All About Circuits Forum How do I 12v to 6v (not 5v) using 7805?
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#1
05-18-2010, 05:13 AM
 Grayham Member Join Date: May 2010 Location: Australia, Victoria Posts: 79
How do I 12v to 6v (not 5v) using 7805?

Hi,
I have been searching for this for a while but found nothing.

I have made the circuit from 12v to 5v using the circuit from this datasheet
http://www.amplebiz.com/spec/L7805.pdf
On page 18 (figure 19), but without the resistors R1 and R2.

But I want 6v instead of the stock 5v this regulator outputs.
I am new to electronics and cannot understand the formula for R1 and R2 to make this 6v.

Can someone please explain to me this formula:
VO = Vxx (1+R2/R1)+Id*R2

What is the 'Id' variable otherwise I might be able to work it out?
#2
05-18-2010, 07:30 AM
 rjenkins Senior Member Join Date: Nov 2005 Location: Sheffield, England Posts: 1,015

Id is the chip's supply current, between positive input and it's ground or common terminal.

It's in Table 4 of the data sheet.

Unlike for a regulator that is designed to be variable, the Id current is not fully specified. The datasheet just says it will not more than 6mA, so you may have a bit of trial and error.

Try a 100 Ohm resistor between the reg common and your circuit 0V. That will need 10mA to drop the one volt you need.

Then add a 470 Ohm resistor in series with a 470 Ohm preset between the regulator output and common, that will allow you to adjust the reference current between roughly 5 and 10 mA. Together with the actual Id of the chip, you should be able to adjust the preset for 6V out.
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#3
05-18-2010, 12:32 PM
 beenthere Senior Member Join Date: Apr 2004 Location: Missouri, USA (GMT -6) Posts: 15,815 Blog Entries: 10

You can use an LM317 variable regulator or get an LM7806 in place of the 7805.
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#4
05-18-2010, 01:07 PM
 Bill_Marsden Super Moderator Join Date: Mar 2008 Location: Dallas, TX (GMT-5 w/ DST) Posts: 19,044 Blog Entries: 5

You can also use a similar configuration the LM317 on the 7805 to bump its voltage up. Personally I'd buy a new regulator.

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If you have a question, please start a thread/topic. I do not provide gratis assistance via PM nor E-mail, as that would violate the intent of this Board, which is sharing knowledge ... and deprives you of other knowledgeable input. Thanks for the verbage Wookie.
#5
05-18-2010, 02:19 PM
 SgtWookie Expert Member Join Date: Jul 2007 Location: In the vast midwest of the USA; CST Posts: 22,038

Quote:
 Originally Posted by rjenkins Id is the chip's supply current, between positive input and it's ground or common terminal. It's in Table 4 of the data sheet. Unlike for a regulator that is designed to be variable, the Id current is not fully specified. The datasheet just says it will not more than 6mA, so you may have a bit of trial and error. Try a 100 Ohm resistor between the reg common and your circuit 0V. That will need 10mA to drop the one volt you need. Then add a 470 Ohm resistor in series with a 470 Ohm preset between the regulator output and common, that will allow you to adjust the reference current between roughly 5 and 10 mA. Together with the actual Id of the chip, you should be able to adjust the preset for 6V out.
That's close.

Id can vary due to temp, Vin and Iout.

If you figure that Id will be around 5mA +/- 5% or so, that works to get a "ballpark" output voltage with just a resistor from the GND terminal to GND, so roughly 200 Ohms per volt increase on the output, +/-10%.

However, if you use a 1K resistor from Vout to the ground terminal, then you have ID= 5mA +/-10%, plus 5mA +/- the tolerance of the regulator, which is roughly 4.8% up to around 750mA out - so you increase the precision somewhat - because the regulator's output voltage needs a 5mA load before regulation is guaranteed.

With the 1k resistor from VOUT to the GND terminal, then adding a 100 Ohm resistor to GND will give you roughly 6v out.

If any degree of precision is required, I'd go with an LM317 instead.
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#6
05-18-2010, 08:24 PM
 Bill_Marsden Super Moderator Join Date: Mar 2008 Location: Dallas, TX (GMT-5 w/ DST) Posts: 19,044 Blog Entries: 5

You could also put 1 or 2 diodes on the ground lead of the 7805, which would raise the voltage by .6 to 1.2V.
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"Good enough is enemy of the best." An old engineering saying, Author unknown.

General info:
If you have a question, please start a thread/topic. I do not provide gratis assistance via PM nor E-mail, as that would violate the intent of this Board, which is sharing knowledge ... and deprives you of other knowledgeable input. Thanks for the verbage Wookie.
#7
05-18-2010, 08:53 PM
 SgtWookie Expert Member Join Date: Jul 2007 Location: In the vast midwest of the USA; CST Posts: 22,038

A 1N4148/1N914 will have a Vf of about 0.7v @ 5mA current at room temperature.
A 1N4002 will have a Vf of about 6.5v @ 5mA current at room temp.

The trouble is that changes over temp will result in more variation than if a resistor were used; and you only have adjustments in rather coarse steps.
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#8
05-19-2010, 03:48 AM
 Grayham Member Join Date: May 2010 Location: Australia, Victoria Posts: 79

Quote:
 Originally Posted by SgtWookie With the 1k resistor from VOUT to the GND terminal, then adding a 100 Ohm resistor to GND will give you roughly 6v out. If any degree of precision is required, I'd go with an LM317 instead.
I tried that and it does work Vout is +6v

But I also tried ONLY a 200ohm resistor from common to GND and that produced the same results.

What is the different between using 2 resistors (like your example) and a single resistor just on IC's common to GND?
#9
05-19-2010, 04:44 AM
 SgtWookie Expert Member Join Date: Jul 2007 Location: In the vast midwest of the USA; CST Posts: 22,038

Quote:
 Originally Posted by Grayham I tried that and it does work Vout is +6v
Assuming 1k from the OUT terminal to the GND terminal, and then 100 Ohms from the GND terminal to actual ground.

Quote:
 But I also tried ONLY a 200ohm resistor from common to GND and that produced the same results. What is the different between using 2 resistors (like your example) and a single resistor just on IC's common to GND?
That's a very good question. I'm afraid that I cannot give you a very good answer (exception below), as the data in the datasheets does not indicate that you will receive any better result from using a single resistor from GND to ground, than using two resistors (for example, 1k and 100 Ohm vs a single 200 Ohm resistor).

The only exception is, that with a 1k resistor to the GND terminal and a 100 Ohm resistor from the GND terminal to ground, is that the minimum 5mA load is satisfied for guaranteed regulation. If your load was less than 5mA, then using the two-resistor method would be beneficial.

What the datasheet was attempting to imply is that the output voltage regulation was better than the GND terminal current regulation. Were that the case, it would be beneficial. However, the output voltage regulation is no better than the GND terminal current regulation, so it's a moot point.

Reading and interpreting datasheets can be difficult, and requires a lot of practice. If it's any consolation to you, I make my fair share of mistakes trying to read and interpret them properly.
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 The Following User Says Thank You to SgtWookie For This Useful Post: Grayham (05-19-2010)
#10
05-19-2010, 07:33 AM
 Grayham Member Join Date: May 2010 Location: Australia, Victoria Posts: 79

StgWookie thanks I will stick with the proper datasheet method of 2 resistors that being said.

Would you mind explaining (I still can't work it out) how you came up with the 1K and 100ohm resistor values?
What was your formula? (my 200ohm was just trial and error on Multisim).

 Tags 12v, 7805

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