A power supply must be quick enough to see the output voltage drop a little when it is loaded or rise a little when it is unloaded then correct the error as soon as possible before it gets very bad. An integrator will slow it down too much.

It is called Transient Response.

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Uncle $crooge

 

Ah I see, how about using a mixture of a standard error amplifier and an integrator ?

The integrator would be setup to be as fast as possible with a low pass on the output to try and stop any oscillation, the standard amp would be setup as normal.
Both amps would then be fed into a summing amp, which would be biased towards the standard amp.

If the standard amp got 3/4 of the bias and the integrator got 1/4 of the bias, the maximum Transient Response would be 125% of the time that it would normally be. Slightly slower but that's in a worse case scenario. Surly that would be acceptable ?

Of course it means that the integrator needs 4 times the voltage swing in order to be 100% useful, again in a worse case scenario.

Am I looking at this wrong ?

 

You need to analyse the speed and phase shift of the opamp, the driver transistor, the output transistor and the output filter capacitor to see the transient response or oscillation, then add series and shunt capacitors in the negative feedback loop to make it quick without ringing or oscillation..

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Uncle $crooge

 

So you think it would be worthwhile if the time was spent to analyse and optimse it ? I'm just wondering why a modified version of this method hasn't been used before - In controller IC's I mean, because I'm yet to find a PSU whose output doesn't change from a large range of loads.

 

The error amplifier in a power supply has an extremely high voltage gain so it reduces the output voltage change from no load to full load to a very small amount. The wiring to the load has a higher voltage drop than the power supply regulator.

The error amplifier and the high current output transistors are slow to react so there is a transient at the output when the load current is quickly changed. Good design minimises the transient. Poor design causes oscillation.

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Uncle $crooge

 

Ah see, so the best ways to reduce voltage drop from high loads is:
1) Use a higher gain error amplifier.
2) Connect the voltage feedback as close to the load as possible.

Is this correct ? What other methods could be used to help ? (Not taling about rapidly changing loads here, I mean the different output voltages when a static high resistance load is used and a static low resistance laod is used.)

 

A 7805 regulator has a typical output voltage drop of only 15mV when its load changes from 5mA to 1.5A. That is pretty good voltage regulation.

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Uncle $crooge

 

Although its output impedance goes up with frequency, it's effectively like an inductor connected in series with the output.

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Ah fair enough. How about with higher currents. Let's pick say .. 0 - 10A.

I know it would be too inefficent and should be done through switching, but I'm interested. :P

 

Look at the datasheet, it has circuits that boost the output from the regulator to up to 10A.

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I ment what would the regulation be like, I didn't think it told you in the datasheet ?

Edit: Just checked and it doesn't. Also how did you know that it's a 10A regulator example, in the datasheet it doesn't give any values for the examples ?

 

The regulation of a regulator with a boost transistor won't be different because the regulator is working at much less than its max current rating so its regulation will be better.

The datasheet shows an old boost transistor with a max current of 7A. If you find a PNP transistor with a max current rating of 10A and if it has enough gain then it can be used instead. The series current-sensing resistor can have its value reduced so that the base current for the transistor is higher for low gain transistors.

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Uncle $crooge

 

Fair enough, thank you for explaining that.

If you was to use a darlington pair for it's higher gain, would I be right in assuming that the responce and regulation would be better (Due to the higher gain), however the regulator would have a higher dropout voltage ?

 

A darlington transistor has a high current gain. Its voltage gain (or loss) is the same as any ordinary transistor so the response of the circuit won't be any different except darlington transistors are usually much slower than ordinary transistors.
Yes, a darlington transistor has a higher dropout voltage.

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Uncle $crooge

 

Ah I see, so it's voltage gain that's required.

Some of the examples have a PNP which then drives an NPN. Is this just because it's easier to find NPNs that have a higher current rating than PNPs ? Am I right in assuming that these will also suffer from a higher dropout voltage than the ones using a single PNP ? I guess this would be the reason why most low dropout voltage regulator ICs have a lower current rating than their "normal" dropout voltage counterparts.

Would a higher current gain have any benifits, such as less current lost through the resistors before the regulator IC it's self ? I guess that because more current would go through the transistor, it would theoretically allow an even higher over-all current rating for the circuit ? (As you're now limited by the rating of the transistor, not the regulator IC.)

Thank you for taking the time to walk me through this, I can be a bit slow sometimes. (Most of the time )

Edit: On a slightly different note, seems as we were talking about transistors.

Would an NPN setup with the Collector going to the load and the Emitter going to Ground (So placed after the load) be less efficient thank a PNP which is setup with the Emitter going to +VS and the Collector going to the load ? (So placed before the load)

Edit2: I've just looked at a couple of datasheets for high power transistors and it seems that they only have a gain of "Min: 20, Max: 70". I'm used to little transistors with gains of 200-500. Is this right or am I looking at shite transistors ? Am I right in assuming that with the correct base resistor and the correct resistor going into the Vin of the regulator (Current sense resistor ?) that these transistors would be fine ?

Would the max theoretical current through the circuit (Assuming the transistor and resistors could handle infinate current) be the current gain multiplied by the max current of the regulator IC it's self ?

Finally how are those two resistors before the regulator IC calculated and how is the PNP transistor dimensioned before the NPN ? Is the PNP calculated through the current rating of the NPN multiplied by the current gain of the NPN ?

Cheers.