Voltage regulator (7805) and voltage divisor problem - Any good ideas? - Electronic Circuits Projects Diagrams Free

johankj · 9th September 2007, 05:25 PM

I'm having a problem with my voltage divisor network, which is basically set up as in the drawing (simplified).

I want to read the voltage across R2 (Vr), but allas...

My problem is that there is a resistance across the regulated output of the 7805 and ground of roughly 4k. My R1 value is 100k, and the R2 is a 100k variable resistor.

When assembled, R1 and R2 reads ~50k, I suspect because of the lower resistance across the 7805.

Is there a way to fix this? Or even better, can I completely ignore this?

Papabravo · 9th September 2007, 05:56 PM

If you want to measure the voltage just do it. It sounds from your description that you want to measure the resistance. You typically can't do that with the circuit powered up, since resistance is measured by passing a small current through a circuit and measuring the voltage drop.

johankj · 9th September 2007, 06:40 PM

Thanks for the reply.

I am trying to measure the resistance, yes, but indirectly by measuring the voltage (Vr).

So when I power up the 7805, it doesn't matter that there is a a 4k resistance across it? I've enclosed another drawing that shows what I am measuring the resistances to be, where R3 represents the resistance across ground and output of the 7805.

Leftyretro · 9th September 2007, 06:49 PM

Quote:

Originally Posted by johankj

Thanks for the reply.

I am trying to measure the resistance, yes, but indirectly by measuring the voltage (Vr).

So when I power up the 7805, it doesn't matter that there is a a 4k resistance across it? I've enclosed another drawing that shows what I am measuring the resistances to be, where R3 represents the resistance across ground and output of the 7805.

Not sure how you determined that there is a 4k resistance in parallel with your R1/R2 string, however the 7804 does consume some current for it's overhead use. This really does not change the calculated current through your R1/R2 series string as the regulators job is to keep the voltage constant to this string regardless of any other load on it's output, up to it's rated 1 amp max current rating.

Lefty

ericgibbs · 9th September 2007, 06:51 PM

[quote=johankj]I'm having a problem with my voltage divisor network, which is basically set up as in the drawing (simplified).

I want to read the voltage across R2 (Vr), but allas...

My problem is that there is a resistance across the regulated output of the 7805 and ground of roughly 4k. My R1 value is 100k, and the R2 is a 100k variable resistor.

When assembled, R1 and R2 reads ~50k, I suspect because of the lower resistance across the 7805.

Is there a way to fix this? Or even better, can I completely ignore this? Yes![/QUOTE]

Hi,
Its quite normal,nothing to worry about.

[ its a little feedback from the 'flux capacitor'

]

johankj · 9th September 2007, 07:32 PM

Quote:

Or even better, can I completely ignore this? Yes!

Thanks for the input guys. :-) I'll just assume that it's nothing to worry about.

Quote:

Not sure how you determined that there is a 4k resistance in parallel with your R1/R2 string

As I measured the resistance across the variable resistor R2, it had dropped from 100k to ~50k after assembly. After a bit of head scratching I found there was a 4k resistance across the output of the 7805, measured with my trusty multimeter.

Quote:

its a little feedback from the 'flux capacitor'

:-P LOL...

Hero999 · 9th September 2007, 08:25 PM

It's exactly the same as the formula for the LM317.

$V_{OUT} = V_{REF}(1+\frac{R2}{R1})+I_{ADJ}\times R2$

This can be re-arranged to:
$R2 = R1 \frac{V_{OUT}-V_{REF}}{{R1} \times I_{ADJ}+V_{REF}}$

The difference is:
Vref is your regulator's output voltage.
Iadj is the quiescent current.
Your problem might be due to the fact you've ignored Iadg which is a lot bigger on an LM7805 than an LM317.

johankj · 10th September 2007, 01:22 PM

Quote:

Originally Posted by Hero999

It's exactly the same as the formula for the LM317.

$V_{OUT} = V_{REF}(1+\frac{R2}{R1})+I_{ADJ}\times R2$

This can be re-arranged to:
$R2 = R1 \frac{V_{OUT}-V_{REF}}{{R1} \times I_{ADJ}+V_{REF}}$

The difference is:
Vref is your regulator's output voltage.
Iadj is the quiescent current.
Your problem might be due to the fact you've ignored Iadg which is a lot bigger on an LM7805 than an LM317.

Jupp, you're right, I've completely ignored Iadj. I had it down to Vout/R2 = Vref/(R1 + R2) => Vout = Vref(R2/(R1 + R2)).

(Edit: Corrected my own formulæ)

neon · 24th August 2008, 11:28 AM

Quote:

Originally Posted by johankj

I'm having a problem with my voltage divisor network, which is basically set up as in the drawing (simplified).

I want to read the voltage across R2 (Vr), but allas...

My problem is that there is a resistance across the regulated output of the 7805 and ground of roughly 4k. My R1 value is 100k, and the R2 is a 100k variable resistor.

When assembled, R1 and R2 reads ~50k, I suspect because of the lower resistance across the 7805.

Is there a way to fix this? Or even better, can I completely ignore this?

you will get 5v at the output not at the divider. can it be fixed no way. but if you use lm117 you may change the output from 1.2 v to close -3v of the input by changing the reg. bias at the adj. pin.

ericgibbs · 24th August 2008, 12:33 PM

Quote:

Originally Posted by neon

you will get 5v at the output not at the divider. can it be fixed no way. but if you use lm117 you may change the output from 1.2 v to close -3v of the input by changing the reg. bias at the adj. pin.

This post is a year old.!

Willbe · 24th August 2008, 02:48 PM

Quote:

Originally Posted by johankj

I'm having a problem with my voltage divisor network, which is basically set up as in the drawing (simplified).

I want to read the voltage across R2 (Vr), but allas...

My problem is that there is a resistance across the regulated output of the 7805 and ground of roughly 4k. My R1 value is 100k, and the R2 is a 100k variable resistor.

When assembled, R1 and R2 reads ~50k, I suspect because of the lower resistance across the 7805.

Is there a way to fix this? Or even better, can I completely ignore this?

You know, you are defeating the purpose of having a regulator (it's supposed to give ~ 0 ohm output impedance) by putting these high value resistors on the output.

Hero999 · 24th August 2008, 08:02 PM

I didn't realise that he wanted to take the power from across the resistor - what a silly idea.

I suppose I resurected this thread by linking it in another thread.

neon · 26th August 2008, 06:58 PM

Quote:

Originally Posted by Hero999

I didn't realise that he wanted to take the power from across the resistor - what a silly idea.

I suppose I resurected this thread by linking it in another thread.

SOMETIMES IT PAYS TO READ OR AT LEAST UNDERSTAND.

Hero999 · 26th August 2008, 07:12 PM

neon · 27th August 2008, 10:25 AM

Quote:

Originally Posted by Hero999

it got your attention din't it. served the porpuse correctly.

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9th September 2007, 05:56 PM	(permalink)
Papabravo	If you want to measure the voltage just do it. It sounds from your description that you want to measure the resistance. You typically can't do that with the circuit powered up, since resistance is measured by passing a small current through a circuit and measuring the voltage drop.

9th September 2007, 06:51 PM	(permalink)
ericgibbs	[quote=johankj]I'm having a problem with my voltage divisor network, which is basically set up as in the drawing (simplified). I want to read the voltage across R2 (Vr), but allas... My problem is that there is a resistance across the regulated output of the 7805 and ground of roughly 4k. My R1 value is 100k, and the R2 is a 100k variable resistor. When assembled, R1 and R2 reads ~50k, I suspect because of the lower resistance across the 7805. Is there a way to fix this? Or even better, can I completely ignore this? Yes![/QUOTE] Hi, Its quite normal,nothing to worry about. [ its a little feedback from the 'flux capacitor' ] __________________ Eric "Good enough is Perfect" PIC tutorials: Gramo's: www.digital-diy.net/ Bill's: www.blueroomelectronics.com/ Last edited by ericgibbs; 9th September 2007 at 06:55 PM.

9th September 2007, 08:25 PM	(permalink)
Hero999	It's exactly the same as the formula for the LM317. $V_{OUT} = V_{REF}(1+\frac{R2}{R1})+I_{ADJ}\times R2$ This can be re-arranged to: $R2 = R1 \frac{V_{OUT}-V_{REF}}{{R1} \times I_{ADJ}+V_{REF}}$ The difference is: Vref is your regulator's output voltage. Iadj is the quiescent current. Your problem might be due to the fact you've ignored Iadg which is a lot bigger on an LM7805 than an LM317. __________________ I also post at the following sites: http://www.stop-microsoft.org http://www.heated-debates.com Screen name: Aloone_Jonez And http://www.silicontronics.com, same screen name as here.

24th August 2008, 08:02 PM	(permalink)
Hero999	I didn't realise that he wanted to take the power from across the resistor - what a silly idea. I suppose I resurected this thread by linking it in another thread. __________________ I also post at the following sites: http://www.stop-microsoft.org http://www.heated-debates.com Screen name: Aloone_Jonez And http://www.silicontronics.com, same screen name as here.

26th August 2008, 07:12 PM	(permalink)
Hero999	__________________ I also post at the following sites: http://www.stop-microsoft.org http://www.heated-debates.com Screen name: Aloone_Jonez And http://www.silicontronics.com, same screen name as here.