# Voltage Regulator

Discussion in 'General Electronics Chat' started by roberto.drago, Apr 20, 2010.

1. ### roberto.drago Thread Starter New Member

Apr 20, 2010
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Hi,

We would like to convert a 9V voltage to a 5V voltage using a voltage regulator as discussed in http://forum.allaboutcircuits.com/showthread.php?t=25761. Yet out concern is that we need about 2A of current. Could the regulators be simply connected in parallel? or should there be addition circuitry in the case that the regulators do not balance the current amongst themselves.

Thanks,
Roberto

2. ### SgtWookie Expert

Jul 17, 2007
22,194
1,764
Are you considering using a 9v "transistor" battery as your source of power?

If so, look for a different kind of battery. A 2A load on a 9v battery will kill it within a few minutes, and it's likely that the battery will explode.

3. ### roberto.drago Thread Starter New Member

Apr 20, 2010
6
0
Hi,

Thanks for you reply. No actually we will be using a power supply which can handle this current.

Thanks again,
Roberto

4. ### ELECTRONERD Senior Member

May 26, 2009
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5. ### roberto.drago Thread Starter New Member

Apr 20, 2010
6
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Hi,

This is what we need!! Thanks a lot. Just one last question, what is the purpose of the NPN transistor and TTL logic base input shown on the datasheet? (see the example circuit : 5V Logic Regulator with Electronic Shutdown)

Thanks,
Roberto

6. ### SgtWookie Expert

Jul 17, 2007
22,194
1,764
There are a lot of applications schematics in the datasheet.

The transistor shown causes the output to go to almost zero volts.

You will need a large heat sink for the regulator, as 4/9ths of the total power, or 8 Watts will be dissipated in the regulator itself.

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7. ### kingdano Member

Apr 14, 2010
377
19

This. Definitely.

Please go through the calculations in the datasheet to properly size your heatsink.

8. ### roberto.drago Thread Starter New Member

Apr 20, 2010
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So should I leave the base TTL connector floating (or connected to ground) if I want a constant 5V output?

9. ### ELECTRONERD Senior Member

May 26, 2009
1,146
16
That's for a logic regulator, if you want it. To simply get 5V out, you use the equation:

$V_O_U_T = V_R_E_F(1 + \frac{R_2}{R_1}) + I_A_D_JR_2$

So since you have a 9V input, and you want a 5V output, we can find out the R2 resistor value:

Parameters:

• $V_O_U_T = 5V$
• $V_I_N = 9V$
• $V_R_E_F = 1.25V$
• $I_A_D_J = 50\mu A$
• $R_1 = 120\Omega$
Substitute and solve Algebraically:

$5V = 1.25V(1 + \frac{R_2}{120\Omega}) + 50\mu AR_2$

Now we may conclude that R2 = 358.3Ω ≈ 360Ω

Your circuit schematic for the voltage regulator is shown on page 5.

Austin

Last edited: Apr 20, 2010
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10. ### kingdano Member

Apr 14, 2010
377
19
that's well done.

*applause*

11. ### roberto.drago Thread Starter New Member

Apr 20, 2010
6
0
Hi

We found out that our power supply gives 18V, what values change in order to get the 5V from the same LM338?

Thanks,
Roberto

12. ### kingdano Member

Apr 14, 2010
377
19
You have seen the equations done - no one here is interested in doing your work for you.

I think the point of the forums is to help you figure it out, not to do it for you.

13. ### roberto.drago Thread Starter New Member

Apr 20, 2010
6
0
Thanks for all your help kingdano. Appreciate it, yet we are still students and learning electronics.

The thing is that the input voltage does not seem to be related to the output voltage. With our given supply, we are not getting the required output voltage.

14. ### ELECTRONERD Senior Member

May 26, 2009
1,146
16
According to the equation It isn't. It has to do with the voltage divider resistors. As long as you have a voltage supply within the constraint limits, and 18V does, you'll be fine. The regulator will have a voltage drop across it's input and output in order to power itself (Vref = 1.25V).

Austin

15. ### Jaguarjoe Active Member

Apr 7, 2010
770
91
With an 18 volt supply and a 5 volt output, the regulator will be dissipating 26 watts. I don't think it will survive.

16. ### kingdano Member

Apr 14, 2010
377
19
this is why you should go through the datasheet equations for calculating an appropriate heatsink for your application.

i have recently used an LM317T to regulate 36V down to 8V and found a heatsink which dissipated enough heat to satisfy the device.

17. ### ELECTRONERD Senior Member

May 26, 2009
1,146
16
With an adequate heatsink and probably using the TO-3 package, the LM338 should work with 18V. Also keep in mind that this regulator is capable of 5A, while the op is only using 2A.

Austin

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18. ### ifixit Distinguished Member

Nov 20, 2008
647
113
Hi,

Alternately, you could put a power resistor in series with the regulator to dissapate some of the un-wanted heat in the resistor instead of the regulator.

Can you calculate what value and power rating you might need?

Regards,
Ifixit

19. ### retched AAC Fanatic!

Dec 5, 2009
5,201
315
Thats right. The power has got to go away somewhere.. It doesn't all have to be in the regulator. You can continually drop before the regulator as long as you are still in range for proper regulator operation.

20. ### ELECTRONERD Senior Member

May 26, 2009
1,146
16
Indeed, that would work. He could have power dissipated through a resistor or a series of resistors in which they create an overall voltage drop. This in turn will make the LM338 voltage drop very low.

Excellent Idea!

Austin