5V and 3.3V with Changing Source

[R_C]

I am a hardware novice and need some advice. My circuit needs both 5V and 3.3V. Most of the time the circuit will be powered by an 11-14V source but occasionally I will disconnect from this source and it will be powered by a 9V battery. First, what techniques should I be looking at to allow the circuit to run uninterrupted when disconnecting from the main source and switching to the 9V battery then back to the main power source? Would a barrel power jack that includes a switch work? That is, inserting the barrel with the main power source opens the switch disconnecting the 9V battery?

Second, I get confused looking at the wealth of available voltage regulators. I have successfully prototyped using an LM317 to get my 5V. It has an input-to-output voltage differential of 3-37V so that appears adequate for my 9-14V input. I am using an LM2937-3.3 to get 3.3V from the 5V. The 5V side of the circuit will require no more than 150mA and the 3.3V side no more than 100mA. Are these regulators good solutions? At these currents and input voltages will I need to consider heatsinks? Thanks for any ideas.

 

[bananasiong]

Hi,
I've forgotten where did I got that picture, it is used for battery back up, maybe it helps.
And, you don't need any heatsink for such low current and not much of dropout voltage.

 

[Hero999]

You got it from me, I posted it awhile ago.

 

[Tomble]

Second, I get confused looking at the wealth of available voltage regulators. I have successfully prototyped using an LM317 to get my 5V.

The 7805 is probably the best-known choice of regulator for 5V; Unlike the 317, it'll give you that voltage without needing any sort of calibration You're still supposed to put an input and output cap on it though, as described in the datasheets. Haven't tried without so I couldn't say if they were 100% vital, but even if they aren't they're a good idea. But other than that, no extra components needed. Oh, and they're common as muck, so you could find them anywhere and cheap I think there's a few different varieties, with different packages and max. current ratings.

It would probably also give better performance than LDO-style regs too, seeing as you have a reasonable amount of headroom for it... Although then again it might not be too good on the battery I suppose after the voltage starts to drop Meh. There's also "quasi-LDO" regs though that are a compromise between the two types: medium performance and medium amounts of dropout if it turns out to be a concern.

Switch-mode regulators would be much more efficient but maybe a bit noisier, but I know too little about those to comment at all beyond that.

 

[Sceadwian]

Fixed voltage regulators need the caps to avoid over/under shooting if the load current changes quickly. If the load is essentially fixed the caps might not be needed, but for any digital circuits at least something would be better than none at all.

 

[audioguru]

A 9V alkaline battery's voltage quickly falls to 7.2V then keeps falling slower which is too low for the input of an LM317 set to 5V or a 7805 regulator.

A "9V" rechargable battery is actually 7.2V (6 cells) or 8.4V (7 cells) and also has a voltage that is too low.

Use a low-dropout 5V regulator.

 

[Hero999]

A LM317 doesn't need calibrating, unless a very precise voltage is required, you just calculate the resistor values and away you go. The LM317 does have better characteristics over all but it comes at the cost of having to add a couple of resistors.

 

[R_C]

Thanks everyone for all of your responses.

A 9V alkaline battery's voltage quickly falls to 7.2V then keeps falling slower which is too low for the input of an LM317 set to 5V or a 7805 regulator.

A "9V" rechargable battery is actually 7.2V (6 cells) or 8.4V (7 cells) and also has a voltage that is too low.

Use a low-dropout 5V regulator.

I had seen the LDO term but never understood it before. After more reading I think I understand. The dropout voltage of a 7805 regulator is 2V and 3V for the LM317, which means I wouldn't run very long at 250mA on a 9V battery before the regulator shuts down. I found an LM2937 LDO with a 0.5V dropout that can still handle input voltages up to 26V so that should be a good solution.

I am still concerned about dissipating heat. If Vin is 14 and Vout is 5 and I have a 250mA load don't I need to dissipate 2.25W ((14-5) *.250)? According to the LM2937 datasheet that appears to be above the power dissipating capability of a TO-220 package when the ambient temperature is above 0 degrees C. Is a heatsink required here?

 

[ericgibbs]

Thanks everyone for all of your responses.



I had seen the LDO term but never understood it before. After more reading I think I understand. The dropout voltage of a 7805 regulator is 2V and 3V for the LM317, which means I wouldn't run very long at 250mA on a 9V battery before the regulator shuts down. I found an LM2937 LDO with a 0.5V dropout that can still handle input voltages up to 26V so that should be a good solution.

I am still concerned about dissipating heat. If Vin is 14 and Vout is 5 and I have a 250mA load don't I need to dissipate 2.25W ((14-5) *.250)? According to the LM2937 datasheet that appears to be above the power dissipating capability of a TO-220 package when the ambient temperature is above 0 degrees C. Is a heatsink required here?

hi,
It is possible to connect a series power resistor in the +14V line before the regulator.
For example if, the the +14V was regulated and you required never more than 250mA, you use a 20hm: 2Watt resistor to drop from 14V to 9V.

This would mean the Vreg would have 4V across it at 250mA, which is about 1Watt.

Using a 32R 3Watt resistor would give a drop of 8Volts, that is 6V into the Vreg.
That is 1V across the Vreg at 250mA, which is 0.25Watt for the Vreg.

If you do it this way I would suggest that the smoothing cap is connected between the junction of the resistor and the Vreg, to 0V.

 

[eng1]

If Vin is 14 and Vout is 5 and I have a 250mA load don't I need to dissipate 2.25W ((14-5) *.250)?

Correct.

Is a heatsink required here?

Yes, use a heatsink. For a given level of power, its size depends on the ambient temperature you're considering.

 

[R_C]

hi,
It is possible to connect a series power resistor in the +14V line before the regulator.
For example if, the the +14V was regulated and you required never more than 250mA, you use a 20hm: 2Watt resistor to drop from 14V to 9V.

This would mean the Vreg would have 4V across it at 250mA, which is about 1Watt.

Using a 32R 3Watt resistor would give a drop of 8Volts, that is 6V into the Vreg.
That is 1V across the Vreg at 250mA, which is 0.25Watt for the Vreg.

If you do it this way I would suggest that the smoothing cap is connected between the junction of the resistor and the Vreg, to 0V.

What if the +14V is unregulated? Would this still work?

 

[Hero999]

Yes, it would still work, providing you don't make the resistor too higher value so the voltage drop is too high for the regulator to work on full load when the voltage is at its lowest level.

You could just use a switching for the 14V supply input.

A Black Regulator would be perfect.
https://www.romanblack.com/smps/smps.htm

Set the output voltage for 6V and connect it before the low drop-out regulator which will eliminate the ripple and noise on the output. This arrangement is know as a hybrid regulator and is often used for high quality high powered power supplies and gives you the best of both worlds: the reasonable efficiency and compact size of a switching regulator with the smooth noise free output and excellant regulation of a linear supply.

 

[ericgibbs]

What if the +14V is unregulated? Would this still work?

hi,
What is the unreg 14V going to be when you are drawing 250mA?
Measure/calculate it and calculate the optimum value for the resistor.

 

[Hero999]

R_C,
From your origaional post you said it varied from 11V to 14V, just calculate your resistor assuming 11V or even 10V, that way you can be sure it'll work on the lowest input voltage.

I'd prefer the switching regulator option for the 11V to 14V supply voltage though.

 

[R_C]

Recall, that I also want a 9V battery backup to the circuit. Using Hero999's schematic and searching these forums I see it can be done easily with a couple of rectifier diodes. But I must keep the regulator's input voltage above that of a 9V battery. So the power resistor could only drop the input voltage to something above the voltage of a new 9V non-rechargable. Maybe the heatsink is a simpler solution?

 

[Hero999]

You only use the resistor for the 11V to 14V supply.

The 9V battery is connected after the series resistor so it doesn't do anything.