Retrovert
Addicted Member
Additional data from National Semiconductor, the original source for the LM317 and friends:
https://www.ti.com/lit/ug/snoa826/snoa826.pdf?ts=1616616335945
LM317, LM340, LP2975: A User’s Guide To Compensating Low-Dropout Regulators
by Chester Simpson
National Semiconductor
The NPN Darlington pass transistor configuration requires that at least 1.5V to 2.5V be maintained from input-to-output for the device to stay in regulation. This minimum voltage “headroom” (called the dropout voltage) is given by:
...
Feedback is used in all voltage regulators to hold the output voltage constant. The output voltage is sampled through a resistive divider (Figure 5), and that signal is fed back to one input of the error amplifier. Since the other input of the error amplifier is tied to a reference voltage, the error amplifier will supply current as required to the pass transistor to keep the regulated output at the correct DC voltage.
It is important to note that for a stable loop, negative feedback must be used. Negative feedback (sometimes called degenerative feedback) is opposite in polarity to the source signal (see Figure 6).
Because it is opposite in polarity with the source, negative feedback will always cause a response by the loop which opposes any change at the output. This means that if the output voltage tries to rise (or fall), the loop will respond to force it back to the nominal value.
Positive feedback occurs when the feedback signal has the same polarity as the source signal. In this case, the loop responds in the same direction as any change which occurs at the output. This is clearly unstable, since it does not cancel out changes in output voltage, but amplifies them.
It should be obvious that no one would intentionally design positive feedback into the loop of a linear regulator, but negative feedback becomes positive feedback if it experiences a phase shift of 180°.
...
POLES
A pole (Figure 8) is defined as a point where the slope of the gain curve changes by -20 dB/decade (with reference to the slope of the curve prior to the pole). Note that the effect is additive: each additional pole will increase the negative slope by the factor “n” X (-20 dB/decade), where “n” is the number of additional poles.
The phase shift introduced by a single pole is frequency dependent, varying from 0 to -90° (with a phase shift of -45° at the pole frequency). The most important point is that nearly all of the phase shift added by a pole (or zero) occurs within the frequency range one decade above and one decade below the pole (or zero) frequency.
NOTE: a single pole can add only -90° of total phase shift, so at least two poles are needed to reach -180° (which is where instability can occur).
ZEROES
A zero (Figure 9) is defined as a point where the gain changes by +20 dB/ decade (with respect to the slope prior to the zero). As before, the change in slope is additive with additional zeroes.
The phase shift introduced by a zero varies from 0 to +90°, with a +45° shift occurring at the frequency of the zero.
The most important thing to observe about a zero is that it is an “anti-pole”, which is to say its effects on gain and phase are exactly the opposite of a pole.
This is why zeroes are intentionally added to the feedback loops of LDO regulators: they can cancel out the effect of one of the poles that would cause instability if left uncompensated.
...
The plot of Phase Shift shows how the various poles and zeroes contribute their effect on the feedback signal. To produce this plot, the phase shift at each frequency point was calculated based upon summing the contributions of every pole and zero at that frequency. The phase shift at any frequency “f” which is caused by a pole located at frequency “fp” can be calculated from:
The phase shift resulting from a zero located at frequency “fz” can be found using:
Is this loop stable? To answer that question, we need only know the phase shift at 0 dB (which is 1 MHz in this case). Finding this does not require complex calculations:
As stated in the previous sections, a pole or zero contributes nearly its full phase shift in the frequency range one decade above and below the center frequency of the pole (or zero). Therefore, The first two poles and the first zero contribute their full phase shifts of -180° and +90°, respectively, resulting in a net phase shift of -90°.
The final pole is exactly one decade below the 0 dB frequency. Using the formula for Pole Phase Shift, this pole will contribute -84° of phase shift @ 1 MHz. Added to the -90° from the two previous poles and the zero, the total phase shift is -174° (which means the phase margin is 6°). This loop would either oscillate or ring severely.
...
To reduce the negative phase shift (and prevent oscillations), a zero must be added to the loop. A zero can contribute as much as +90° of positive phase shift, which will cancel out the effects of one of the two low frequency poles.
All monolithic LDO regulators require that this zero be added to the loop, and they derive it from a characteristic that is inherent in the output capacitor: equivalent series resistance (usually referred to as ESR).
...
The ESR of the output capacitor puts a zero in the loop gain which can be used to reduce excess negative phase shift. The frequency where the zero occurs is directly related to the value of the ESR and amount of output capacitance:
Using the example in the previous section (Bode plot shown in Figure 12), we will assume that the value of COUT = 10 µF and the output capacitor ESR = 1Ω, which means a zero will occur at 16 kHz.
Figure 14 shows how this added zero will change the unstable plot into a stable one:
The bandwidth of the loop is increased so that the 0 dB crossover frequency moves from 30 kHz to 100 kHz.
The zero adds a total of +81° positive phase shift at 100 kHz (the 0 dB frequency). This will reduce the negative phase shift caused by the poles PL and P1. Since the pole PPWR is located at 500 kHz, it adds only -11° of phase shift at 100 kHz.
Summing all poles and zeroes, the total phase shift at 0 dB is now -110°. This corresponds to a phase margin of +70°, which is extremely stable.
This illustrates how an output capacitor with the correct value of ESR can generate a zero that stabilizes an LDO.
...
HIGH ESR
Using the example developed in the previous sections, we will change the conditions and assume the ESR of the 10 µF output capacitor is increased to 20Ω.This will decrease the frequency of the zero to 800 Hz (Figure 16). Reducing the frequency of the zero causes the loop bandwidth to increase, moving the 0dB crossover frequency from 100 kHz to 2 MHz.
This increased bandwidth means that the pole PPWR occurs at a gain value of +20 dB (compared to -10 dB in Figure 14).
Analyzing the plot (Figure 16) for phase margin, it can be assumed that the zero cancels out either P1 or PL. This means the loop has a two-pole response with the low frequency pole contributing -90° of phase shift and the high frequency pole PPWR contributing about -76° of phase shift.
Although this appears to leave a phase margin of 14° (which might be stable), bench test data shows that ESR values > 10Ω usually cause instability because of phase shifts contributed by other high-frequency poles which are not shown in this simplified model.
...
LOW ESR
An output capacitor with a very low ESR value can cause oscillations for a different reason.
Continuing the example developed in the previous section, we will now reduce the ESR of the 10 µF output capacitor to 50 mΩ, increasing the frequency of the zero to 320 kHz (Figure 17).
When the plot is analyzed for phase margin, no calculations are required to see that it is unstable.
The -90° phase shift from each of the two poles P1 and PL will produce a total phase shift of -180° at the 0 dB frequency.
For this system to be stable, a zero is needed that would provide positive phase shift before the 0 dB point. However, since the zero is at 320 kHz, it’s too far out to do any good (and is cancelled out by PPWR).
OUTPUT CAPACITOR SELECTION
Since the output capacitor is the user’s tool for compensating a monolithic LDO regulator, it must be selected very carefully. Most cases of oscillations in LDO applications are caused by the ESR of the output capacitor being too high or too low.
When selecting an output capacitor for an LDO, a solid Tantalum capacitor is always the best choice. Tests performed on an AVX 4.7 µF Tantalum showed an ESR of 1.3Ω @ 25°C, a value that is almost perfectly centered in the stable region (Figure 15).
Also very important, the ESR of the AVX capacitor varied less than 2:1 over the temperature range of -40°C to +125°C. Aluminum electrolytic capacitors are notorious for exhibiting an exponential increase in ESR at cold temperatures, and are not suitable for use as an LDO output capacitor.
It must be noted that large (≥ 1 µF) ceramic capacitors typically have very low ESR values (< 20 mΩ), and will cause most LDO regulators to oscillate if connected directly to the output. A ceramic capacitor can be used if some external resistance is added in series with it to increase the effective ESR. Large value ceramics also have a poor tempco (typically Z5U) which means the capacitance will drop in half as the temperature is increased or decreased to the operating limits.
LM317, LM340, LP2975: A User’s Guide To Compensating Low-Dropout Regulators
by Chester Simpson
National Semiconductor
The NPN Darlington pass transistor configuration requires that at least 1.5V to 2.5V be maintained from input-to-output for the device to stay in regulation. This minimum voltage “headroom” (called the dropout voltage) is given by:
VDROP = 2VBE + VSAT (NPN REG)
...
Feedback is used in all voltage regulators to hold the output voltage constant. The output voltage is sampled through a resistive divider (Figure 5), and that signal is fed back to one input of the error amplifier. Since the other input of the error amplifier is tied to a reference voltage, the error amplifier will supply current as required to the pass transistor to keep the regulated output at the correct DC voltage.
It is important to note that for a stable loop, negative feedback must be used. Negative feedback (sometimes called degenerative feedback) is opposite in polarity to the source signal (see Figure 6).
Because it is opposite in polarity with the source, negative feedback will always cause a response by the loop which opposes any change at the output. This means that if the output voltage tries to rise (or fall), the loop will respond to force it back to the nominal value.
Positive feedback occurs when the feedback signal has the same polarity as the source signal. In this case, the loop responds in the same direction as any change which occurs at the output. This is clearly unstable, since it does not cancel out changes in output voltage, but amplifies them.
It should be obvious that no one would intentionally design positive feedback into the loop of a linear regulator, but negative feedback becomes positive feedback if it experiences a phase shift of 180°.
...
POLES
A pole (Figure 8) is defined as a point where the slope of the gain curve changes by -20 dB/decade (with reference to the slope of the curve prior to the pole). Note that the effect is additive: each additional pole will increase the negative slope by the factor “n” X (-20 dB/decade), where “n” is the number of additional poles.
The phase shift introduced by a single pole is frequency dependent, varying from 0 to -90° (with a phase shift of -45° at the pole frequency). The most important point is that nearly all of the phase shift added by a pole (or zero) occurs within the frequency range one decade above and one decade below the pole (or zero) frequency.
NOTE: a single pole can add only -90° of total phase shift, so at least two poles are needed to reach -180° (which is where instability can occur).
ZEROES
A zero (Figure 9) is defined as a point where the gain changes by +20 dB/ decade (with respect to the slope prior to the zero). As before, the change in slope is additive with additional zeroes.
The phase shift introduced by a zero varies from 0 to +90°, with a +45° shift occurring at the frequency of the zero.
The most important thing to observe about a zero is that it is an “anti-pole”, which is to say its effects on gain and phase are exactly the opposite of a pole.
This is why zeroes are intentionally added to the feedback loops of LDO regulators: they can cancel out the effect of one of the poles that would cause instability if left uncompensated.
...
The plot of Phase Shift shows how the various poles and zeroes contribute their effect on the feedback signal. To produce this plot, the phase shift at each frequency point was calculated based upon summing the contributions of every pole and zero at that frequency. The phase shift at any frequency “f” which is caused by a pole located at frequency “fp” can be calculated from:
Pole Phase Shift = - arctan (f / fp)
The phase shift resulting from a zero located at frequency “fz” can be found using:
Zero Phase Shift = arctan (f / fz)
Is this loop stable? To answer that question, we need only know the phase shift at 0 dB (which is 1 MHz in this case). Finding this does not require complex calculations:
As stated in the previous sections, a pole or zero contributes nearly its full phase shift in the frequency range one decade above and below the center frequency of the pole (or zero). Therefore, The first two poles and the first zero contribute their full phase shifts of -180° and +90°, respectively, resulting in a net phase shift of -90°.
The final pole is exactly one decade below the 0 dB frequency. Using the formula for Pole Phase Shift, this pole will contribute -84° of phase shift @ 1 MHz. Added to the -90° from the two previous poles and the zero, the total phase shift is -174° (which means the phase margin is 6°). This loop would either oscillate or ring severely.
...
To reduce the negative phase shift (and prevent oscillations), a zero must be added to the loop. A zero can contribute as much as +90° of positive phase shift, which will cancel out the effects of one of the two low frequency poles.
All monolithic LDO regulators require that this zero be added to the loop, and they derive it from a characteristic that is inherent in the output capacitor: equivalent series resistance (usually referred to as ESR).
...
The ESR of the output capacitor puts a zero in the loop gain which can be used to reduce excess negative phase shift. The frequency where the zero occurs is directly related to the value of the ESR and amount of output capacitance:
FZERO = 1 / (2π X COUT X ESR)
Using the example in the previous section (Bode plot shown in Figure 12), we will assume that the value of COUT = 10 µF and the output capacitor ESR = 1Ω, which means a zero will occur at 16 kHz.
Figure 14 shows how this added zero will change the unstable plot into a stable one:
The bandwidth of the loop is increased so that the 0 dB crossover frequency moves from 30 kHz to 100 kHz.
The zero adds a total of +81° positive phase shift at 100 kHz (the 0 dB frequency). This will reduce the negative phase shift caused by the poles PL and P1. Since the pole PPWR is located at 500 kHz, it adds only -11° of phase shift at 100 kHz.
Summing all poles and zeroes, the total phase shift at 0 dB is now -110°. This corresponds to a phase margin of +70°, which is extremely stable.
This illustrates how an output capacitor with the correct value of ESR can generate a zero that stabilizes an LDO.
...
HIGH ESR
Using the example developed in the previous sections, we will change the conditions and assume the ESR of the 10 µF output capacitor is increased to 20Ω.This will decrease the frequency of the zero to 800 Hz (Figure 16). Reducing the frequency of the zero causes the loop bandwidth to increase, moving the 0dB crossover frequency from 100 kHz to 2 MHz.
This increased bandwidth means that the pole PPWR occurs at a gain value of +20 dB (compared to -10 dB in Figure 14).
Analyzing the plot (Figure 16) for phase margin, it can be assumed that the zero cancels out either P1 or PL. This means the loop has a two-pole response with the low frequency pole contributing -90° of phase shift and the high frequency pole PPWR contributing about -76° of phase shift.
Although this appears to leave a phase margin of 14° (which might be stable), bench test data shows that ESR values > 10Ω usually cause instability because of phase shifts contributed by other high-frequency poles which are not shown in this simplified model.
...
LOW ESR
An output capacitor with a very low ESR value can cause oscillations for a different reason.
Continuing the example developed in the previous section, we will now reduce the ESR of the 10 µF output capacitor to 50 mΩ, increasing the frequency of the zero to 320 kHz (Figure 17).
When the plot is analyzed for phase margin, no calculations are required to see that it is unstable.
The -90° phase shift from each of the two poles P1 and PL will produce a total phase shift of -180° at the 0 dB frequency.
For this system to be stable, a zero is needed that would provide positive phase shift before the 0 dB point. However, since the zero is at 320 kHz, it’s too far out to do any good (and is cancelled out by PPWR).
OUTPUT CAPACITOR SELECTION
Since the output capacitor is the user’s tool for compensating a monolithic LDO regulator, it must be selected very carefully. Most cases of oscillations in LDO applications are caused by the ESR of the output capacitor being too high or too low.
When selecting an output capacitor for an LDO, a solid Tantalum capacitor is always the best choice. Tests performed on an AVX 4.7 µF Tantalum showed an ESR of 1.3Ω @ 25°C, a value that is almost perfectly centered in the stable region (Figure 15).
Also very important, the ESR of the AVX capacitor varied less than 2:1 over the temperature range of -40°C to +125°C. Aluminum electrolytic capacitors are notorious for exhibiting an exponential increase in ESR at cold temperatures, and are not suitable for use as an LDO output capacitor.
It must be noted that large (≥ 1 µF) ceramic capacitors typically have very low ESR values (< 20 mΩ), and will cause most LDO regulators to oscillate if connected directly to the output. A ceramic capacitor can be used if some external resistance is added in series with it to increase the effective ESR. Large value ceramics also have a poor tempco (typically Z5U) which means the capacitance will drop in half as the temperature is increased or decreased to the operating limits.

