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Reduce DC voltage, How? please..


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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.

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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!

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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!

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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

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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.

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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
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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)

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  • 3 weeks later...

If you look at the datasheet you'll find that the IADJ 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
VOUT = 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:

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

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


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You can also minimize the effect of IADJ 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.

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I'vejust noticed I posted the LM117 section instead of the LM317 section but this doesn't make any differance here as IADJ 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 IADJ 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 VREF is far greater.

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  • 2 weeks later...

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 
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