (the result has units of time). For example, in the circuits in Fig.8 and 9 we get RC = 1.0kΩ × 1.0µF = 0.1ms. When a capacitor is charged via a resistor from a fixed voltage, the voltage across the capacitor will be 63% of the applied voltage after time RC has elapsed (as seen in Fig.10). Similarly, when a capacitor is discharged through a resistor the voltage will reach 37% of the initial value after the time constant time has elapsed. The time constants of each of the set of RC circuits in the equivalent circuit in Fig.6 typically increase by a factor of ten along the sequence from 1 to N. Time constants for dielectric absorption can be very long – implying many hours or more to approach the final recovery voltage. It is also worth Fig.11. Simulation example. 54 Practical Electronics | November | 2021 noting that a dielectric absorption means that for real capacitors the charge and discharge curves do not follow the perfect exponential characteristic of ideal devices. Simulation example The circuit in Fig.11 is an LTspice schematic which can be used as a starting point to explore the equivalent circuit in Fig.6. The switches (S1 and S2) are controlled by the pulse waveforms from V2 and V3 to operate the circuit in a similar way to that described earlier for Fig.3. The simulation is configured with very idealised switches (very high off and very low on resistance, as defined by the SW1 model), so this is not a realistic circuit in terms of what is done to the capacitor. The switch model can be changed, or resistors added (as in Fig.3) to make this more like a real test circuit. Just three absorption RC circuits are used to keep the simulation simple, with their time constants increasing by a factor of five, so the simulation does not have to be too long. The longest of these is R3C4 which is just under one second. Thus, it is reasonable in this context that charging the capacitor for a ‘long time’ has a duration of five seconds – an RC circuit charges to more than 99% of the applied voltage in 5RC. V2 is configured to close switch S2 from simulation time 1s to 6s to charge all the capacitors in the equivalent circuit (see the simulation waveforms in Fig.12). Immediately after this, C1 is discharged very quickly by closing S1 for 200µs, after which both switches are open. The short discharge duration and relatively large values of R1 to R3 compared to S1’s resistances mean that C2 to C4 will lose very little charge while S1 is closed – this can be confirmed by zooming in on the waveform. It is difficult to see the details of the recovery voltage in Fig.12, so Fig.13 provides a zoom-in. The final voltage is 100mV, which is about 10% of the applied voltage. We can also obtain this figure by analysing the charge sharing – the total absorption capacitance is 3 × 37nF = 111nF. The overall total capacitance with C1 is 1111nF. With Fig.13. Zoom in to show recovery voltage. Practical Electronics | November | 2021 Fig.12. Simulation results from the circuit in Fig.11. C1 starting at 0V and C2 to C4 at 1V, the four capacitors sharing charge will result in a final voltage of (111/1111) × 1V = 100mV. The recovery time is dominated by the R3C4 time constant, and we simulate the recovery for 5 × R3C4, so we are close to the final value. The simulation can be modified to study the effect of different dielectric absorption characteristics or measurement scenarios. For example, Fig.14 shows the effect of adding a 10MΩ resistor across C1 – which could be the resistance of a voltmeter. This effect is to discharge C1 at the same time as it is being charged from the absorption capacitors, so the recovery voltage peaks and then decreases. Fig.14. Recovery voltage with 10MΩ resistor across C1. 55