What is the function of the R1 resistor connected in series with the capacitor in the input power supply circuit?

安圣基

What is the function of the R1 resistor connected in series with the capacitor in the input power supply circuit? [Copy link]

 

What is the function of the R1 resistor connected in series with the capacitor in the input power supply circuit?

chunyang

Suppress the charging surge of the filter capacitor at the moment of power-on.

led2015

The resistor R1 may play a filtering role in the input power circuit, which can effectively filter out high-frequency noise and improve the quality of the input signal.

安圣基

Doesn't the inductor connected in series also play a role in suppressing charging surge current?

dwdsp

Would it be better to use NTC instead?

bobde163

Adding a resistor here is equivalent to increasing the ESR of the capacitor. I think it should be used to suppress the capacitor from absorbing large current when powered on. After stable operation, it can absorb some high-frequency glitches.

tagetage

Was this circuit diagram drawn by a foreigner?

Fastron is a German company that mainly manufactures inductors. This R1 should be an inductor with a DC resistance of 1 ohm.

安圣基

chunyang posted on 2023-3-29 22:25 Suppress the charging surge of the filter capacitor at the moment of power on.

I think the best explanation is to prevent the power-on surge current. But I don't understand why the inductor has been added before, which can play a role in limiting the surge current.

chunyang

An Shengji posted on 2023-3-30 13:29 I think the best explanation should be to prevent power-on surge current. But I don’t understand why the inductor has been added before, which can play a role in limiting the surge current...

Pay attention to the inductance. The ability of the inductor to suppress surges is positively correlated with the inductance.

Whether R1 is necessary requires analysis of the previous circuit. In most cases, it is not necessary, after all, the voltage is not high and the capacity of the filter capacitor is not large. There are many technical means to suppress surges now. Series resistance or inductance is the most primitive method, but either the negative impact is too great, or the cost and space occupancy rate are too high, so it is rarely used. If you want to use inductors to suppress power-on surges, the inductance must be carefully calculated. At present, the main occasion for using inductors to suppress surges is to deal with transient electromagnetic induction, such as the induction of lightning strikes, but multiple means are still required, and inductance alone is not enough.

chunyang

led2015 Published on 2023-3-29 23:30 The R1 resistor may play a filtering role in the input power circuit, which can effectively filter out high-frequency noise and improve the quality of the input signal.

Let’s talk about how R1 “plays a filtering role”.

chunyang

dwdsp posted on 2023-3-30 08:27 Is it better to use NTC instead?

The increase of NTC impedance depends on the rise of temperature, and this hysteresis can be very large, so it must be carefully analyzed whether it can be used. Moreover, if the impedance of NTC is the same as R1, then what is the point of using NTC? If the value is less than R1, then transient analysis must be performed.

It can be said that, as far as the original poster's circuit is concerned, NTC is not better, but worse, because its characteristics are not fully utilized and the cost is greatly increased.

chunyang

bobde163 posted on 2023-3-30 10:37 Adding resistance here is equivalent to increasing the ESR of the capacitor. I think it should be used to suppress the capacitor from absorbing large current when powered on. After stable operation, it can absorb a...

Let's talk about how using resistors in this way can "absorb some high-frequency glitches."

chunyang

tagetage posted on 2023-3-30 12:39 Was this circuit diagram drawn by a foreigner? ? ? Fastron is a German company that mainly produces inductors. This R1 should be a DC resistor...

Inductance would never be expressed this way.

tagetage

chunyang posted on 2023-3-30 14:02 Inductors will never be expressed like this.

You can try to analyze, if R1 is an inductor with a DC resistance of 1 ohm, is it possible? ? ? Is it reasonable? ? ?

tagetage

C1 is a tantalum capacitor.

Tantalum capacitors

2. DC leakage current

1) Ripple current & surge current

The main mechanism of surge current and ripple current destruction is to cause overheating of the device, resulting in device burnout. The relationship between power loss (P) and ripple current (Irms) is expressed by the following formula: P = V-·I leak + Irms2·R ≈ Irms2·Rs,

Where: V-: DC bias voltage (V);

Ileakage: leakage current (A);

Rs: equivalent series resistance (Ω);

Irms: ripple current.

It can be seen from the above formula that when Rs increases or when Irms increases, the power loss increases. Therefore, in high-frequency circuits, it is required that the ripple current passing through the tantalum electrolytic capacitor is small and a tantalum electrolytic capacitor with a small equivalent series resistance is selected. In all applications, it should be noted that the ripple current and surge current in the circuit should not be too large. Generally speaking, the surge current and ripple current are not written in the tantalum capacitor specification book. It is necessary to consult the supplier, or use I=U/R for a rough estimate, where U is the rated voltage and surge voltage. R is the equivalent series impedance. The national military standard and suppliers generally recommend connecting a resistor in series in a low-impedance circuit to reduce the impact of its surge current (generally 1V/ohm or 3V/ohm is recommended in series)

maychang

tagetage posted on 2023-3-30 19:14 You can try to analyze, if the DC resistance of R1 is an inductor of 1 ohm, is it possible? ? ? Is it reasonable or not? ? ?

[If R1 is an inductor with a DC resistance of 1 ohm, is it possible? Is it reasonable or not? ? ? ]

Rice is sold by kilogram, and cloth is sold by foot. If you go to a cloth shop and buy eight kilograms of cloth, do you think it is reasonable?

bobde163

chunyang posted on 2023-3-30 14:02 Let's talk about how the resistor is used in this way to "absorb some high-frequency glitches".

Thanks for the reply from the version. I carefully analyzed it from a theoretical perspective. Combined with the device parameters of the above circuit, the main function of this resistor is to suppress the capacitor voltage overshoot when powered on. I used simulation software to verify it, and also built an actual circuit to capture the waveform. Adding this resistor in series with the large-capacity C1 capacitor is equivalent to increasing the ESR, which can effectively suppress the capacitor voltage overshoot when powered on. Second, after the circuit enters steady-state operation, if there is high-frequency interference at the input end, as shown in the above-mentioned inductance and capacitance parameters, the low-pass filter itself can filter most of the high-frequency interference. The role of the series resistor here is very small, but since it is a resistor device, it will consume energy as long as there is current flowing through it. Although the absorbed energy is very little compared to the inductance and capacitance, it still has a little inhibitory effect. The above is my own understanding based on theory and practical conclusions. If there are any errors, please correct them.

chunyang

bobde163 posted on 2023-3-31 10:38 Thank you for your reply. I carefully analyzed it from a theoretical perspective. Combined with the device parameters of the above circuit, the main function of this resistor is to suppress the power supply when it is powered on...

Connecting resistors in this way will not reduce high-frequency interference. Not only will it not be filtered out, it will also reduce the filtering effect of the capacitor. The capacitive reactance of a capacitor is a function of frequency. The higher the frequency, the lower the capacitive reactance. This is the principle of high-frequency decoupling of capacitors. When the influence of distributed parameters is not considered, the impedance of the resistor is a constant and has nothing to do with the frequency. Therefore, connecting it in series to the capacitor loop will only increase the loop impedance, and the filtering/decoupling effect will naturally deteriorate.

As for voltage overshoot, this is also wrong. Note that the previous power supply is a regulated power supply. Moreover, the fluctuation of power supply output is normal, as long as it is within the range required by the load, there is no need to deal with it specifically. What you should pay attention to is the transient change of current.

chunyang

tagetage published on 2023-3-30 19:21 C1 is a tantalum capacitor. It is the key point of tantalum capacitor 2. DC leakage current 1) Ripple current & surge current Surge current and ripple current destroy the machine...

Well said, ignore that C1 is tantalum electrolytic.

If the ripple of the front-stage power supply is too large, the tantalum electrolytic will have a large loss and there will be reliability risks. However, more solutions are to use it in parallel with the aluminum electrolytic to control the ripple within the level that the tantalum electrolytic can tolerate.

The front stage of the circuit in the original poster's post comes from a switching type voltage regulator, which has large ripple and high frequency. The original poster's picture shows an additional filter circuit, which uses the good high-frequency characteristics of tantalum electrolytics to reduce ripple. R1 is used to protect tantalum electrolytics. In addition to enhancing high-frequency decoupling, C2 can also reduce voltage fluctuations on tantalum electrolytics, further ensuring reliability.

However, it is necessary to examine whether the circuit posted by the OP is necessary, because the information in the OP's post is incomplete and it is impossible to know the output noise characteristics of the previous power supply.

bobde163

chunyang posted on 2023-3-31 13:26 Connecting the resistor in this way cannot reduce high-frequency interference. Not only can it not be filtered out, but it will also reduce the filtering effect of the capacitor. The capacitive reactance of the capacitor is a function of frequency, and the frequency...

Thanks for the version correction. Regarding your first point about filtering high frequencies, I indeed overlooked the position of the series resistor. The resistor position in the figure will reduce the effect of high-frequency filtering. The filtering effect is relatively poor at high frequencies below 20K. Regarding the second point, this resistor is added to suppress the output voltage overshoot at power-on. My analysis is this: When the circuit is powered on in the initial state, it is assumed that the inductor current and capacitor are analyzed first. The inductor current is 0, the capacitor voltage is 0, and the front-stage power supply charges the capacitor through the inductor. The charging current starts from 0 and slowly increases. When the capacitor voltage is charged to the input power supply voltage, the voltage across the inductor drops to 0, and the current reaches the maximum value. At this time, a reverse electromotive force will be generated on the inductor and superimposed on the input voltage. Continue to charge the capacitor, the capacitor voltage will overshoot, but in reality, real inductors and capacitors are not ideal devices, and they still have a certain ESR. In addition, the electrolytic capacitors used in the figure have a relatively large ESR, so there is an actual overshoot, but it may not be obvious. If this resistor is added, theoretically, the overshoot voltage value can be further reduced, allowing the output voltage to rise more slowly. Another point is that if the front-stage power supply is a voltage-stabilized power supply with good transient response, and has the ability to provide a large current to the series inductor in time when the circuit is powered on, then the output is more likely to have a voltage overshoot. Therefore, I think adding this resistor can suppress the output voltage overshoot until a certain power-on time.