# Dropper resistor

Discussion in 'Homework Help' started by hackers99, Nov 18, 2009.

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1. ### hackers99 Thread Starter Member

Nov 18, 2009
36
0
sorry...........

Last edited: Nov 18, 2009
2. ### hackers99 Thread Starter Member

Nov 18, 2009
36
0
hey guyz plz help...

3. ### blueroomelectronics AAC Fanatic!

Jul 22, 2007
1,758
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You're going to have to learn patience. Are you writing the exam while texting?

Nov 18, 2009
36
0
sorry ya...

5. ### Papabravo Expert

Feb 24, 2006
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I am by no means sure that this is the correct approach. No one resistor value will work for all combinations of load. Where did you get the idea that it was?

If you want to split the power dissipation among several chips then I recommend an emitter follower preregulator made from an NPN, a resistor and a zener diode. You can easily extend this to provide a measure of protection from polarity reversal and loss of ground. Checkout the following which is easy to google with "Zener Diode Preregultor"

http://sound.westhost.com/project102.htm

If I was doing this I'd pick about at least an 18V Zener to make sure the 7815 had plenty of headroom.
The 7805 will draw 0.045*(15-5) = 450 milliwatts
The 7815 will draw (0.045+0.020)*(18-15) = 195 milliwatts
The Emitter Follower will draw 0.065*(30-18) = 780 milliwatts

A higher value zener will distribute the dissipation between the Emitter Follower and the 7815 more evenly if that is worthwhile to you. You might try a 22V zener, I don't think they come in 22.5V models.

Theta(JA) for a TO-220 package is 65 degrees C per watt, so limiting the dissipation to a half watt seems like a good idea. So your three devices at 500 milliwatts would be at 20 + 32.5 = 52.5 degrees C or 126.5 degrees F; warm but not finger burning hot.

Last edited: Nov 18, 2009
6. ### blueroomelectronics AAC Fanatic!

Jul 22, 2007
1,758
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Wonder when the exam ends? Isn't using the internet on your cell a form of cheating?

7. ### hackers99 Thread Starter Member

Nov 18, 2009
36
0
papa thnks man for ur ideas...

8. ### Papabravo Expert

Feb 24, 2006
10,340
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Ya mean I wasted my time giving him an answer he can't use? Can I get the A instead? Maybe an attaboy? Heavy Sigh....

9. ### hackers99 Thread Starter Member

Nov 18, 2009
36
0
is the dropper part of circuit consume more input power...........

10. ### hackers99 Thread Starter Member

Nov 18, 2009
36
0
can u tell me what is the meaning of "in addition to the current drawn by 7805"...

11. ### Papabravo Expert

Feb 24, 2006
10,340
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Which part consumes what power is up to you. I started with a minimum zener value of 18V and calculated the PD for each part. Since they were unbalanced, I suggested a zener value of 22V or 23V since the ideal value of 22.5 V probably does not exist.

You yourself said that the 7805 draws 45 mA. This current must also flow through the 7815, and there is an additional 20 mA required by the 15V parts. So we have 65 mA flowing through the 7815 and this same current must flow through the Emitter Follower.

The problem with dropping resistors is that they only work for a single set of load conditions. You never answered the question. Where did you get the idea that dropping resistors are an appropriate way to distribute power?

Nov 18, 2009
36
0

13. ### beenthere Retired Moderator

Apr 20, 2004
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No thanks - this is just strange enough as is. Your instructor probably won't look for your posts anyway.

14. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
What? Why would you want this thread deleted? I expended some effort on those responses. They contain some useful information. Fortuneately, thread deletion is not one of those things that you get to decide.