I need a power supply that will give me 12v, 9v, and 5v. I have a 15v transformer already, with 7812, 7809, and 7805 regulators. Now, should I hook the regulators up in series (i.e., with the output of the 7812 feeding the input of the 7809 which would feed the input of the 7805), or in parallel (i.e., inputs of all 3 regulators connected to the secondary of the transformer)?
I can't see that one way or the other would make much difference, but I thought I should check to see if there are advantages/disadvantages to either method.
On a sunny day (Tue, 12 Dec 2006 13:03:17 -0500) it happened "tempus fugit" wrote in :
Putting them in series will have the first one dissipate more, as now all current goes to that one, then the next one. Depending on the current this may not be a good idea. Safest is in parallel.
So would this also give better isolation of the 3 supplies? I was wondering if I should use a resistor from secondary to each input to help with isolation, or if this is even necessary.
The difference - and the key to selecting the better method - is the current draw on each output voltage, and the thermal consideration for each regulator that flows from that information.
The 12v will draw a maximum of 200mA, the 9v maybe 200 also, and the 5v is powering CMOS logic, so in the order of a few mA, maybe less. My transformer is rated at 500mA, and I am estimating on the high side for the 12 and 9v supplies, and these will be maximums if everything is on at once (which it virtually never is).
Given this information, which method would be preferable?
Firstly you need to get a handle on the trough voltage on the output of the rectifier/filter system you intend. That will need to be high enough to avoid dropout in the 7812 at load. But for thermal reasons, you don't want it any higher than it has to be, and dissipation in a series element(reg) will depend on the amount of ripple above that trough value. More filter cap = less ripple = less heat in follwing linear regs.
For the sake of the following comparison, assume we have a 15vDC input - no ripple.
Thermally, running the 7805 off either the raw DC input or either other reg would make no difference to them or the 7805. Direct.
If you run the 7809 off the input, it will run the hottest and may require a heatsink depending on the "on" time.
If you run it off the 7812, THAT will become the hotspot and will require a sink.
Assume steady 200mA on 12v and 9v lines then:
7809 fed by 7182: 7812 dissipation 1200mW 7809 dissipation 600mW
7809 off DC in: 7812 dissipation 600mW 7809 dissipation 1200mW.
Six and two threes.
BUT, as the effective DCin voltage increases above 15, the first scheme becomes the worse. I'd go with the second scheme and put a small sink on the 7809.
OK, for the purpose of argument, let's allow 50 mA for the +5.
It depends on which regulator you want to dissipate the most power. In cascade, given a 15V supply, the 12V will have 450 mA plus the idling current of the 7809 & 7805; so, maybe 475 mA (on the high side), at 3V, so you know how many watts that is, and so on.
The other way around, the 7812 only dissipates 200 mA * 3V, but then the 7809 dissipates its 200mA while dropping 6 V; in cascade it would be 200 mA * 3V, and so on.
Of course, in either case, you need all of the recommended capacitors.
You'll have to dump heat somewhere, regardless of how you connect. Others have computed based on 15 volts DC, but you'll actually have about 19.8 volts at the filter cap. The heatsinkless solution is to use a power resistor to dump the heat. Use 2 watt resistors. At a few mA 5V load, the resistor is not needed - but is added anyway, and based on a 100 mA load. The 12 and 9 volt regulators will dissipate about 1/2 watt at the 200 mA loads you specified - the 7805 will be far below that at the "few mA" you specified, and about .4 watt at 100 mA. Add the input and output caps to the regs that I omitted for simplicity.
Arrgh! Forgot to check for mains +/- 10% variation. You would need to change the resistance values to 10 ohms for the 12 volt reg, and 25 ohms for the 9V reg to accomodate a mains down 10% at 108V. That puts the
9V reg at ~ 1.36 watts & the 12V reg at ~ 1.6 watts worst case when the mains voltage increases to 132v. So you'll have to use heatsinks, and the resistors won't buy you much.
Others have commented on the power dissipation differences; two other differences are in the power sequencing (series connection guarantees the +15 is always higher voltage than +12, and the
+12 always higher than +5, during power-up and power-off events), and in the output impedance.
Linear regulators, when lightly loaded, have somewhat higher noise and output impedance. So, if regulation and noise are critical, either loading all three regulators (maybe 20 mA or so) or connecting them in series and loading the +5 regulator only, is sometimes beneficial. If you look in the specs for linear regulators, the dynamic/AC characteristics are always measured with a current of ~500 mA , and load ripple rejection isn't really guaranteed at all for an unloaded circuit.
Given that I'd not need simultaneous high current from multiple supplies, I'd go for series connection. If you go parallel, clamp rectifiers can deal with the power sequencing issue.
OK guys thanks for all the tips. I already switched it over to parallel by the time I read your post, whit, so I guess I'll keep it like that for now and see how it goes. Also, I did a more accurate measurement of the 9v rail, and the maximum current is more like 100-115 mA (my meter only reads to 2 decimal places, so i don't know if a reading of 0.02 is 20mA or 29 mA or anywhere in between).
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