DC Voltage Multiplyer - Electronic Circuits Projects Diagrams Free

Gregory · 6th March 2008, 02:51 PM

How do I go about making a DC voltage multiplyer.
I want to go from 12V DC battery to 46 v DC
I am Building a simple field strength meter
The current that I require is not very high.

ronsimpson · 6th March 2008, 03:04 PM

Go to national.com
find datasheet for LM5002
Look at page 16
There is a 16V to 48V 125mA example.
Change R1/R2 and I believe it will work with a much lower input voltage. (with 100mA output)
Change R5/R6 to get 46 volts.

Hero999 · 7th March 2008, 08:54 AM

Whay do you need 46V for a signal strength meter?

Why 46V anyway?

It's not a standard voltage like 48V.

Gregory · 9th March 2008, 09:24 PM

I apologize for a mis type as the voltage is 12 Volt Input with a 42 Volt output.
I have attached a Circuit digram of the field strength meter I am building and a circuit of the Voltage multiplyer using LM5002.
I have Changed the value of the resistors to suit the requirements as above.
Could you check the Resistor values as I am a bit rusty in this area.
Thank you greg

Analog · 9th March 2008, 09:46 PM

Linear tech has some simple switching PS chips that you can configure directly from their website. These are switching power supplies DC/DC converters.

Diver300 · 9th March 2008, 09:56 PM

If you omit the 3.6k in series with the 10k potentiometer you can run from about 32 V instead of 42 V.

You don't seem to have understood how R1 and R2, and R5 and R6 are used.

R1 and R2 form a voltage divider which takes a fraction of the input voltage. When that gets to 1.25 V, the LM5002 turns on. If you want to run it off 12 V you should change the ratio of R1 and R2. You seemed to have changed their value and left the ratio the same. R1 and R2 have nothing to do with the current.

Similarly R5 and R6 form a voltage divider on the output voltage. The divider has to be arranged so that when the output gets up to 42V (or 32V), the divider outputs 1.25V. That makes the LM5002 turn off, therefore limiting the voltage.

In all voltage dividers, the ratio of the resistances is the most important feature. The value of the resistances is not important and can change over a wide range as long as the ratio is correct.

Gregory · 17th March 2008, 11:37 PM

I have worked out R5 & R6
Is this correct
When working out R1 & R2 Do I use the input voltage of 12 Voltage
The Current in the formular what current do I use.
I would like to know this ,if I am going to use the formular down the track.
Thank yo for your Help.
I enjoy the forum and the help that is provided.
Greg

Diver300 · 18th March 2008, 10:06 PM

I think that is nearly right.

The reference voltage for the LM5002 is 1.26 V

That makes the current 1.26/1470 = 857 µA

The voltage across R5 is 32 - 1.26 = 30.74 V

so R5 = 30.74 / 857 µ = 35.17 k

hm:

If you want the circuit to turn off if the supply voltage is too low, then you need to fit R1 and R2. If you want it to work when the voltage gets beyond 11.5 V, you could use R2 = 1.47 k

hm:

Then the voltage across R1 is 11.5 - 1.26 = 10.24 V

R1 is 10.24 / 857 µ = 11.95 k

hm:

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6th March 2008, 02:51 PM	(permalink)
Gregory	DC Voltage Multiplyer How do I go about making a DC voltage multiplyer. I want to go from 12V DC battery to 46 v DC I am Building a simple field strength meter The current that I require is not very high.

6th March 2008, 03:04 PM	(permalink)
ronsimpson	Go to national.com find datasheet for LM5002 Look at page 16 There is a 16V to 48V 125mA example. Change R1/R2 and I believe it will work with a much lower input voltage. (with 100mA output) Change R5/R6 to get 46 volts.

7th March 2008, 08:54 AM	(permalink)
Hero999	Whay do you need 46V for a signal strength meter? Why 46V anyway? It's not a standard voltage like 48V. __________________ I also post at the following sites: http://www.stop-microsoft.org http://www.heated-debates.com Screen name: Aloone_Jonez

9th March 2008, 09:46 PM	(permalink)
Analog	Linear tech has some simple switching PS chips that you can configure directly from their website. These are switching power supplies DC/DC converters. __________________ "Everything that is done in the world is done by hope." -Martin Luther "There are two ways to live your life. One is as though nothing is a miracle. The other is as though everything is a miracle."-Albert Einstein

9th March 2008, 09:56 PM	(permalink)
Diver300	If you omit the 3.6k in series with the 10k potentiometer you can run from about 32 V instead of 42 V. You don't seem to have understood how R1 and R2, and R5 and R6 are used. R1 and R2 form a voltage divider which takes a fraction of the input voltage. When that gets to 1.25 V, the LM5002 turns on. If you want to run it off 12 V you should change the ratio of R1 and R2. You seemed to have changed their value and left the ratio the same. R1 and R2 have nothing to do with the current. Similarly R5 and R6 form a voltage divider on the output voltage. The divider has to be arranged so that when the output gets up to 42V (or 32V), the divider outputs 1.25V. That makes the LM5002 turn off, therefore limiting the voltage. In all voltage dividers, the ratio of the resistances is the most important feature. The value of the resistances is not important and can change over a wide range as long as the ratio is correct.

18th March 2008, 10:06 PM	(permalink)
Diver300	I think that is nearly right. The reference voltage for the LM5002 is 1.26 V That makes the current 1.26/1470 = 857 µA The voltage across R5 is 32 - 1.26 = 30.74 V so R5 = 30.74 / 857 µ = 35.17 khm: If you want the circuit to turn off if the supply voltage is too low, then you need to fit R1 and R2. If you want it to work when the voltage gets beyond 11.5 V, you could use R2 = 1.47 khm: Then the voltage across R1 is 11.5 - 1.26 = 10.24 V R1 is 10.24 / 857 µ = 11.95 khm: