7
\$\begingroup\$

I am looking at an older circuit, and I was wondering why the designer added these two resistors in series with the input voltage of this 5 V regulator. The regulator would be powering a system that pulls 30 mA maximum.

I think I had seen that some people add resistors to try to dissipate heat, but the specified resistors are 1/4 W 0805 resistors. Is that really what they are for?

The regulator is MCP1703-5002E/DB (datasheet).

Enter image description here

\$\endgroup\$
2
  • 6
    \$\begingroup\$ Run thermal calculations on that regulator package with and without those resistors at full load. Add your findings to the question ... or as an answer if it helps you. \$\endgroup\$
    – user16324
    Aug 13, 2020 at 18:04
  • 1
    \$\begingroup\$ 5V/12V = BEST CASE EFFICIENCY oops \$\endgroup\$ Aug 13, 2020 at 22:33

4 Answers 4

27
\$\begingroup\$

I was wondering why the designer added these two resistors in series with the input voltage of this 5v regulator.

There are obvious possible (and interconnected) reasons: -

  1. The two resistors will limit the output current that the regulator can provide to the load. This may be regarded as important by the designer. The current limit will be about 120 mA and this will also...
  2. Reduce the power dissipation that the regulator has to deal with. This may be an important consideration regards potential fault conditions in the load. The normal running current may be 30 mA but, under fault, current will rise.

Drop-out voltage from data sheet: -

enter image description here

So, at a load current of 120 mA the drop-out voltage is about 0.16 volts and the minimum voltage needed to run the regulator is 5.16 volts hence, the voltage across the resistors will be around 6.9 volts at a current of 115 mA.

There could be another reason too: -

If the load is fairly stable at around 100 mA there would be a guaranteed volt drop across the resistors of 6 volts and this means that if the input rail (called 12v_ISO) rose too high, the regulator would be somewhat better protected. The regulator has a maximum input voltage of 16 volts but the designer may be aware that this voltage might spike up to over 16 volts in certain situations.

A trick that the designer may have missed is not applying more input capacitance. With 100 nF in series with 60 ohms, this acts as a 27 kHz low pass filter but the main improvement to the device's PSRR is to be made at frequencies starting at a few hundred Hz: -

enter image description here

So, based on that I'd be making the input capacitor more like 10 uF resulting in a cut-off of 265 Hz and this will improve PSRR quite a bit. The data sheet tends to favour an input capacitor of 1 uF for most of its performance graphs so this might be another little thing that the designer missed.

Data sheet quote: -

For most applications (up to 100 mA), a 1 µF ceramic capacitor will be sufficient to ensure circuit stability. Larger values can be used to improve circuit AC performance.

\$\endgroup\$
2
  • 4
    \$\begingroup\$ OMG, There are so many designer tricks that I didn't know that I didn't know! :) \$\endgroup\$
    – tlfong01
    Aug 14, 2020 at 0:49
  • 2
    \$\begingroup\$ +1. Two ha'penny resistors are probably cheaper than a regulator in a better package. \$\endgroup\$
    – user16324
    Aug 15, 2020 at 11:57
10
\$\begingroup\$

There are a couple of reasons a designer might do this.

One is to spread out the power dissipation.

With linear regulators, the input current is approximately equal to the output current.

With the specified 30 mA flowing from 12 V to the 5 V output there is 0.03 A * 7 V = 210 mW power to be dissipated. Without resistors the entire 210 mW is dissipated by the regulator.

With resistors, they dissipate some, dropping the input voltage to the regulator, so it dissipates less.

The SOT-223-3 package is perfectly capable of dissipating 210 mW, so it’s unlikely this was the primary reason for the addition of the resistors.

Another reason is it’s a cheap way to get a little extra filtering. The 60 ohms resistor with the 0.1 µF capacitor yields a 26 kHz filter to help knock out high frequency crap from the 12v_ISO rail which may otherwise pass through the regulator relatively unattenuated. By the way, the 0.1 µF capacitor is a bit small – a capacitor is needed for regulator stability regardless of resistors. The Datasheet suggests something closer to 1 µF would be safer.

Lastly, even though the expected max draw is only 30 mA, the designer would like to make sure nothing blows up in an unexpected fail, like short-circuiting the output. The regulator has built-in short circuit protection at 400 mA. This may be more current than the designer would like to see occur. By adding 60 ohms, if more than about 110 mA flows, the voltage to the input of the regulator starts falling below its dropout voltage and it will shut down. At this maximum current, the power a 60 ohm resistor would have to dissipate is 0.11 A * 0.11 A * 60 Ω = 73 mW which maybe the designer wasn’t comfortable with. By putting in two 30 ohm resistors, each would dissipate only 36 mW.

\$\endgroup\$
2
  • \$\begingroup\$ So I am glad to meet another designer, or actually another engineer (making all sorts of enggr trade offs) this morning. \$\endgroup\$
    – tlfong01
    Aug 14, 2020 at 0:52
  • 2
    \$\begingroup\$ With 2 resistors to drop the voltage before the regulator, this is all about spreading the heat in a linear circuit. More and more regulators are SMD package with no huge metal tab to dissipate heat. At low current levels 2 resistors is space efficient.+1 \$\endgroup\$
    – user105652
    Aug 14, 2020 at 4:07
7
\$\begingroup\$

I agree with @down3db, this was likely done as a way to filter out noise from the 12V supply. The chosen LDO does not have amazing PSRR and so this may be a way to improve upon it by filter the noise before it gets to the LDO.

I disagree that the LDO will be getting very hot in this application, since the load is only 30mA the power dissipation will only be 210mW and with the thermal resistance of this device in the SOT-233 package being 62°C/W that means the junction temp will only rise 13°C above the ambient temperature which is pretty good. So this seems to eliminate the theory of trying to dissipate the power/heat through the resistors.

Another theory I saw in another answer is that the resistors could act as a current limit. While this is technically true that it would further limit the current, it could lead to some very non-ideal behavior. The startup could be compromised as the LDO, like all circuits, needs a minimum amount of voltage to turn on/operate but you also need to charge the output cap during start-up which means you'll draw extra current from the input supply. This could be a problem as the resistors will cause more I*R drop with that extra current and that could drop the input voltage below the minimum requirement, causing the LDO to turn off until the input charges back up (which gives the output cap time to discharge and you could get in nasty cycle of on/off behavior). So this may be why they did implemented it this way, but you should be very careful any time you increase the impedance of the source to a regulator as the system can become unstable.

(+10 years characterizing and debugging LDOs and the circuits people build with them.)

\$\endgroup\$
2
\$\begingroup\$

60 ohms is a very stiff pull-up.

Judging by the capacitor values, they were added "just in case" you have noise on the 12 V line (so that later you can come back and tweak the values), or to prevent instability (reduce the slew rate on the input ensure a nice stable start or switch).

But as a note: Using a linear regulator to drop 7 V will give you a very hot part that wastes a lot of power. You should step down with another stage (a buck) if you can first, because your overall efficiency will go way up.

\$\endgroup\$
8
  • 3
    \$\begingroup\$ "step down with another stage" - or maybe put some resistance in front of it? \$\endgroup\$ Aug 13, 2020 at 18:46
  • 1
    \$\begingroup\$ If you're a hobbyist, sure. \$\endgroup\$
    – down3db
    Aug 13, 2020 at 18:55
  • 4
    \$\begingroup\$ @AndrewMorton that doesnt actually reduce the power wasted and the heat produced though, it just spreads it around to more parts (which is useful, but not nearly as good as using a more efficient device like a buck converter) \$\endgroup\$
    – BeB00
    Aug 13, 2020 at 18:56
  • 2
    \$\begingroup\$ @BeB00 It could be that the low quiescent current of the part in question was an overriding factor, but we may never know. \$\endgroup\$ Aug 13, 2020 at 19:05
  • 3
    \$\begingroup\$ TI has some really nice tools that will give you a suggested circuit design, and you can set which thing you'd like to optimize for (size, weight, area, power, cost - SWAPC). ti.com/design-resources/design-tools-simulation/… Might want to give them a shot; they're pretty dang useful. \$\endgroup\$
    – down3db
    Aug 13, 2020 at 20:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.