Can I connect two voltage regulators in series, to reduce power consumption?

Thread Starter

onlyonce18

Joined May 8, 2016
46
I was thinking about making a 5V power source to charge 2 cellphones in my car. Using a couple of LM7805 connected in parallel to the 12V lighter would be easy, but if each cellphone consumes 1A, the power dissipated by each LM7805 would be around (12V - 5V) * 1A = 7W

If I connect both LM7805 in series, can I assume there will be a 6V drop on each, and therefore a power dissipation of (6V - 5V) * 1A = 1W? If so, the capacitors would be at the same places in both regulators?

Please see the attached diagram and the LM7805 datasheet.

I'm leaving aside all other components (like the security fuse and resistors) because my question is focused on the feasibility of using the regulators in series instead of parallel configuration. Also, probably a switching regulator would be a better choice to reduce the dissipation, I´m just curious if the series circuit would work.

Thanks for your comments
 

Attachments

kubeek

Joined Sep 20, 2005
5,608
The second circuit is not in series, it is just wrong.
If I connect both LM7805 in series, can I assume there will be a 6V drop on each, and therefore a power dissipation of (6V - 5V) * 1A = 1W? If so, the capacitors would be at the same places in both regulators?
I am not sure whre you get that 6V, but it makes no sense.
Anyway, there is no way around that 7W figure as long as you are using linear regulators. That amount of power has to be burned away in order to get 5V at 1A output. The only solution that will not waste so much heat is a switching regulator.
 

LesJones

Joined Jan 8, 2017
2,326
Why not use a switching regulator such as this one. You will find many simailar ones on Ebay. A switching regulator does not waste as much power as a linear regulator.

Les.
 

AlbertHall

Joined Jun 4, 2014
8,385
You could use a 7808 or 7809 to reduce the 12V to 8V or 9V then feed the 7805 with that reduced voltage to spread the power between the two regulators.
 

dl324

Joined Mar 30, 2015
8,968
Circuit in question.
upload_2018-1-19_8-58-44.png
You can't get 5V from the top regulator unless you ground it's ground pin.

If you use linear regulators, the power still needs to be dissipated somewhere.
 

Thread Starter

onlyonce18

Joined May 8, 2016
46
The second circuit is not in series, it is just wrong.
I am not sure whre you get that 6V, but it makes no sense.
Anyway, there is no way around that 7W figure as long as you are using linear regulators. That amount of power has to be burned away in order to get 5V at 1A output. The only solution that will not waste so much heat is a switching regulator.

I was wondering if the regulator will act as a "resistor", so 2 regulators in series connected to 12V would have a the same voltage drop, 12V / 2 = 6V each.

If I connect 2 LEDs in series I'll have the first LED anode to the source's (+), then the cathode connected to the second LED´s anode, and finally the second LED's cathode to ground. That's why I started wondering about the power regulator in this configuration.
 

Thread Starter

onlyonce18

Joined May 8, 2016
46
It works and it doesn't. You need some pretty heafty capacitors to make it work without significant oscillation. Also, charging will stop as soon as the first phone is charged.

Thanks! I'm assuming the regulator will stop conducting electrons between its input pins once the load on the 5V output dissapear? (phone's battery is charged)
 

WBahn

Joined Mar 31, 2012
24,709
It works and it doesn't. You need some pretty heafty capacitors to make it work without significant oscillation. Also, charging will stop as soon as the first phone is charged.
So how does the "it works" portion of that work?

Let's say that the capacitors you recommend are put in and that there is no significant oscillation.

What is the input voltage to the second 7805?

Doesn't it need to be at least 2 V greater than the output voltage, so at least 7 V?

What is the output voltage of the first 7805?

Wouldn't it be 5 V greater than the input voltage to the second, which would make it at least 12 V in order to be in regulation (which it couldn't be unless it's input voltage was at least 14 V).

Where is the current that is powering the second 7805 coming from?

It is coming OUT of the ground pin of the first one. But this is not a power pin, it is the reference for the regulating circuitry.

How will the first phone completing its charge affect the circuit?
 

WBahn

Joined Mar 31, 2012
24,709
Thanks! I'm assuming the regulator will stop conducting electrons between its input pins once the load on the 5V output dissapear? (phone's battery is charged)
Try hooking it up on a breadboard or some other suitable prototype and measure the voltages everywhere. Don't connect up an actual cell phone, but instead use resistors as loads, starting with some fairly large ones that will only draw 50 mA or so. That will let you use 1/4 W resistors, at least long enough to take some measurements, though 1 W resistors would be better (or several larger 1/4 W resistors in parallel).
 

Thread Starter

onlyonce18

Joined May 8, 2016
46
Try hooking it up on a breadboard or some other suitable prototype and measure the voltages everywhere. Don't connect up an actual cell phone, but instead use resistors as loads, starting with some fairly large ones that will only draw 50 mA or so. That will let you use 1/4 W resistors, at least long enough to take some measurements, though 1 W resistors would be better (or several larger 1/4 W resistors in parallel).
You are correct about the 2V at least above the 5V needed at the input, so the power source should be at least 14V and not 12V. Still it was just curiosity, if I can get spare parts I'll make the test. I guess I can use just one regulator and a 1W resistor between the regulator's ground pin and power source's ground and see what happens when I put a load at the 5V output.

Thanks!
 

dl324

Joined Mar 30, 2015
8,968
I guess I can use just one regulator and a 1W resistor between the regulator's ground pin and power source's ground and see what happens when I put a load at the 5V output.
I can save you the bother; the answers are in the datasheet. The quiescent current from the regulator will increase the output voltage by the voltage drop of the resistor you insert in the ground connection.
upload_2018-1-19_10-33-9.png
upload_2018-1-19_10-34-8.png
upload_2018-1-19_10-34-31.png
Maximum quiescent current is 9.5mA.
 

WBahn

Joined Mar 31, 2012
24,709
You are correct about the 2V at least above the 5V needed at the input, so the power source should be at least 14V and not 12V. Still it was just curiosity, if I can get spare parts I'll make the test. I guess I can use just one regulator and a 1W resistor between the regulator's ground pin and power source's ground and see what happens when I put a load at the 5V output.

Thanks!
More important is that the voltage at the ground of the first (input of the second one) is not defined, which means that the output of the first one is not defined.

You can make voltage regulators above 5 V using a 7805 by using a topology similar to what you would do with a LM317. It's not intended for that application and so you need to be much more aware of the quiescent current coming out of the ground pin, but at the end of the day the regulator just tries to maintain 5 V between the output and whatever is at it's ground pin.

Do the tests and discover for yourself how it behaves and then spend some time with the data sheet until you are comfortable that what you observe is what you should expect. You'll learn a lot in the process.
 

Thread Starter

onlyonce18

Joined May 8, 2016
46
More important is that the voltage at the ground of the first (input of the second one) is not defined, which means that the output of the first one is not defined.

You can make voltage regulators above 5 V using a 7805 by using a topology similar to what you would do with a LM317. It's not intended for that application and so you need to be much more aware of the quiescent current coming out of the ground pin, but at the end of the day the regulator just tries to maintain 5 V between the output and whatever is at it's ground pin.

Do the tests and discover for yourself how it behaves and then spend some time with the data sheet until you are comfortable that what you observe is what you should expect. You'll learn a lot in the process.

I'll make the tests as soon as I get the spares, thank you!
 

GopherT

Joined Nov 23, 2012
7,983
You are correct about the 2V at least above the 5V needed at the input, so the power source should be at least 14V and not 12V. Still it was just curiosity, if I can get spare parts I'll make the test. I guess I can use just one regulator and a 1W resistor between the regulator's ground pin and power source's ground and see what happens when I put a load at the 5V output.

Thanks!

This gets you half way to your goal of more efficiency (top 7805 doesn't drop the full 7 volts but bottom one does). The good news is that the bottom one only drops the full 7 volts if the bottom phone needs more current that the top phone.

D391A5E1-8F8E-4E0B-8397-E9EF641DE3AB.jpeg
 
Last edited:

Thread Starter

onlyonce18

Joined May 8, 2016
46
This gets you half way to your goal of more efficiency (top 7805 doesn't drop the full 7 volts but bottom one does). The good news is that the bottom one only drops the full 7 volts if the bottom phone needs more current that the top phone.

View attachment 144042
Thanks, that´s another way to make the circuit. In this case the power source could be 12V isn´t it?
 

AnalogKid

Joined Aug 1, 2013
8,111
I get what you are trying to do, and there actually is a way to do it by placing the two phones in series:

+12 V > 5 V shunt regulator with phone #1 across it (in parallel with it) > 5 V series regulator > phone #2 > GND

The shunt regulator does not exist as a single power IC. You would have to grow the circuit from scratch. But overall this arrangement would dissipate less power because the current used to charge the first phone would then go through the second phone and charge it. The problem is that phone #1 is charged by whatever current phone #2 draws. When phone #2 is fully charged, all charging of phone #1 stops.

Ampex broadcast video recorders in the 60's had massive shunt regulators for the signal system. Analog with balls.

ak
 
Top