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?

  • \$\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 '19 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 '19 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 '19 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 '19 at 15:46

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.

  • \$\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 '19 at 5:25
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    \$\begingroup\$ Does clean DC truly exist in the Real world? \$\endgroup\$ – ijuneja Jun 21 '19 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 '19 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 '19 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 '19 at 6:57

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.


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.


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 '19 at 6:45

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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