Selecting input voltage for 7805

Question

I generally use switching 12V power supplies for my circuits since they are so common, but I understand the 7805 can take a lower input voltage.

If I gave the 7805 a lower input voltage (i.e. 9V), I understand there will be less heat dissipated as it would be a smaller voltage drop. 6V would be even less, assuming my regulator can go that low.

But, is there a disadvantage to giving less input voltage? Performance? More or less noise? And on the power supply side, are lower-voltage or higher-voltage supplies typically better made?

(I've got a 100uF aluminum cap on the input, and 10uF aluminum and 0.01uF ceramic caps on the output.)

Answer 1

The 7805 will drop out typically around 1.6V above the output voltage. At 1A it's guaranteed to not drop out with 2V of input-output differential. Most likely you'll not be running anywhere near 1A or you'd be using a switching regulator, but even at low current the dropout is not so low- that's because the 7805 is not an LDO regulator and there are Vbe drops in there.

One could guess that the input ripple rejection probably deteriorates as you get close to the dropout voltage and the gain drops. The datasheet specification is at 5V input-output differential, so they sidestep that issue. If you have a sensitive analog circuit like an RF module you may wish to use a higher input voltage than the absolute minimum.

If you're using a 7805 with unregulated (transformer, rectifier and filter) input voltage probably needs to be something like 10V to be safe and account for line voltage tolerance, ripple and so on). If you're using it with a regulated supply (like a switching wall wart) 9V is good, 7.5V is okay, but 6V is not high enough. There are LDO regulators that have very low dropout (so 6V would be fine) but they have other disadvantages (they are only conditionally stable- pay careful attention to the output capacitor value, ESR and type), they are more expensive, less sources, and generally have much lower input voltage capability so they're easier to fry with input transients. Much modern electronics uses LDO regulators and/or switching regulators, there are literally thousands of possible parts to use, but none yet has quite the staying power of the 7805/78M05/78L05.

I would say that if you need a heat sink on the 7805 it's time to move to a switching regulator in most cases. There's no problem using the 7805 or 78M05 at 10, 50 or 100mA, and it's better than a 78L05 (more expensive, but the circuit is different and has better guaranteed performance). The trade-off of an LDO vs. a 78xx regulator is a bit more complex and it is heavily dependent on the input voltage and how much control you have over it.

Answer 2

Switching regulators do not disapate power the same way a linear regulator (e.g. LM7805) does. The typical (old style) LM7805 needs a minimum of 7v to guarantee 5v out, and that would dissapate continuosly with at least 2v x I out, with a linearly higher dissapation as the input voltage goes higher.

A well designed switching regulator dissapates power only during the breif time that the power components are switching, (some is lost in ineffiecient switching and some lost to Rds or Rce). Typical switching regulators are more efficient with higher input voltages.

Lower input voltages will give lower dissapation in linear regulators. But one disadvantage to using a lower input voltage is that your input supply, or battery, might be bogged down in high load situations, and go below the regulator's drop out level. With higher input voltages some switching regulators will actually switch at higher frequencies making capacitor noise filtering more efficient.

Answer 3

But, is there a disadvantage to giving less input voltage? Performance? More or less noise?

No, as long as

Vin > Vo + Vdrop

where Vdrop is the internal voltage drop of the regulator. The drop is normally proportional to the output current and can be as low as a few hundred millivolt for low drop regulators up to several volt.

Answer 4

Assuming that you want a regulated 5 volt source which can supply 1 ampere into a load, and that you want to use use a 7805 connected to a 9 volt supply to get that 5 volt output, the 7805 would have to dissipate: \$\ Pd =(Vin - Vout)\times Iload \ \ =\ \ (9V-5V) \times 1A = 4\ watts\$, while the load was dissipating \$\ 5V \times 1A = 5\ watts \$

The input power would, therefore, be 9 watts and, since efficiency = \$ Pout/Pin \$, we'd wind up with an efficiency of about 55%.

A switcher would run, typically, at about 80%, so it'd be dissipating 1.25 watts regardless of the input voltage - within limits, of course - and, compared to the 7805's 4 watts, would run much cooler.

The advantage of using a lower voltage on the input of a linear regulator is that the regulator will run cooler but with, say, 2 volts of headroom needed for a 7805 to run properly, it'll still be wasting 2 watts compared to the switcher's 1.25.

The downside for the switcher is that its output will be a little noisier then the linear's and it'll be more expensive. Maybe. By the time you get finished with the heat sink, the mounting hardware, and the the heat sink compound, who knows???

Answer 5

No matter the less input voltage, on giving low voltage the regulator can deliver max of 5V so your ip volatge shld be above 5V, As per 7805 chip LM7805 fairchild 7V is desirable.