Related voltage divider circuit design

[Akshay2waghamare]

I am designing the circuit with voltage divider and loads after it but i am not getting how to design the circuit of voltage divider with less power dissipation in resistors

 

[frankrose]

Use bigger resistors.

 

[KlausST]

Hi,

I am designing the circuit with voltage divider and loads after it but i am not getting how to design the circuit of voltage divider with less power dissipation in resistors

There is no question, so how can we help you?

If you want us to help you with the calculations, you first need to priovide the requirements, specifications, mainly as values with units..
And (hand-) draw a simple schematic.
Try to give all informations at once.

Klaus

 

[Akshay2waghamare]

Hi,


There is no question, so how can we help you?

If you want us to help you with the calculations, you first need to priovide the requirements, specifications, mainly as values with units..
And (hand-) draw a simple schematic.
Try to give all informations at once.

Klaus

Hello sir,

I am designing voltage divider circuit with input voltage of 42vdc, and output i need 15vdc but the whole power is dissipating in resistors only and getting damage, so how solve this problem?
sir, please check the circuit attached file

 

[FvM]

That's a power conversion rather than voltage divider problem. Resistors may be used to drop a supply voltage, but power losses are respectively high.

Most people would implement the high voltage input converter as efficient SMPS. Next best solution, if you don't want it, would be a linear high voltage regulator, e.g. using discrete transistors and zener diodes.

But a 42V input, 5V/500 mA output regulator will at least dissipate 18.5 W, a resistor divider even more.

 

[wwfeldman]

I assume the resistor divider in your circuit is supposed to being the 42VDC below the maximum DC input voltage of the 7805, about 35V.

One problem is that the 7805 looks like a variable resistor in parallel with R2, so the divider isn't fixed anymore.
Second problem is that all of the current to the regulator and R2 must pass through R1, which will likely require a high wattage resistor.
Third problem is that the more current your load draws from the regulator, the more voltage drop you get in R1.

Suggested alternatives:
1) Use a buck type switching regulator.
Best start point is to search for a buck regulator (DC to DC step down).

2) Build your own pre-regulator.
Assume input of about 60V DC and output of about 25V, 2A. These provide sufficient margin
to operate from your 42V DC and work with the 7805.

3) find a linear regulator that meets your voltage and current output requirements and can use about 60 VDC input

 

[KlausST]

Hi,

your requirement is:

Voltage: 42V DC --> 5V DC
Current: I assume varying from 0mA to 500mA.

Now you have linear (analog) circuits.
This are:
* diodes, zeners
* resitors
* transistors
* linear regulator (7805)

All in common to these parts is, that they will dissipate the "unused" voltage x current as heat.
Thus it does not matter whether you use Zeners, resistors, transistors --> the amount of heat will be the same.

In your case the "usueful" ouput power is 500mA x 5V = 2.5W.
But you need to drop a lot of voltage: 42V - 5V = 37V. (in resistor or other linear devices) this means the dissipated heat is 37V x 0.5A = 18.5W

***
Solution:
Don´t use those "linear" parts to drop the voltage.
There are so called "switching regulators" or "SMPS", or "step down switcher", or "buck regulators"
--> go to your distributo´s internet site and do a search on them.

They have a good power conversion factor. Let´s say 85% efficiency.
This means if the 5V part draws 0.5A (5V x 0.5A = 2.5W output power)
It draws only 2.5W / 90% = 2.94W at the input. This means the input current is not 0.5A anymore but only 2.78W/42V = about 66mA only.
And the dissipated power (heat) is 2.94W - 2.5W = 0.44W only (in opposite to 18.5W before)

***
One question:
Why do you use that high 42V DC to generate 5V DC?
--> Use a much lower input voltage if available. If you have about 9V DC, then you may use the 7805 and avoid the use of a step down regulator.

Klaus

 

[kripacharya]

I second KlaussST suggestions and analysis. Read it carefully to get best benefit out of it.

 

[d123]

Hi,

To address question of PD in resistors, people tend to use instead of e.g. 1 * 1k, 10 * 100R in series - but it's still the same overall PD, just distributed over smaller value resistors.

This answer is just for info., not helpful in a practical way like posts #5 - 8.

 

[Akanimo]

Switching regulator is the best bet but Just a try:

What about cinnecting a 15V zener diode to GND immediately upstream of the linear regulator and then connecting this combination to in series with a low value resistor?

The zener diode will conduct for sure but the current through it should be dependent on the impedance looking into the input of the linear regulator. If that impedance is much lower than that of the zener, then most of the current will prefer to flow through the linear regulator.

Just thoughts though.

 

[Akanimo]

...
But you need to drop a lot of voltage: 42V - 5V = 37V. (in resistor or other linear devices) this means the dissipated heat is 37V x 0.5A = 18.5W
...
Klaus

You're right Klaus. 37V would have to be dropped in the resistor contributing to a minimum of 37/42 = 88% power loss.

It's a bad idea altogether.

 

[kripacharya]

Use an lm2596 based buck dc-dc converter in series with a 10ohm 10watt resistor.

Even better is if you can use a different transformer with 6v or 9v output.

 

[Akanimo]

What's upstream of the 30Vac supply? Is it a line -frequency transformer connected to mains? If it is, then just go for an flyback converter. With 5V/500mA output, a DCM offline flyback would suffice, else a 5V/500mA output dc/dc buck converter would do.

 

[kripacharya]

What's upstream of the 30Vac supply? Is it a line -frequency transformer connected to mains? If it is, then just go for an flyback converter. With 5V/500mA output, a DCM offline flyback would suffice, else a 5V/500mA output dc/dc buck converter would do.

Brilliant. Just go and buy an off-the-shelf 5v/1a usb charger module and adapt it.
No need to 'design' anything.