Reduce DC voltage, How? please..

[steveggz]

Hi all, I have a 12v 1 amp dc transformer that I would like to make 9.6v. I have never searched the internet more than I have to come up with nothing. Please help. I dont know much but im a quick learner.
I thought adding resistors to the transformer(after the transformer) would reduce voltage but it seems that resistors bring down the amprege. Then I read about diodes.
If I add silicon diodes to the line after the transformer will it reduce the voltage by 0.6 for each diode I add? If so, what kind of silicon diodes do I need and do they sell them at radio shack?
I also heard of zener diodes, seems more complicated.
The reason I would like to reduce the voltage is to use it for an old video camera that uses 9.6v.
I dont want to buy a special transformer, Id really like to modify this one.

[Staigen]

Hi there steveggz, velcome to this forum! :)

I have a 12v 1 amp dc transformer

Transformers aren't DC, maybee it is a 12 Volt 1 Amp DC Wallwart or something similar?

The easiest way to power your camera i belive, is to use a 7809 regulator between your "transformator" and camera, it have 9 Volt out, but your camera will work just fine on 9 Volt! Dont forget to put your 7809 on a heatsink of proper dimensions!

//Staigen

[steveggz]

Hi Staigen, thanks for the reply. Sounds good, I didnt come across voltage regulators when searching. Just for the sake of knowing, would connecting silicon diodes reduce the voltage by 0.6v?
And is it true that resistors dont reduce the output voltage anly the amprege?
Thanks

[Staigen]

Hi

ust for the sake of knowing, would connecting silicon diodes reduce the voltage by 0.6v?

Yes, but it is not a good way of reducing the voltage of a wallwart!

And is it true that resistors dont reduce the output voltage anly the amprege?

No, its not true, when increasing the current the voltage to the load is decreasing and vice versa and this is a very bad way of reducing voltage to a load!

//Staigen

[steveggz]

Thanks Staigen, you helped me a lot. Im going to buy a voltage regulator and heatsink.

[Staigen]

Hi

Wait a minute, at the same time buy 2 caps, one .22 uF plasic foil and one 10 uF 16V electrolytic! The .22 one is connected between the input terminal and ground and the 10 uF one is connected between output terminal and ground! Than it is ok! ;D

//Staigen

[ante]

Hi Guys,

If you put a standard diode between the gnd pin and negative rail (cathode to negative rail) on the 7809 you will get 9.6Volts. This will also reduce the heat in the regulator. 8)

[ante]

Something like this:

[Guest Alun]

You about using the LM317 instead? It has better ripple rejection and better regulation than its fixed counterparts.



Look, I'll even re-arange the formula for you. ;D

[ante]

Hi Alun,

Will work too, but more work, more parts and more expensive! ;D
Steve like it as simple as possible. 8)

[Guest Alun]

There may be more parts Ante but it isn't more expensive as resistors are much cheaper than diodes.

[ante]

Hi Alun,

Less parts is simpler, OK? A pot costs more than a diode where I live; I don’t know the proportions where you buy your stuff. The regulators are the same prize if I chose the LM7809 for 1.5A, if I settle for 1A (which would be enough here) it’s cheaper than the LM317T.

But this is splitting hairs, don’t you think? ;D

[steveggz]

Thanks for the replies, I went to radio shack today but had no luck finding the parts I needed. The employee had no idea what I was asking for. I found a site called mouser electronics, il try to see if I can find it there.
Thanks Ante, 9.6 is exactly what I need. I have couple silicon diodes so il use one like you described.

I have another question, il post it as another topic for easy searching.

[Guest Alun]

Less parts is simpler, OK? A pot costs more than a diode where I live

Pots? ;D
Why would you need to use a pot?
I know the circuit suggests one, but if you substitute it for an ordinary fixed resistor then it'll work a treat - just put the numbers in to the formula to get the value.

But this is splitting hairs, don

[ante]

Hi Guys,

Steve; I am surprised that they where not familiar with the LM7809 at Radio Shack??? I am sure you will find your parts though.

Alun; Well, debateing a circuit can sometimes be useful and yes I know you (at least to some degree) and I did not expect anything less from than you interfering as usual! ;D ;D ;D
BTW; I say the diode stays, it makes the regulator run just a little cooler and the voltage is correct too! 8)
 

[V8meathead]

Alun:  I kind of understand how you got your equation. 
I don't really get the concept behind it setting the Iadj*R2 = 0.  Could you explain this in more detail?

[Guest Alun]

If you look at the datasheet you'll find that the I_ADJ is only 100uA maximum so will make very little differance to the output.

For example:
We want a 20V regulator so using my formula:
R1 = 240
V_OUT = 20
R2 = 240(20/1.25 - 1) = 3600 = 3k6.

Now lets put that back into the origional formula to see what the actual voltage will be:

V_OUT = 1.25(1+3600/240)+100*10^-6*3600 = 20.36V

So we're only 360mV out, from the datasheet V_REF can vary from 1.2V to 1.3V and this will make far more differance than I_ADJ. Try calculating the values of V_OUT at both extreems of R_REF to see for yourself. ;D

[V8meathead]

Thanks for clearin this up I appreciate your help ill try that  :D

[Guest Alun]

You can also minimize the effect of I_ADJ by making R2 smaller but to do this R1 needs to be proportionally smaller, if you make R1 120R and R2 1k8 then the ideal output voltage will be 20.18V. The bad thing about this is the lower resistor values will increase the quesent current which will now be 10.42mA instead of 5.21mA.

[audioguru]

The more expensive LM117 uses 240 ohms for R1, the cheaper LM317 is supposed to use 120 ohms. Its current holds down the output voltage without any other load.

[Guest Alun]

I'vejust noticed I posted the LM117 section instead of the LM317 section but this doesn't make any differance here as I_ADJ is the same for both. I didn't know that R2 is supposed to be 120ohms for the LM317 I've always used 240 and had no problems at all. I understand that R2 provides the minimum load current and with the LM317 it's 10mA maximum and 3.5mA minimum so I must've been lucky. Hang on the only project I've built that doesn't have a minimum load anyway was my LM217 power supply and the LM217 specifies 5mA so I was alright anyway.

This wasn't my point though my point was that using smaller values of R2 reduces the I_ADJ error, but to do this you need to reduce R1 by the same factor. In theory if R2 was 9k ato get 20V R1 would need to be 60R so the real output voltage would be 20.09V, however as I said before this doesn't matter as the error caused by V_REF is far greater.

[MP]

The data sheet does not specify a resistor (R1) between Vout and Vref for either LM117 or LM317. The resistors used in the various examples are just that. "Examples".
The formula is Vout=1.25V(1+R2/R1)+Iadj(R2).

MP

[audioguru]

The data sheet does not specify a resistor (R1) between Vout and Vref for either LM117 or LM317.

Alun is correct. The minimum load current rating for a premium LM117 is 5mA. The R1 value is calculated 1.25V/5mA= 250 ohms. 240 ohms is the closest 5% resistor value and is shown in the application circuit on the 1st page of the datasheet. The cheaper LM317 has a min load current rating of 10mA, so needs a 120 ohm resistor to hold down the output voltage to the regulated voltage without any other load.
The minimum load current is spec'd with a voltage from input to output of 40V and is lower with lower voltages. The datasheet shows a curve of a typical load current being half with 10V from input to output when it is hot. ;D