# How does this circuit work?

Discussion in 'General Electronics Chat' started by jkcobain, Apr 9, 2016.

1. ### jkcobain Thread Starter Member

Jun 8, 2014
40
5
I found it in a forum. It reduces the voltage, but I don't know how it works. I just don't get it.

Also, if I do that, how can I calculate voltage, current, and power in the transistor and resistors?

I need to reduce voltage in order to connect the lm7824, and most important, the lm7805. My source (after a transformer, capacitors, and a bridge) is about 36v dc. How could I achieve that without using circuits like lm2678 (they're way too expensive!).

Any help will be appreciated!

2. ### tracecom AAC Fanatic!

Apr 16, 2010
3,879
1,396
R1 and R2 form a voltage divider that sets the bias on the transistor; the voltage at the base of the transistor will be somewhere between 0V and 40V depending on the values of R1 and R2.

rudha13 likes this.
3. ### Lestraveled Well-Known Member

May 19, 2014
1,957
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Another name for this circuit is a voltage follower. As Trace said, R1 and R2 are a voltage divider that sets the base at a voltage. The emitter will follow that voltage.

4. ### jkcobain Thread Starter Member

Jun 8, 2014
40
5
I understand the voltage dividing, but what I don't get is why is the emitter "following" the voltage?

5. ### jkcobain Thread Starter Member

Jun 8, 2014
40
5
Ok, thanks for providing the name. I'm watching a video about it, I hope I get it after that

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6. ### #12 Expert

Nov 30, 2010
16,704
7,351
As I remember, the 7824 chip is good to 40 volts of input, so you aren't going to break it with 36 volts.
As for the voltage divider, 36 times R2/(R1+R2) is the base voltage.
The current in those two resistors should be 1/10th of the collector current.
That is what I will call the bias current.
Having established a voltage for the base, the emitter voltage (output voltage) will be that bias voltage minus about 0.6v to 0.7v.
The power dissipated by each resistor will be the bias current times the voltage across each resistor.
The power dissipated by the transistor has 2 parts. The first part will be the base to emitter voltage (0.6v to 0.7v) times the base current. This part is usually so small you can ignore it. The second part is your (36 volts minus the output voltage), times the load current.

All these things are derived from Ohm's Law and Watt's Law.
E=IR and P=IE

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7. ### tracecom AAC Fanatic!

Apr 16, 2010
3,879
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I have used LM78xx regulators in a cascaded fashion. For example, you could put your 36V into an LM7824, and the output from that into an LM7816, and the output from that into an LM7812, etc. Just remember that each regulator is carrying the load for all the others that are "downstream" from it.

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8. ### wayneh Expert

Sep 9, 2010
12,389
3,245
Did it click? The strategy relies on the cool property of a transistor, that it will be non-conducting if the base voltage is not 0.7V above the emitter voltage. It will switch off at any voltage below that. Above that voltage, it conducts like crazy and offers a low resistance and voltage drop.

So, if you set the base at, say, 20V, the transistor looks like a wire - highly conductive - until the emitter voltage reaches 19.3V. At that point, the transistor starts to shut off. Whatever the load does, this circuit attempts to hold the output emitter voltage at that knife's edge where it just 0.7V below the base voltage.

Note that the base current is not zero and will distort the calculations of the set point.

Last edited: Apr 10, 2016
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9. ### jkcobain Thread Starter Member

Jun 8, 2014
40
5
#12, and wayneh. Thank you very much! your explanations are very clear. I get it now.

And as for the problem I have, I think I have no other choice than using DC/DC converters. Because I read that the 78xx series require a voltage not much higher than the output voltage, the datasheet says that the maximum voltage is 35/40v, but in the practice, that would require a huge heat sink (at least for the LM7805, not sure about the LM7824). Using the reducer in the image above, will result in a lot of power disipation required by the transistor (I need about 2A in the outpput), and also by the resistors.

ebay has crazy prices and I can't understand why:

The LM2678 is around 4-5 USD
But a power supply using LM2596 (already designed, and mounted incluiding capacitors and inductors) is around 1 USD

That confuses me a lot. Because I first searched for LM2678-5/24 and they're really expensive, and in my city everything is even more expensive than the items in ebay, but then I found the LM2569 is much cheaper, but I don't know why there is such a difference in the prices. Anyway, I will try to find a cheap DC/DC converter here in my city (where there are not much electronics stores).

Thank you a lot, guys!

10. ### wayneh Expert

Sep 9, 2010
12,389
3,245
Yes, it's unlikely you will save money by building your own, especially if your time as any value at all. The only reason to build, in my humble opinion, is when you need something sufficiently different from what's already available.

And your strategy to use a DC-DC converter is the right one. Those devices can provide an efficiency over 90%. You can't get anywhere close to that with a "linear" solution that burns off excess voltage in a resistor, transistor, or regulator.

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11. ### #12 Expert

Nov 30, 2010
16,704
7,351
One of my rules for this site is that I refuse to design anything you can buy as a finished product because that product will be cheaper than the parts required to build it, and it will be designed and produced with every consideration for safety and reliability. I can't see any point in starting from scratch to design and build something you can get cheaper and better without spending my time on it.

12. ### wayneh Expert

Sep 9, 2010
12,389
3,245
Make-or-buy is one of the most basic business management questions and it's almost always made wrong. It seems to be a reliable and predictable human failing.

13. ### jkcobain Thread Starter Member

Jun 8, 2014
40
5
I wanted to build it because it would be the supply for a PCB with more components in it, and I didn't want my board to be "in parts" (or however it's said in english). I mean, I didn't want my board to have its supply apart, but I think I will have to chose between aesthetics and price... and I think price wins this time.

14. ### rudha13 New Member

Sep 6, 2015
14
0
wow! that's a good video..! thanks for sharing it..!