I am building an application for the TMS3705 transponder base station IC with a PCB. The datasheet specifies it needs 5 volts as input power in order for the system to work:

Example given by the datasheet

I want to supply this voltage via USB using my computer.

Should I input the 5 volts my laptop outputs directly to this circuit, or should I use a 5V voltage regulator?

I remember one time in an electronics course I had, the professor stated that, even if the input will be the same as the output, one should normally use a voltage regulator regardless.

This is what the USB connector and voltage regulator I plan to build looks like:

enter image description here

  • \$\begingroup\$ I think you should use an LDO regulator in place of AMS1117 ... You need ~ 6.5V as input voltage. You need something as this st.com/resource/en/datasheet/ld56020.pdf \$\endgroup\$
    – Antonio51
    Jun 7, 2022 at 20:41
  • \$\begingroup\$ The 5V coming from y our USB is already regulated. You don't need any of this extra stuff \$\endgroup\$
    – Kyle B
    Jun 8, 2022 at 5:58

2 Answers 2


the professor stated that, even if the input will be the same as the output, one should normally use a voltage regulator regardless.

Nope. A linear voltage regulator can't produce an output voltage higher than the input. In fact, output will always be lower, and the minimum input-to-output voltage where the regulator still regulates is called "dropout voltage". Here's a simplified schematic.

enter image description here (image link)

A low dropout regulator is basically a PMOS or PNP transistor acting as a controlled current source. The error amplifier adjusts current to keep the output voltage stable. Since the PMOS has non zero resistance when fully on, there is always a dropout voltage.

If input voltage is too low, for example 5V input on a 5V regulator, then the error amp will turn on the transistor fully in an attempt to bring the output voltage up. Since that's not possible in this case, it just sits there, being useless and wasting a bit of voltage.

There is a type of switching regulator that can produce an output voltage equal, lower or higher than the input: the buck-boost converter. That's what you'd use if you had a battery voltage going from 4.2V to 3V as it discharges, and you absolutely needed 3V3, for example. But in this case, it's not necessary, because the datasheet of your chip says:

enter image description here

This means with a normal USB port delivering close to 5V, you got quite a bit of margin to accomodate voltage drop in the cable, ferrite bead, etc.

What you do need is a decoupling cap on your board, to provide a low impedance power supply at high frequency, but your shematic already has those.

Note that your ferrite bead can ring with the low ESR ceramic capacitors and create a voltage spike when you will plug the USB cable. That can fry your chip. One solution is to add a low cost general purpose aluminium electrolytic cap on the right side of the ferrite bead, 100µF or so, in parallel with the ceramic decoupling caps.

Also, if you need a regulator for something else, AMS1117 is not the best choice. It has a very slow transient response, so if you use it with a load that draws chopped current, like a micro that goes to sleep, output voltage will not recover quick enough when the micro wake up.

  • \$\begingroup\$ About the solution provided regarding the ferrite bead ringing with the ESR cap, you mean I should place the 100µF or so capacitor in series, and at the right side, with the ferrite bead? I am sorry if it sounds like a stupid question, but I want to make sure I understood the problem here. \$\endgroup\$
    – CKiamy
    Jun 7, 2022 at 21:09
  • \$\begingroup\$ Electrolytic cap should be parallel with C16 on your schematic. You can reduce C16 to 10µF, remove regulator, then C17 is no longer necessary. \$\endgroup\$
    – bobflux
    Jun 7, 2022 at 21:30

You can't get 5V outof a linear regulator supplied with 5V.

There is always a voltage difference between input and output for a linear regulator.

The AMS1117 you specify in your schematic requires a difference of over 1 volt between input and output:

enter image description here

You need at least 6.1V in to get 5V out from the AMS1117.

The datasheet doesn't say what will happen if the input voltage is too low. You might actually get about 4V out of it.

It is almost certain that the output won't be regulated.

In this case, you can't use a regulator. You don't really need one, either.

Use the USB 5V to power your IC. The TMS3705 draws less than 10 mA, so you shouldn't have any trouble powering it from USB. Make sure to include decoupling capacitors.

  • \$\begingroup\$ I could not find an english version for the AMS1117 regulator, so thank you so much for providing it here! Since I already have the bypass/decoupling capacitors in my circuit, and taking into account your tips, ideally the circuit would only have the filter between the USB and the output, right? \$\endgroup\$
    – CKiamy
    Jun 7, 2022 at 21:04
  • \$\begingroup\$ @CKiamy Strange... I can literally put "AMS1117" into Google and the first result is the English datasheet. (And my device's locale is not English.) \$\endgroup\$
    – TypeIA
    Jun 7, 2022 at 21:21

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