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How to get high voltage and high current with LM317

I am working on a design which I need about 5 amp and 8 V output. I decided to use the LM317 for this purpose. I followed the circuit diagram in the datasheet as attached. As seen in my circuit diagram as attached, it was difficult for me to get up to 5 amp and 8 V. I tried tweaking the different values, but when the voltage increase, the current will decrease. 

Any help from anyone with related experience will be much appreciated.  


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Message 1 of 8

"За деньги маму родную продам,"- сказал один из героев фильма (республика "Шкид") про беспризорников ...

-Редактирование моделей в вашем случае допускается?

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Message 2 of 8

Yes please, you can edit the model.

Thank you.

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Message 3 of 8

Hi Gbouna,



You may post the image (PNG format) of the circuit that you constructed so that others or anyone with related experience but do not have Multisim (Desktop) application might be able to provide any help which you will appreciate much. What is your input voltage? How did you wire the circuit? I (and many others) can't see it because I'm/we're only using Multisim Live (Free Subscription).


Since you only need a fixed 8 V output, you may also consider using the 7808 or LM340-08.



Best regards,

G. Goodwin


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Message 4 of 8

Ниже 10 вольт на нагрузке не получить, так как выполняется закон Ома 2 ом * 5 а = 10 в
Чтобы получить 5 а, увеличиваем



еапряжение источника до 12 вольт и Bf модели Q1

Message 5 of 8


Hi tipa,



The input voltage is the first design parameter that came into my mind when I read Gbouna's (the author's) description of the behavior of the circuit, especially since I can't view the circuit created in Multisim. It's noteworthy that current-boosting the popular three-terminal IC voltage regulators increases the overall minimum dropout voltage. However, when I asked "What is the input voltage?", it's more than just about the DC value. If the circuit is intended just for simulation in order to verify the concept then a DC source is enough. If the circuit will be employed in real-world application then some other factors should be considered, for example, ripple voltage if the input is derived from AC source. 


I appreciate that you included an image of the circuit constructed in Multisim.



Best regards,

G. Goodwin


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Message 6 of 8

Спасибо за 1 кудо!

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Message 7 of 8


The 2 Ω load resistor can be replaced by a 1.6 Ω resistor in series with a rheostat.


With an input voltage of 12 V, the 50 kΩ output voltage adjustment rheostat is too high and will cause improper operation of the circuit. The ratio of the full resistance of this rheostat and 120 Ω should be carefully determined to keep the circuit within operational limits.


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Message 8 of 8