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  #1  
Old 05-18-2010, 05:13 AM
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Grayham Grayham is offline
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Default 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?
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Old 05-18-2010, 07:30 AM
rjenkins rjenkins is offline
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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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Old 05-18-2010, 12:32 PM
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You can use an LM317 variable regulator or get an LM7806 in place of the 7805.
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Old 05-18-2010, 01:07 PM
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You can also use a similar configuration the LM317 on the 7805 to bump its voltage up. Personally I'd buy a new regulator.

If you don't understand what I'm talking about, ask.
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Old 05-18-2010, 02:19 PM
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Quote:
Originally Posted by rjenkins View Post
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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Old 05-18-2010, 08:24 PM
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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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Old 05-18-2010, 08:53 PM
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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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Old 05-19-2010, 03:48 AM
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Quote:
Originally Posted by SgtWookie View Post
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?
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Old 05-19-2010, 04:44 AM
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Quote:
Originally Posted by Grayham View Post
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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  #10  
Old 05-19-2010, 07:33 AM
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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).
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