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I'm trying to further my understanding of electronics, so I decided to try to design a fixed voltage regulator capable of supplying an amp or so. I put this together from first principles without referring to any kind of reference on how voltage regulators are usually designed.

My thoughts were:

  • Zener and resistor to provide a fixed voltage reference.
  • Comparator to detect when the output voltage was above the target threshold.
  • Transistor to switch the supply on and off.
  • Capacitor to act as a reservoir.

With that in mind, I designed this fixed 5V regulator, which appears to work:

Voltage regulator design

What I did notice, however, is that it has certain limitations which I can't quite derive the cause of:

  • The current from V1 (input) roughly equals the current at R2 (output), despite differing voltages. This seems to match the behaviour of linear voltage regulators (is that what I just created?) but I'm not sure why it happens. Why is so much power dissipated from Q2 considering it's just switching on and off?
  • When V1 is less than about 7.5V, the output voltage never hits the 5V threshold, but instead hovers around 4V. I have tried this with varying loads but it simply does not function below that input voltage. What is the cause of this?
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  • \$\begingroup\$ The existing answers already address the reasons for what you're seeing. Try introducing a little positive feedback around your opamp 'comparator' to force it to behave a little more like a switcher - just as an exercise ... \$\endgroup\$ – brhans Mar 1 '17 at 14:19
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    \$\begingroup\$ "Comparator to detect ..." - There is no comparator in your circuit, just an op-amp. If you replace it with an actual comparator, you might see different (not necessarily better) behaviour. \$\endgroup\$ – marcelm Mar 1 '17 at 15:10
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    \$\begingroup\$ Note that even if the transistor was only ever fully on or fully off, it would still be a linear regulator - you'd just be using the resistance of the wires instead of making the transistor have resistance. \$\endgroup\$ – immibis Mar 2 '17 at 3:42
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I put this together from first principles without referring to any kind of reference on how voltage regulators are usually designed.

Not a good start, but you've actually ended up with almost the exact design of most linear regulators. But the "first principle" you've forgotten about is the MOSFET linear region. Have you tried this thing in a simulator? The system will settle at a point where the transistor is half-on, dissipating power as a resistor.

When V1 is less than about 7.5V, the output voltage never hits the 5V threshold, but instead hovers around 4V. I have tried this with varying loads but it simply does not function below that input voltage. What is the cause of this?

This is called the "dropout voltage". It's due to limitations in how close to the input rails the opamp is capable of driving; you lose approximately 0.7V in the output transistor of the opamp and another 0.7V because of the threshold voltage of the MOSFET.

You might be able to do better with a better op-amp than the ancient, obsolete 741. Otherwise, you're trying to design what's called an LDO: low-dropout regulator.

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  • \$\begingroup\$ facepalm - these are all things I knew, but failed to apply in context. Thank you. \$\endgroup\$ – Polynomial Mar 1 '17 at 14:03
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    \$\begingroup\$ I should mention that this was purely designed in a simulator, and yes, that's exactly what happens. I'm not quite mad enough to put something together like this with real parts without referring to a reference. \$\endgroup\$ – Polynomial Mar 1 '17 at 14:04
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    \$\begingroup\$ A linear regulator is basically a smart resistor - the transistor plays the part of the resistor here. \$\endgroup\$ – Ecnerwal Mar 1 '17 at 14:17
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    \$\begingroup\$ Why is it not a good start? (assuming this is a hobby/learning project not for production) \$\endgroup\$ – immibis Mar 2 '17 at 20:23
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Why is so much power dissipated from Q2 considering it's just switching on and off?

Because it isn't a switching regulator circuit - it's a linear regulator that you have designed.

The current from V1 (input) roughly equals the current at R2 (output), despite differing voltages. This seems to match the behaviour of linear voltage regulators (is that what I just created?)

Yes, you have.

When V1 is less than about 7.5V, the output voltage never hits the 5V threshold

You need about a couple of volts on the gate (with respect to source) to begin turning on the MOSFET. This has to come from the op-amp and it probably "loses" about a volt on its output compared to the incoming power rail. So, if you want an output voltage of 5 volts then you need an input supply of about 8 volts and that will be on light loads.

On heavy loads, the gate-source voltage might need to be 3 or 4 volts. Now you will probably need an incoming supply that is about 10 volts to keep the regulator output at 5 volts.

Have respect for the simple regulator, especially those that are low drop-out types!!

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  • \$\begingroup\$ Additionally the zener current is very low even at 10v its only 5ma, device is specified at closer to 50ma. Zener voltage will drop with lower reverse currents. If you are expecting such a wide range I'd use a voltage reference device instead. \$\endgroup\$ – Trevor_G Mar 1 '17 at 14:22
  • \$\begingroup\$ "Have respect for the simple regulator" - indeed! I really didn't appreciate how much engineering goes into the humble LDO! \$\endgroup\$ – Polynomial Mar 1 '17 at 14:30
  • \$\begingroup\$ Yup, there's a lot of engineering. We've not even begun to talk about stability, PSRR or noise here. \$\endgroup\$ – pjc50 Mar 1 '17 at 14:49
  • \$\begingroup\$ You might try a P_channel power MOSFET. Since that runs in INVERTING_MODE, compared to how the N_channel IRFP054 is used, you'll need to flip the inputs of the OpAmp. \$\endgroup\$ – analogsystemsrf Mar 1 '17 at 16:50
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    \$\begingroup\$ It's perhaps worth noting that even if the MOSFET were to be used as a switch rather than in its linear region, it would still have to dissipate significant heat, because you'd be trying to charge a capacitor from a voltage source, which can never be more than 50% efficient. \$\endgroup\$ – pericynthion Mar 1 '17 at 17:46
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Design is OK except the dropout of the FET LDO can be lower than the BJT LDO, but FET compensation may demand a limited range ESR for stability and allow some ripple for feedback.

You can make it up to 98% efficient by the good choice of inductor with a low RDSOn switch and low DCR choke. Now you have a buck regulator. Simulation here

enter image description here

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  • \$\begingroup\$ This is a really old answer, but I'm not really convinced this is a buck regulator. It only has one switching element, and the transistor still dissipates significant amounts of power. \$\endgroup\$ – Hearth Oct 18 '18 at 14:18
  • \$\begingroup\$ @Felthry Why doubt my simulation, check the Zener Vz with mouse, add Tranny to scope, change scope to display Watts max,min for Vce,Ice, note the variable input triangle input V and pulsed load from 0.7 to 1.9A then change NPN to NFET (delete, draw FET) change gm to 1 to 5 and add to Scope , change to Watts min,max, add DCR to L , drag corner of part with shift or ^ key? to rubber band mode to stretch shrink or rotate. Prove it works or how much better you can make it. Change cap to add low ESR then add 0.1uF with lower ESR. \$\endgroup\$ – Sunnyskyguy EE75 Oct 18 '18 at 16:26
  • \$\begingroup\$ Well for one thing I can see just hovering over the transistor that it's dissipating upwards of 20 watts in short bursts and regularly dissipating multiple watts, which shouldn't happen in a switching converter. Strangely, you can't graph power dissipation in transistors on the falstad simulator. \$\endgroup\$ – Hearth Oct 19 '18 at 13:03
  • \$\begingroup\$ You can see Watts in scope scale , but plot Power in FETs , here with PFET tweaked for 90% efficiency 125W load pulse full step 50% with 2V ripple input and 5 mV ripple output. tinyurl.com/ya5gyufe. Some parts include ESR, FET choice is important. @Felthry \$\endgroup\$ – Sunnyskyguy EE75 Oct 19 '18 at 18:58
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The power is dumped in the transistor because it is the series element, so all the current for the load has to go through it, whilst at the same time it has to drop the difference between the input voltage and the output voltage.

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What is the cause of this?

With the supply to thee opamp at v1, the max output voltage on the opamp and thee MOSFETs gate is v1. The MOSFET will need some vgs to work, sand that's typically 2 to 5v, depending on the MOSFET used. 0.7v for bits and 1.3v for Darlington.

That means the max the MOSFETs source can see is v1 - 2 to 5v. That's exactly what you saw.

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