4
\$\begingroup\$

So I have a 48 V supply and want 5v output at sub 100 mA. The 78L05 is a super low cost 3-pin voltage regulator, but has a maximum input voltage of 30 V.

I was wondering if it is a cost effective / reasonable idea to supply this regulator with some voltage (say 20 V) from a simple zener/transistor series or shunt regulator. So we have 48 V into a simple regulator with all of its temperature sensitive warts and poor regulation, cascaded into a well behaved, low-power 3-pin IC regulator to give a nice stable 5 V.

Edit: Linear regulation down this much of a voltage drop is wasteful, however, this 48 V circuit is only activated for very short epochs at infrequent intervals.

Just for illustration from this source.

Fig 1. Series reg

series reg

Fig 2. Shunt reg

enter image description here

Question: Should I do this? If I should do this, then what input voltage should I give the 2nd stage regulator for best efficiency, and would shunt or series first stage regulation be preferred and why?

\$\endgroup\$
7
  • \$\begingroup\$ I wasn't going to actually write an answer. Just a comment. But John's answer is pretty much what I wanted to say in a comment. Yes, the use of a zener+BJT to reduce the voltage for the linear IC regulator will work fine -- consistent with dissipation issues you must study, too. If this is only 100 mA then you are talking about 5 watts dissipation (about) out of which 1/2 watt is being used. You need to find ways to throw away all that excess dissipation. But you can spread it around a bit. (Use 2 or 3 resistors instead of one, etc.) \$\endgroup\$ Sep 2, 2023 at 23:32
  • 3
    \$\begingroup\$ Very wasteful. I hope you have a good reason to make a regulator with less than 10% efficiency. \$\endgroup\$ Sep 2, 2023 at 23:34
  • 1
    \$\begingroup\$ It is incredibly wasteful, but its part of a circuit that is only pulsed on for a fraction of a second once or twice per minute. A switching regulator is overkill here \$\endgroup\$
    – learnvst
    Sep 2, 2023 at 23:40
  • 2
    \$\begingroup\$ Then you definitely want a series regulator, and maybe you can do with minimal heat sinking. \$\endgroup\$ Sep 2, 2023 at 23:45
  • \$\begingroup\$ I probably should have added that critical info to question (will edit) \$\endgroup\$
    – learnvst
    Sep 2, 2023 at 23:46

7 Answers 7

7
\$\begingroup\$

Below, everything in the blue box is an active zener diode. I use it to dispose of 25V or so of the total supply voltage, leaving 23V for the downstream regulator to deal with:

schematic

simulate this circuit – Schematic created using CircuitLab

This boxed area develops ("drops") about \$24V + 0.7V\$, and only the remaining \$48V-25V=23V\$ appears at OUT. You could use any zener diode you like, up to say 39V, leaving a little over 8V for the 7805, which would minimise the power dissipation there.

With no load at all, the output will rise to 48V. Even an unloaded 7805 at the output should prevent that, but just in case, I include R2 to draw a tiny current.

Power dissipated by the dropping element (the transistor here) is 2.5W (at 100mA load), which is a little high for a lone zener diode, and is why I suggest this active equivalent.


While you could use a single 3W 24V zener diode to do the same job, how about three 2W 12V diodes in series:

schematic

simulate this circuit

The remaining 12V is for the 7805, which would dissipate \$100mA \times (12V-5V) = 0.7W\$. No heatsinks necessary.


Having so much voltage to dispose of is usually considered to be problematic, but we can take advantage of it, by employing some series resistance.

Presumably you'll have some capacitance at the 7805 input, which will cause quite a bit of inrush current. We can mitigate that with series resistance, as long as we choose a value that won't drop too much voltage under full load. Let's say we can afford to lose another 3V:

$$ R = \frac{V}{I} = \frac{3V}{100mA} = 30\Omega $$

schematic

simulate this circuit

I used 27Ω for R3, close enough. This resistor limits inrush current to a maximum of:

$$ I_{INRUSH} = \frac{48V-3\times 12V}{27\Omega} = 440mA $$

The 7805 and its cousins have terrible supply noise rejection at frequencies in the kilohertz, but with R3 and C1 in place, we have a low-pass filter, which will attenuate input noise upwards of \$\frac{1}{2\pi \times 27\Omega \times 10\mu F} = 590Hz \$.


All the above solutions assume that the 48V is stable and will never droop significantly below 48V. Their problem is that they all develop a fixed voltage across some "pass element", regardless of the supply voltage, and any supply droop will appear directly at the 7805 input.

The solution to that is a shunt regulator, or some derivative that regulates its output with respect to ground (0V), instead of the positive supply. Below left is the classic (and terribly inefficient) zener shunt design:

schematic

simulate this circuit

Both of those circuits will work, but the zener shunt design on the left will always pass well over 100mA, whether that current is demanded by the load or not. It also means that when there's no load, the poor zener diode must bear all of that current.

The transistor offloads the current sourcing responsibility of R1, which means we can reduce the quiescent current flowing in R1 and D1, but retain the benefits of having a ground-based reference. Now it's transistor Q1 that dissipates all the power, up to 3.7W, but only under load.

Don't forget that the output will be at least one \$V_{BE}\$ (0.7V) below the 12V reference potential at Q1's base.

Since the zener diode will remove much of the noise present on the 48V supply, the output of the 7805 following this stage should be lovely and smooth. You could even use the series resistor from the previous idea, to improve higher frequency noise rejection further still.

This last circuit would be my personal choice of solutions.

\$\endgroup\$
1
  • \$\begingroup\$ The series zeners work a charm with the series resistor alone. I'll experiment combining this with the series transistor as the ground based ref is attractive. \$\endgroup\$
    – learnvst
    Sep 5, 2023 at 0:16
7
\$\begingroup\$

I'd just throw a 20V, 3W (or more) Zener in series with the 78L05 input to drop it down some and call it a day. By having the Zener controlling the base of a pass element, you're just shifting the dissipation from a Zener to a transistor. Something still has to dissipate all that power.

A 78L05 will have to dissipate 2.5W - that's not feasible for that part IIRC. Use a 7805 instead - at least it'll be able to shed that power. They are available in surface-mount power packages that dissipate heat well to the ground plane they'd be mounted on. And dissipate heat they'll need.

The Zener should also be soldered to large copper areas to dissipate heat through the pins into the board.

\$\endgroup\$
2
  • \$\begingroup\$ Could still use a 78L05 if a higher value series zenner was used though, right? \$\endgroup\$
    – learnvst
    Sep 3, 2023 at 9:42
  • 1
    \$\begingroup\$ @learnvst If that 48V source supply is not constant, and/or has ripple, then a high-voltage "zener" method as outlined by Kuba still risks overheating a 78L05. The larger-package 7805 is a safer approach. Your circuit #1 (Fig,1) is OK if you use a big transistor, an 8.2V zener, and Rs~8k ohm. The transistor will likely need a heatsink. \$\endgroup\$
    – glen_geek
    Sep 3, 2023 at 13:09
6
\$\begingroup\$

The preregulator (transistor + zener or simply a big series zener) may prevent the 7805 getting a lethal input of more than 30 volts, but the whole idea is poor energy-wise. A linear series regulation system (single or cascaded) will dissipate current of approximately (Vin - Vout) * Load which in your case means 43V * 100 mA i.e. 4.3 watts. That needs some real heatsink to keep it cool.

The parallel regulator scheme does not give any efficiency advantage. The opposite: it will dissipate continuously. A series regulator has high dissipation only when the load takes much current.

If you can afford nearly 90% of the power to get dissipated in the regulator and you can keep it cool, it is of course a working solution. But there's a caveat: Even short high current peaks can destroy transistors if the design skips their existence. Beware charging capacitors and going out of the safe Vce vs. Ic operating area.

A simple buck regulator would be much more effective, but it also has disadvantages like needing filters to stop noise emissions.

\$\endgroup\$
1
\$\begingroup\$

If your 48V supply is nicely regulated and guaranteed free from major transients maybe just consider putting a 24V 1W zener diode in series with a TO252 or TO220 78M05. Both are cheap as chips, and may well be able to handle your brief power impulses effortlessly (but do run the numbers on transient power dissipation of both).

\$\endgroup\$
1
\$\begingroup\$

As others have said, at your power levels it makes more sense to use a series zener diode. However, your approach has an advantage at higher power levels.

The pre-regulator (that's the name for your application genre) can be designed to dissipate a significant amount of power, removing the thermal load from the downstream regulator. Besides the voltage drop, this can be done to move some of the heat to a more convenient cooling location, to reduce temperature coefficient errors, reduce cost (an extra power transistor is much cheaper than a high-power 3-terminal regulator), increase reliability, etc.

I prefer the first schematic approach over the second. In terms of efficiency, shunt regulators are the worst. In some applications they can present a lower output impedance than a series regulator, and they do prevent load current steps and transients from going upstream and potentially corrupting other power systems. Waaay back I ran into this in Ampex video recorders. Super-clean power, but tons of heat.

IMO in your application they are all cons and no pros.

\$\endgroup\$
1
\$\begingroup\$

If your 48V supply is stable then this is possibly the simplest and cheapest solution available, and as effective as any of the alternatives suggested in your application.

Add a series input resistor that will drop voltage to about 8 volts at the absolute maximum current that will be required.
Assume Imax = 100 mA.

Resistor maximum dissipation = V x I = (48-8) x 0.1 = 4 watts.
Use a 10 watt air cooled resistor.

Add a 8V (or similar) Zener across the regulator input.
The zener will carry all the current worst case.
Zener dissipation = V x I = 8 x 0.1 = 0.8 watt.
Use a 2+ Watt zener.

The regulator will see 8V input consistently.
Regulator maximum dissspatio = V x I = (8 - 5) x 0.1 = 0.3 Watt.
You can change Vin_regulator to closer to Vout for even less dissipation.
Look at the worst case dropout in the datasheet.
For even lower regulator dissipation use a suitable LDO (low dropout) voltage regulator and a Vin_regulator closer to Vout_regulator.

Using an under-rated series resistor (as above) to drop excess voltage and absorb excess energy is likely to be more reliable than using semiconductor solutions.
The zener works at low voltage so needs lowish power dissipation rating.
Keeping power disspation low in all components relative to ratings helps reliability.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
0
\$\begingroup\$

48V to 5V at 100mA is 4.3W of heat to get rid of. So you need a 5°C/W heat sink.

Maybe something cheap like this one, check datasheet... oops, the 6.2°C/W requires a fan blowing on it. Maybe this one... same problem.

This one should work, it's pretty large and heavy. If you factor in all the costs, a bigger enclosure to fit it (with airflow), strong board standoffs so the board doesn't crack under the weight of the heat sink when the device is dropped, etc...

It will end up much more expensive, larger and heavier than one of the readymade DC-DC converters that will do the job in less than 1 cubic centimeter.

Now if the device is powered constantly, these 4.3W of wasted power in the linear solution add up to 38 kWh annually. At 30c/kWh this is about 11€ per year of electricity, so the buck converter pays for itself.

\$\endgroup\$

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.