# Circuit Explanation

Discussion in 'General Electronics Chat' started by doug3460, Feb 21, 2009.

1. ### doug3460 Thread Starter Active Member

Oct 19, 2008
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0
I need some electronics 101 help understanding how this circuit works (see attached). I can't seem to get my mind around what is happening so I can do the math for the entire circuit.

My explanation is that when power is applied, the transistor is biased high, sending a current out of the base to the diode, resistor & LED. But my mind wants to say this is what should happen with an NPN because it's in an Emitter-Base arrangement - but the circuit doesn't work with an NPN.

As an aside, I also don't understand the pulldown 10k resistor - why is it connected to ground & just not in series with the emitter? Guess I don't quite get the "pulldown/pullup" concepts either, lol. But for now I just want to understand why this PNP circuit works.

As always, appreciate any help & thanks for your time.

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2. ### beenthere Retired Moderator

Apr 20, 2004
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I don't understand either circuit. Just attaching the LED to the 5 volt source through the 470 ohm resistor works fine. Both circuits are non-functional with respect to the transistor. They are just a LED, a resistor, and two diode drops to source.

3. ### mik3 Senior Member

Feb 4, 2008
4,846
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What do you want to achieve?

4. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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It's a "Trick Circuit". The LED is powered by 3 PN Junctions: E-B, D1, LED, with the 470 Ohm limiting current. A few un-needed components, the transistor isn't doing anything but providing a voltage drop.

5. ### mik3 Senior Member

Feb 4, 2008
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Yes, but if the transistor has high gain then it will blow.

6. ### ifixit Distinguished Member

Nov 20, 2008
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The 2N3906 will turn on because of ~4mA of base current. With Hfe=100(guess) then the +5V will have a load of ~400mA. The 10K provides a minimum load of 500uA.

7. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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Connect the grounds. The collector and resistor are bypassed.

8. ### doug3460 Thread Starter Active Member

Oct 19, 2008
87
0
lol, i was afraid i might have to explain more but didn't want to get long winded in the original post. oh well, here goes:

The circuit provides the load to make an bench ATX PSU latch on. It replaces the traditional sandbar resistors. Those never made sense to me because I haven't see a motherboard yet with those one them. I figured there had to be a way to provide a safe load & avoid all that durn heat! So, in my learnings so far I know a transistor "eats" .6V and a diode "eats" about the same. I've also seen that a resistor, usually somewhere between 10k & 100k, is used on the power lead to the transistor. Sooo, I built the circuit.

This circuit works dandy, I just don't know why. lol. Been running on the bench now for about 15 hours. The unit latches on just fine. The POST is solid. All the outputs are nominal. And although the transistor gets warm, I can still hold it (not the case with the sandbar resistors traditionally used).

I originally tried an NPN, but without success. That's why I was asking for some feedback because I wanted to really understand it. In my mind, I should have been able to work the mathmatics before the circuit ever hit the board. Real time measurement are:

Vs= 5.12
Voltage at the transistor base=4.91
Voltage coming out of the diode=4.26
Voltage out of R2=2.01

But I don't know how many amps the LED is pulling. I'm guessing it's the typical Ic=20mA. With Vs=4.26 & a Vf=2.1 for the LED, with an Ic=20mA, then R2 should be around 110Ω. I even tried a 220Ω but the led didn't last long. The 470 is working fine & the LED is well lit, but it shouldn't be since it should only be getting 4.6mA with that resistor.

As I originally posted, I don't understand why this circuit works. I guess I'm having a short between the headsets .

9. ### Cabwood Member

Feb 8, 2009
20
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The base-emitter junction, the diode, and the LED together drop 0.7 + 0.7 + 2.0V, leaving only 1.6V across the 470Ω resistor. That means base current of 3mA or so. The 2N3906 β drops rapidly as collector current rise above 50mA, down to about 40 at 100mA. Assuming β is 40, collector current is only about 120mA. Power is therefore around 0.6W, which is right on its limit. It will get hot, but not that hot.

What's interesting about this circuit is that there is inherent negative feedback; as collector current tries to increase, β drops, thus opposing the increase in collector current.

Also, as temperature rises, base-emitter voltage rises too, reducing the voltage across the 470Ω. So base current will decrease as temperature rises. More negative feedback.

The power dissipation in the transistor stabilises at about 0.5W.

I have no idea what the 10k resistor is for.

10. ### Audioguru New Member

Dec 20, 2007
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In think the transistor is shorted.
The emitter-base voltage is only 0.21V when the datasheet shows it at 0.85V and the transistor will conduct enough current to fry itself.

11. ### Søren Senior Member

Sep 2, 2006
472
28
Hi,

Why not include a microcontroller to PWM the load to ultra-precision

A 47 Ohm 1W resistor would be a much simpler solution and would dissipate around the same amount of heat.

12. ### doug3460 Thread Starter Active Member

Oct 19, 2008
87
0
First, thanks to all for input (except Soren, lol, I'll get to you )...

Audioguru wrote:

You were correct. I was wondering why everywhere I researched said the drop is between ~.6V to .7V & I wasn't getting near that. On the positive side, after extensive testing (read that: 3 more 3906 PNPs), I now know how to make 3 legged diodes (pun intended).

ifixit wrote:

This answer was more in keeping with the explanation I was looking for, however, along with the other replies, I knew I needed to dig further. For one, I still wasn't grasping the 10k pulldown resistor. My research showed that as a consistant number (including a couple sites that went through the math based on 100hfe transistors). So just stuck with 10k since I have no idea how to measure the current in a resistor & my reading led me to understand it's function was voltage control (to stop the in-put from "floating.")

I then went back to an NPN, this time a 3904. I simply tried different configurations (i.e., rotating the pin arrangement 120° until the darn PSU latched on, lol). The attached schematic shows what works. Here are the readings now:

Vs=5; V at the collector=3.9; V exiting the diode=3.1 R2 was changed to 68k because I changed the LED type.

Now to the questions: I still don't understand why this works.

It seems to me to be a non-standard arrangement. Nearest I've found is an Emitter-Followerer, but the load is on the wrong line.

Why is the voltage drop .5 higher than the normal .6? It's constant. The transistor checks good when removed from the circuit (several times).

Any replies/observations are appreciated. Thanks.

P.S. Oh, btw Soren, I just figured if all else fails I'll slap a CPU radiator on the suckers & just liquid cool them - probably UV green through clear pipes.

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13. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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To measure the current in the resistor on a live circuit:

1) Measure the value of the resistor out of circuit (for best results, out of circuit), in this case, if the power is disconnected, and you meter is < 0.4V on resistance test, you should get a correct reading since no PN junctions are forward biased. If working with SMD or other where removing one end of the resistor would be difficult/warranty voiding/bad, you can use the printed value, but add the % tolerance to step 2.

2) Turn on circuit, measure voltage across the two terminals of the resistor. Divide that voltage by the resistance measured in #1, and you have the current going through that resistor. e.g. 5V / 10kΩ = 500μA

The above only works with linear devices such as resistors, not semiconductors.

For a new/bright LED With 2V VF, a 220Ω to 330Ω resistor would be about right for current limiting under 20mA with a 5V Supply and two PN junctions, making the effective voltage on the resistor & LED 3.8V to 4V.

I'm at a loss as to why it would work, unless one or more of the components is faulty.

14. ### mik3 Senior Member

Feb 4, 2008
4,846
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I don't agree. As temperature increases the base-emitter voltage decreases and if it continues uncontrolled the transistor will blow due to overheat. This is the reason BJTs suffer from thermal runaway.

15. ### Audioguru New Member

Dec 20, 2007
9,411
896
Your new circuit with the 2N3904 transistor has its base-emitter diode connected directly to the 5V supply so it is blown out to smithereens (its max allowed voltage is about 0.8V). Then maybe the base shorted to the collector so the transistor is just a piece of wire from the positive supply to the LED. The current is low through the LED.

16. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
Is there some point to all this? A resistor in series with the LED is much more straightforward. 15 yards for misuse of a semiconductor.

17. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
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I am guessing the purpose is to dissipate 2.5 Watts without actually dissipating 2.5W, or spreading out the dissipation so no single item "feels hot".

Maybe use a small filament light bulb instead of the LED/resistor/etc? You get a free workbench light with your power supply.

18. ### doug3460 Thread Starter Active Member

Oct 19, 2008
87
0
lol. An update of The Point:

beenthere wrote:

1. Provide a load to the +5V rail that will allow an ATX bench PSU to latch on.

2. The rules .

• No sandbar resistors.
• Heat output has to be minimal; the circuit can survive independently.
• You get two (2) resistors; one (1) transistor; (1) diode. Their specifications are unknown.
The LED is trivial - it's just an indicator it's working (pretend I can't see the fan & also pretend I didn't wire an LED into the POST line - which I did both, but that's not The Point, lol).

Penalty declined. LOL. I know this can be done. I just am trying to figure out how.

P.S. thanks thatoneguy, I knew Ohms was in there somewhere. For the record, that particular resistor is 519mA (I measured it after your post - thanks: simple, clear, easy to remember. I like it.

Last edited: Feb 22, 2009
19. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
Who is imposing these restrictions? If the supply requires 0.5A minimum to operate, you have to dissipate 2.5W. 2.5W generates the same amount of heat, no matter how exotic (or random) the circuit is. If you don't want all the heat in one place, you can spread it around, as you suggested. With 4 small parts, you will be lucky to find a combination that will not destroy one of them. If they survive, they may not seem to radiate a lot of heat if you separate them physically, but you will probably burn your finger if you touch any one of them.

How do you know this can be done?

20. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
What voltage is across the resistor? For half an amp to be going through it, less than 1V would have to be across it before a 1/2 W resistor fried since P=I*V. Otherwise the resistor is less than 1Ω, since P= I² * R

That alone should be nearly enough to keep the power supply latched.

--ETA:

If I get two resistors and a transistor (all unspecified), Can a TO-3 Transistor such as a 2N3055 be used? It won't be a blocky sand resistor that's hot, just a chunk of metal instead.

Last edited: Feb 22, 2009