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I have to drive a magnetic buzzer whose current is 50mA and max. voltage is 6V. I will use 7805 regulator to drive it as my supply voltage is variable 9V to 24V DC. There will be another small load connected on this 7805 whose current will be 100mA. So total current drawn from 7805 will be less than 200mA.

As the current is small so is it ok to use 7805 without the capacitors at its input and output pins? I want to avoid capacitors to reduce the component count to minimum.

If instead of the two capacitors I only use 1 capacitor then which one to use.. input or output capacitor? and what would be its sufficient value for a current of less than 200mA?

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  • \$\begingroup\$ It might work under some circumstances and then fail after testing. Why do you want to eliminate the capacitors? \$\endgroup\$ – Dean Franks Jun 21 at 5:07
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    \$\begingroup\$ If you want to reduce component count for the reason of space: if you use a TO220, you can solder 0805 SMD right on the legs of the component itself. \$\endgroup\$ – Huisman Jun 21 at 5:32
  • \$\begingroup\$ what are the voltage ratings of 0805 SMD capacitors? If my supply is 24V DC then I think at least 30V rating would be needed. \$\endgroup\$ – alt-rose Jun 21 at 9:42
  • \$\begingroup\$ If your total load is 200 mA (or even 150 mA) and your input voltage is 24 V, you will need a heatsink. \$\endgroup\$ – pericynthion Jun 21 at 15:46
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That has not really anything to do with the maximum current drawn. The 7805 just needs the 100nF at input AND at output to run stable. Without these capacitors there is a high probability of your output voltage to oscillate. Don't save on these two cheap components, in the end it will more likely cost you even more because of failed parts you have to replace.

Also be aware that the regulator will have to dissipate a maximum power of (24-5)V * 0.2A = 3.8W. So you definitly need to use a heat sink.

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  • \$\begingroup\$ how will the output oscillate if the input is a clean DC, kindly elaborate? And how can there be failed parts due to an absence of capacitor, also please clarify this! \$\endgroup\$ – Swagatam Majumdar Jun 21 at 5:25
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    \$\begingroup\$ Does clean DC truly exist in the Real world? \$\endgroup\$ – ijuneja Jun 21 at 6:01
  • \$\begingroup\$ @ijuneja Well, all of my audio amplifiers which are fed with battery power and are still picking up local AM radio station are saying no. \$\endgroup\$ – AndrejaKo Jun 21 at 6:02
  • \$\begingroup\$ if I can afford just 1 capacitor then which one is more important? My input supply voltage is fairly constant DC but with long few meter wires. \$\endgroup\$ – alt-rose Jun 21 at 6:15
  • \$\begingroup\$ As long as you don't have any large current transients I GUESS the output cap might be more important for stability. But if you are switching the load on and off the transient will most likely cause problems with long wires and without an input cap. \$\endgroup\$ – jusaca Jun 21 at 6:57
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Can a 7805 regulator be used without input/output capacitors

Short answer: YES ....but there are conditions

First you should read a datasheet for the regulator such as this one.

The datasheet clearly lays out:

enter image description here

  1. If your DC in supply is not too far away (distance/length not specified) you don't need a capacitor on the input. If your input supply is a small AC/DC walwart it will have a filter capacitor on its output. With a meter or so of cable you could probably do without the input capacitor on the regulator.

  2. You don't need an output capacitor BUT the transient response of the regulator will suffer. In your case you have a load that is quickly turning on/off (your buzzer, so it may well be advisable to add an output capacitor of at least 0.1uF.

  3. It's also worth noting that there is a MINIMUM load current specified for the regulator (5mA) so if your loads turn off then you should consider a resistor of 620 Ohms on the output for a 5volt regulator.

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The simple answer is, the capacitors are not required as long as the input is a clean DC and/or the load does not demand strict transient regulation.

According to the datasheet, the two capacitors are required for the following reasons:

C(in) is required if regulator is located at an appreciable distance from power supply filter. C(out) "improves" stability and transient response.

The above conditions tell us that the capacitor is optional in most cases, as long as the input is a clean DC. But since it's cheap and easy to add them, it's a good practice to include them.

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Capacitor is used for reducing voltage regulation. But if you have a good power source you may avoid the capacitors.

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    \$\begingroup\$ This is not correct. Please read the other answers. \$\endgroup\$ – Transistor Jun 21 at 6:45
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It needs the capacitors because that's the way the 7805 has been designed.

When making an IC, transistors and resistors are very easy to integrate, capacitors of any significant value are much, much harder, so manufacturers can't use them internally.

Now if a regulator circuit was designed to run entirely without capacitance, it would be heavily compromised on speed, regulation, all sorts of good things. So the 7805 is designed faster than that, fast enough to be an oscillator if there's insufficient capacitance on input and output.

Regulated rails often have capacitance on them anyway, so requiring a certain minimum capacitance is rarely seen as a problem, so there's little incentive for the manufacturers to provide products that need no capacitance.

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This is a typical connection for a 7805 IC Voltage regulator. Here, 7805 is used as a variable voltage regulator using the adjustable resistor \$R_2\$.

schematic

simulate this circuit – Schematic created using CircuitLab

The input capacitor is of \$0.1\text{ }\mu\text{F}\$ (MLCC i.e. ceramic disc) or \$1.0 \text{ }\mu \text{F}\$ solid tantalum capacitor. This capacitor \$C_1\$ is used for reducing the sensitivity of the voltage regulator (VR) to the line impedance. It is especially used when the system you're working with is distant enough from the supply. Also, note that the the ceramic capacitor is chosen since it has very low ESL (Equivalent series Inductance) and has a very high frequency response i.e. high frequency ripple rejection upto \$10 \text{ } \text{MHz}\$. It is also common to use Ferrite beads (nonconductive ceramics made of oxides of Ni, Zn, Mn or others) after the decoupling capacitor \$C_1\$. The Ferrite beads at low frequency behave as an inductor and hence they behave as low pass filters and at high frequencies,they are resistive. They are optional but they provide the desired characteristic of high frequency noise isolation and decoupling. If desired, another possibility is that the use of another capacitor in parallel with \$C_1\$ (say \$C_2\$) which is of value say \$1.0 \text{ } \mu \text{F}\$ to \$10.0 \text{ } \mu \text{F}\$ for low frequency decoupling. Solid tantalum capacitor with the former value or Aluminum electrolytic capacitor (switching type not general purpose) of value \$25 \text{ } \mu \text{F}\$ shall be used. Most commonly aluminum electrolytic capacitors are preferred but solid tantalum has replaced them because of their high capacitance-to-volume ratios, higher frequency range and low ESR. But they are expensive and are limited to \$50 \text{ V}\$ and with capacitance of \$ 500 \text{ } \mu \text{F}\$ or less. But a better designed aluminum electrolytic capacitor using organic or polymer electrolytes are more promising. The \$C_\text{adj}\$ (here the 7805's ground is acting as an adjustment pin i.e. like LM317) bypasses the "adjustment terminal" to ground to improve ripple rejection. This prevents the ripple from being amplified if the input voltage is increased. The diode \$D_2\$ discharges the capacitor \$C_\text{adj}\$ if the output is shorted to ground. The diode \$D_1\$ provides the inverse voltage protection (i.e. polarity). The capacitor \$C_\text{adj}\$ of \$10 \text{ } \mu\text{F}\$ improves the ripple rejection by about \$15 \text{ dB}\$ at \$120\text{ Hz}\$ for \$10 \text{ V}\$ applications. Also, another one thing is that \$C_1\$ can also be X7R type owing to its less capacitance change as a function of dc bias voltage, high capacitance change, high K dielectric etc. But high K dielectric formulations are unstable so as Z5U and Y5V which don't have the advantages of the X7R. NP0 (COG) is another possibility which has a stable less K value of dielectric (leading to (unfortunately) a low capacitance range) and nominally 0 temperature coefficient which is highly desired. They ranges are from \$ 0.01 \text{ }\mu\text{F}\$ to \$ 0.1 \text{ }\mu\text{F}\$. Though the output capacitance is not generally required, any external feedback circuit or any external capacitor may cause excessive ringing which can be eliminated by \$C_3\$ of \$ 1 \text{ }\mu\text{F}\$ tantalum or \$ 25 \text{ }\mu\text{F}\$ aluminum electrolytic. \$D_1\$ protects \$C_3\$ from input short-circuit, \$D_2\$ prevents \$C_\text{adj}\$ from output short-circuit and \$D_1\$ and \$D_2\$ both protect \$C_\text{adj}\$ from input short-circuit. \$\$ The following image shows the complete protection circuit for laboratory power supply. [Image Credit: OnSemi] ‘‘Laboratory’’ Power Supply with Adjustable Current Limit and Output Voltage

Also, there is no association of current drawn to the input and output capacitors. Aluminum electrolytic capacitors have a leakage current of tens of \$\mu\text{A}\$ (which is high compared to others but is relatively small considering the circuit in operation). And as denoted in the schematic (first) the \$C_3\$ value is not as stated in it. I couldn't edit the schematic right now. I will correct it later. Sorry for this error.

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