The Sinclair QL Forum

It is worth using the regulator 7805 with 3A at 5V in QL instead of its original?

Yes of course, there’s no problem to use it, in effect I have changed regulator into my QL with a switching one with 3.5 amps capability.

Depends. It is, of course, possible. Whether it is actually needed, may vary.

On a non-expanded QL, an upgrade to a 3Amp type shouldn't be necessary and wouldn't improve anything. On a heavily extended machine, it might be worth thinking about it - Depends a bit on the extensions: The Sandy Super-Q-Board I had in 1990 slurped such an amount of juice I suffered from frequent heat problems and irregular operation and had to upgrade to a better 7805. The Miracle boards I use today seem to be fine with a stock 1A 7805 and run 24 hours even in the hottest summer with no problems. If you don't have any such problems, why change a running system?

Note a 3A version will create the very same (if not more) amount of heat the original one does. Today, we have much better replacements (switching regulators) that can even run without a heat sink without any problems.

Tobias

No. There is zero benefit in increasing the capacity of the internal regulator.

All external expansions run off the 9v expansion port supply, so external expansions do not load the internal 5v voltage regulator. The modern internal expansions like QL-SD are so efficient they make an immeasurable difference to 7805 load.

In fact, replacing the 7805 with a larger one could create problems. If you have a short do you want 1 or 1.5A of current dumped through it, or 3-4 amps? It will also take the 3-4A 7805 longer to react to a short or over current situation and start to throttle, which could massively increase the amount of damage done while the magic smoke is being released.

If you want to cool that end of the QL, replace any 7805 with a 5v buck regulator. If it's ideally sized it would be around 1.2 to 1.5A.

Dave wrote:No. There is zero benefit in increasing the capacity of the internal regulator.
All external expansions run off the 9v expansion port supply, so external expansions do not load the internal 5v voltage regulator. The modern internal expansions like QL-SD are so efficient they make an immeasurable difference to 7805 load.
In fact, replacing the 7805 with a larger one could create problems. If you have a short do you want 1 or 1.5A of current dumped through it, or 3-4 amps? It will also take the 3-4A 7805 longer to react to a short or over current situation and start to throttle, which could massively increase the amount of damage done while the magic smoke is being released.
If you want to cool that end of the QL, replace any 7805 with a 5v buck regulator. If it's ideally sized it would be around 1.2 to 1.5A.

There is one reson why a higher current version of the 7805 was popular as a replacement for the original 7805 1A version, and this has nothing to do with the power consumption of the actual QL, as Dave pointed out - what happens is that additional peripherals load the 9V input voltage to the 7805.
There is a minimum voltage difference between input and output of the 7805 in order to regulate properly, which is known as the 'dropout voltage'. At the full 1A this is around 2.8-3V so the minimum input voltage under load should be at least 8V.
When the original power supply is loaded with extra peripherals, the mean voltage does not drop below this limit. However, because the external supply is not regulated, but just a simple 'peak rectifier' type supply, there is significant ripple superimposed to the mean voltage value, and it increases as the load increases. When observed by an oscilloscope, the voltage coming from the power supply looks like a sawtooth waveform whose maximum value is what one would measure with a DVM, when checking the voltage under load (maximum because in 99% of all circumstances DVMs measure peak value).
What happens then is that the bottom of the sawtooth can drop below the required 8V and when it does, the output of the 7805 drops as the input drops, so you get 'notches' in the normally constant 5V supply 100 times a second. If it goes low enough, the system will become unstable.
The reason why a 3A regulator was recommended is that the same 2.8-3V dropout voltage is specified for it, but at 3A rather than 1A - at 1A it is somewhat lower, so it increased the immunity to this problem.

Today we have more options - basically there are two groups:
1) replace the regulator with a low dropout version of the 7805 (these are pin compatible and the dropout voltage is less than half of the original 7805), which solves the dropout problem but produces exactly the same amount of heat in the heatsink. This is because any linear regulator sort of impersonates a variable resistor - so the heat generated on it is always current through it times voltage across it. From that standpoint, the regulator actually generates LESS heat if other peripherals load the 9V line.
2) Replace the regulator with a switching buck regulator. This type of a regulator acts similarly to what would be a DC transformer (for reference, a real transformer can only operate on AC). The efficiency is very high so very little excess heat is generated, to the point where a heatsink is not needed at all. Also, in many cases the switching regulators have a low dropout voltage and higher current capability, but this has to be a circuit rather than a component.

Nasta wrote:2) Replace the regulator with a switching buck regulator. This type of a regulator acts similarly to what would be a DC transformer (for reference, a real transformer can only operate on AC). The efficiency is very high so very little excess heat is generated, to the point where a heatsink is not needed at all. Also, in many cases the switching regulators have a low dropout voltage and higher current capability, but this has to be a circuit rather than a component.

DC/DC converter modules (switching regulators) are available that are “drop-in” replacements for 7805 series pass voltage regulators.

One other point, the current flowing through a series pass voltage regulator like a 7805 is determined by the load (assuming that the regulator has not entered an overcurrent or thermal limit mode). So for a 7805 the input current is equal to the current out plus the small amount needed for its own internal operation.

But with a DC/DC converter module / switching regulator of the step down type, the current input is less than the output current. How much by depends on the input voltage. So not only are they far more efficient, they actually reduce the demand on the mains PSU / mains adaptor / power brick.

Mark

Thank you for your replies. What type of switching regulator do you suggest?

Careful with the use of 7805 'replacements' using switching regulators. The idea itself is not a problem, but rather how to implement it in the QL.
A fairly seizable low ESR capacitor is mandatory at the input coming from the brick power supply. One big problem is the extension lead (as short as it is) from the PCB to the central pin of the 7805 - while it only passes a fraction of a mA when a 7805 is used, with a switching power supply there are seizable current transients in that piece of wire, easily 3 order of magnitude larger, and usually at a fairly high switching frequency - under these conditions even this short piece of wire is important. So much so that it can prevent proper operation of the switching regulator. In most cases it can be solved by adding a low ESR electrolytic cap between the input and ground pin and also output and ground pin. The first one is usually more important because it's usually missing or only a small value cap is included on the replacement board.