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9-3Vconverter

N

nzstranger

Jan 1, 1970
0
Kia Ora

I'm trying to assemble a pcb with no concrete values from the creators
( a past employee designed it and took the resistor values with him)

We have done most of it but I have been left to deal to the power
supply.

It has a 9 volt battery supplying it and we need 3V out, The pcb looks
like a 3pin regulator, with 2 resistor, one from the center leg to -ve,
the 2nd is in parrallel with output.

How do i work out the resistor values to drop the voltage?

Tim
 
J

John Fields

Jan 1, 1970
0
Kia Ora

I'm trying to assemble a pcb with no concrete values from the creators
( a past employee designed it and took the resistor values with him)

We have done most of it but I have been left to deal to the power
supply.

It has a 9 volt battery supplying it and we need 3V out, The pcb looks
like a 3pin regulator, with 2 resistor, one from the center leg to -ve,
the 2nd is in parrallel with output.

How do i work out the resistor values to drop the voltage?
 
N

nzstranger

Jan 1, 1970
0
That wasn't supplied, I assume something like a LM317/350 or a 78XX.
we have a 7805, could i use this and buck the voltage?
 
N

nzstranger

Jan 1, 1970
0
That wasn't supplied, I assume something like a LM317/350 or a 78XX.
we have a 7805, could i use this and buck the voltage?
 
J

John Fields

Jan 1, 1970
0
That wasn't supplied, I assume something like a LM317/350 or a 78XX.
we have a 7805, could i use this and buck the voltage?

---
When you post to these [electronics] groups it's customary to leave
enough of the message you're replying to in your reply (context) so
that everyone knows what you're referring to. Also, there's no need
to post the same message twice.

Having said that, If you go to:

http://cache.national.com/ds/LM/LM340.pdf

and look at the first circuit on the right hand side of page 1, that
might be what you're looking for. The connections don't match your
description, but I suspect your description is wrong, no offense
intended.
 
R

Rich Grise

Jan 1, 1970
0
That wasn't supplied, I assume something like a LM317/350 or a 78XX.
we have a 7805, could i use this and buck the voltage?

Before I tried that, I'd look up the pinout of the LM317; the data
sheet will give you the resistor values.

Good Luck!
Rich
 
N

nzstranger

Jan 1, 1970
0
John said:
That wasn't supplied, I assume something like a LM317/350 or a 78XX.
we have a 7805, could i use this and buck the voltage?

---
When you post to these [electronics] groups it's customary to leave
enough of the message you're replying to in your reply (context) so
that everyone knows what you're referring to. Also, there's no need
to post the same message twice.

Having said that, If you go to:

http://cache.national.com/ds/LM/LM340.pdf

and look at the first circuit on the right hand side of page 1, that
might be what you're looking for. The connections don't match your
description, but I suspect your description is wrong, no offense
intended.


You quite right about posting the mesages twice, I'm in the habit of
double clicking the post mesage button.

unfortunatly the description was right, however the ic may be a
transistor (mje 13005) in a 3 pin package, this has been used elseware
in the pcb.
could this work?
 
N

nzstranger

Jan 1, 1970
0
nzstranger said:
Kia Ora

I'm trying to assemble a pcb with no concrete values from the creators
( a past employee designed it and took the resistor values with him)

We have done most of it but I have been left to deal to the power
supply.

It has a 9 volt battery supplying it and we need 3V out, The pcb looks
like a 3pin regulator, with 2 resistor, one from the center leg to -ve,
the 2nd is in parrallel with output.

How do i work out the resistor values to drop the voltage?

Tim


Hello again

I have decided to go with the 7805, and keeping the 5 volts as most of
the chips are Cmos.
I'll use a simple voltage divider on vero board and supply the three
components that requires 3V.

I still don't really know what the last designer was thinking, and may
never find out, but I now see why he no longer works with us!

Thank you too all that have helped.
Tim
 
N

nzstranger

Jan 1, 1970
0
nzstranger said:
Kia Ora

I'm trying to assemble a pcb with no concrete values from the creators
( a past employee designed it and took the resistor values with him)

We have done most of it but I have been left to deal to the power
supply.

It has a 9 volt battery supplying it and we need 3V out, The pcb looks
like a 3pin regulator, with 2 resistor, one from the center leg to -ve,
the 2nd is in parrallel with output.

How do i work out the resistor values to drop the voltage?

Tim


Hello again

I have decided to go with the 7805, and keeping the 5 volts as most of
the chips are Cmos.
I'll use a simple voltage divider on vero board and supply the three
components that requires 3V.

I still don't really know what the last designer was thinking, and may
never find out, but I now see why he no longer works with us!

Thank you too all that have helped.
Tim
 
H

Hangglider

Jan 1, 1970
0
It sounds like the original design used an LM317 variable voltage
regulator. I just built a variable voltage battery discharger (to
cycle rechargable batteries) and It is exactly as you described...a
transistor like package with two resistors connected to the center leg.


The important formula you will need to determine the values of the two
resistors (which ultimately determine the output voltage) is: Output
Voltage is equal to 1.25V x (1 + R2/R1)

Check out this link for an incredibly easy to follow explanation of how
an LM317 works:
http://casemods.pointofnoreturn.org/vregtut/tutorial-full.html

I recommend using an R1 value of 100 ohms and calculate the necessary
R2 value based on the desired output voltage.

Just my 2 cents...

Joe

Quote from Jack Handey: It takes a big man to cry...and an even bigger
man to laugh at that man...
 
E

ehsjr

Jan 1, 1970
0
Hangglider said:
It sounds like the original design used an LM317 variable voltage
regulator. I just built a variable voltage battery discharger (to
cycle rechargable batteries) and It is exactly as you described...a
transistor like package with two resistors connected to the center leg.


The important formula you will need to determine the values of the two
resistors (which ultimately determine the output voltage) is: Output
Voltage is equal to 1.25V x (1 + R2/R1)

Check out this link for an incredibly easy to follow explanation of how
an LM317 works:
http://casemods.pointofnoreturn.org/vregtut/tutorial-full.html

I recommend using an R1 value of 100 ohms and calculate the necessary
R2 value based on the desired output voltage.

Just my 2 cents...

Joe

Quote from Jack Handey: It takes a big man to cry...and an even bigger
man to laugh at that man...
I'd use 240 for R1, and a 330 for R2 with the 317. Yields
~2.97 Volts. That should be close enough with 2 standard
resistors. If he wants greater precision, put the 330 in
series with 6 ohms to get 336 total for R2. With R1 at 240,
that computes to 3.00 exactly.

Ed
 
W

Walter Harley

Jan 1, 1970
0
ehsjr said:
I'd use 240 for R1, and a 330 for R2 with the 317. Yields
~2.97 Volts. That should be close enough with 2 standard
resistors. If he wants greater precision, put the 330 in
series with 6 ohms to get 336 total for R2. With R1 at 240,
that computes to 3.00 exactly.

False precision, I think. The LM317's nominal 1.25v reference is only
specced to be somewhere between 1.20 and 1.30v; and unless you pay a lot of
money the resistors are only 1% anyway. Even if you use a trimpot for fine
adjustment, temp drift of the LM317 is 1% over the temp range, and
worst-case load regulation another 1%. Bottom line, if the difference
between 2.97V and 3V (= 1%) matters, this is not the right chip to use.
 
E

ehsjr

Jan 1, 1970
0
Walter said:
False precision, I think.

Absolutely, in the sense I think you mean it.
The post was just addressing computing the values,
not creating a precision voltage regulator.
The issue of a better than 1% precision voltage
regulator wasn't even on the table for discussion.

Ed

The LM317's nominal 1.25v reference is only
 
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