# DIY Power meter



## Vido (Jul 21, 2009)

Hi, I'm new here and I hope I put this in right thread.


I need some help with one DIY project (it's more practice than something useful) of mine.
Well, I would like to make small power meters for my speakers, I must say that it's not for any kind of professional use or home Hi-Fi use, it's for fun (light effect) and it should tell me approx. what power is amplifier outputting in that moment. I found one scheme that is excellent for my project, well, at least some part of it is excellent cause I need more LED's and they should blink at lower power.
I don't want classical VU meter because it measures input signal and not output power and real clip detectors are much complicated.

Here is scheme and part list:




As you can see, it needs no additional power supply because it's connected in parallel with speaker directly on amplifier.
LED's should blink when power is high enough.
D6 10W
D5 25W
D4 50W
D3 100W
D2 150W

I made this (it's my first DIY el. circuit) and it works (as far I can see at leas because 10W is very loud and I can't turn the volume up enough for now but D6 blinks a little on level I think is around 10W so my assumption it that circuit works ok).
I must say that there is no noticeable power loss and no degradation in sound quality.

Problem is that I want more LEDs and some of them to blink on lower power. I would like 5 10LED meters, two for front speakers, two for rear, and one for subwoofer.

For fronts I want approximately these power levels:

150W
75W
40W
20W
10W
5W
1W
0,5W
0,2W
0,05W


For rear I want this:

80W
40W
30W
20W
10W
5W
1W
0,5W
0,2W
0,05W


And for subwoofer this:

500W
250W
100W
50W
25W
10W
2W
0,7W
0,2W
0,05W

I'm interested must I change values of R1, R2, C1 and D1 (and what would they be for my meters) if I want to add more LEDs (and more power for subwoofer) and what should be values of other resistors for power ratings I want (if someone can explain me how to calculate resistor value for expected effect, i tried with standard ohm law equations but I confused myself more than I was before so example would be nice :R).

I imagined it to look something like this:



These are the speaker and reciver/amp if that means anything, as you can see, all speakers are DIY and power rating are approximately 50-60W for fronts, 30W for rears (smaller box) and 250W for subwoofer. They don't stand like this usually , it was just for testing is everything working.




Please help me calculate values of components for power meters I want or tell me how to do it myself couse i really want to see this LEDs blinking like I imagined:help:
Thanks in advance.


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## mechman (Feb 8, 2007)

Vido,

Sounds like a cool project. You need to have at least 5 posts to post pictures. You can put up 3 posts in this thread here and then you can add your pictures.


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## Vido (Jul 21, 2009)

Thanks mechman, I have more than 5 posts now but still can't insert pictures, maybe i can't add photos in posts that were made before i had 5 posts or something like that.

EDIT: everything fixed and pictures are working fine now


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## Vido (Jul 21, 2009)

Nobody knows? :'(
C'mon guys there must be at least one of you that knows how to make something like this.


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## tonyvdb (Sep 5, 2007)

Would you not be better off buying one of these and modifying it. Its designed to connect directly to the speaker terminals.
Here is another one that can also be modified to do the same.


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## Vido (Jul 21, 2009)

Well, I saw similar VU meters but thing is I don't have available 12V power supply and another one is too expensive and neither is first one cheap (+ shipping, I'm from Croatia, not even in EU :R).
All components for original circuit from scheme on photos cost me less than 2$.

Well, in mean time I heard from some guy that if I want for example circuit that measures double power I need R1 and R2 to be double resistance (eg. 100ohm --> 200ohm). And if I want to add more LEDs whole circuit needs to have same resistance as one with fewer LEDs. Is that true? What with D1 and C1? Should they change in some way?



Sorry if I make some grammar and spelling mistakes, my English is not very good


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## glaufman (Nov 25, 2007)

Vido said:


> I'm interested must I change values of R1, R2, C1 and D1 (and what would they be for my meters) if I want to add more LEDs (and more power for subwoofer) and what should be values of other resistors for power ratings I want (if someone can explain me how to calculate resistor value for expected effect, i tried with standard ohm law equations but I confused myself more than I was before so example would be nice :R).
> 
> Please help me calculate values of components for power meters I want or tell me how to do it myself couse i really want to see this LEDs blinking like I imagined:help:
> Thanks in advance.



Here's some info that may or may not help :huh:
First, understand that you won't be truly measuring power. You'll be measuring voltage. and not very precisely at that. That being said, I'm sure you knew that and that you're ok with it. sure, the voltage will be analogous to the power, but anyway...

There's a good chance that R1 and C1 won't have to change to do what you want. The rest of the resistors will/might, depending. I could help more if you give me the current values of all the components, but basically, the way that circuit works:
R1 C1 combine to produce sort of an averager circuit. D1 keeps you from reverse biasing the LEDs. resistors in parallel with the LEDs together set the points at which each LED fires. I'm not familiar with speaker level voltage, but, you need close to 1 volt across a given LED before it'll fire fully, so I'm not sure how low you can go with this type of detection without getting into level translation. 

Give us the values of the original components and I'll be able to tell you more.

I actually would've expected to see C1 on the other side of D1, to make a peak detect... but that's just me...


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## Vido (Jul 21, 2009)

Thanks for answering.

Values of original components (if I understood you correctly) are in first post but here they are again:


What is difference if C1 is before or after D1?
On it's web page http://www.inet.hr/~obacan/clip.htm (on Croatian but scheme is universal ) scheme is with C1 before D1 but that "printing scheme" has C1 after D1 so I'm confused right now, witch is better?


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## glaufman (Nov 25, 2007)

Well, to oversimplify a bit, having C1 before D1 should give a little faster response in the decay, having C1 after D1 should hold peaks just a little longer making them easier to see...
Just ot make sure your circuit is working, you could turn up the volume AFTER disconnecting the speakers... since this circuit measures voltage, for test purposes you don't really care whether the speakers are drawing the power or not... Let me crunch some numbers...


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## glaufman (Nov 25, 2007)

OK. I've only given this a cursory glance so far, but here's my first guess...
the circuit operates pretty much on as a basic voltage divider. Each LED sees a different divider, and therefore a different percentage of the averaged/recitifed input voltage, and therefore fires at a different input voltage. Based on this, you can adjust exactly where each fires by adjusting it's resistor, but because all resistors are in the equations for all the LEDs, changing one will affect the rest as well.
So, what's a boy to do? Don't panic...
Let's start with the easy one, your fronts...
You want the highest LED to fire at the same place. So the overall voltage divider for this one has to remain the same. i.e., we'll try to not change R2, or the sum of R3+R+...R7. Hold on, SWMBO calling... BRB...


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## glaufman (Nov 25, 2007)

OK. I'm back, and I've done some thinking. It' sgetting late, but maybe I can get you 90% of the way there and you can finish or experiment... Sorry fr the verbosity in the interest of thoroughness:
We start by sort of reverse engineering the circuit we have. We start with the power levels the LEDs are supposed to fire at:
P
150
100
50
25
10

We assume we're talking about 8ohm speakers here, so P=V^2/Z, V=sqrt(P*Z), and we see the voltages that resent those powers:
P	V(8ohms)
150	34.64
100	28.28
50	20
25	14.14
10	8.94

Now based on the original values, I see that for 9V input, you will get 2 V across the lowest LED, so I've used that moving forward. So we know that at 9Vin we want 2 V across the lowest LED:
P	V(8ohms)	8.94
150	34.64	
100	28.28	
50	20	
25	14.14	
10	8.94	2

Now that means at 14.14Vin, we should get 3.16V at the same spot:
P	V(8ohms)	8.94	14.14
150	34.64 
100	28.28 
50	20 
25	14.14 
10	8.94	2	3.16

Now we know that at 14.14Vin we need the same 2V across the 25W LED, so at that point we want the anode of that LED at 5.16 V:

P	V(8ohms)	8.94	14.14
150	34.64 
100	28.28 
50	20 
25	14.14 5.16
10	8.94	2	3.16

Carrying this through, we can populate most of this chart:
P	V(8ohms)	8.94	14.14	20	28.28	34.64
150	34.64 20.56
100	28.28 15.15	18.56
50	20 9.3	13.15	
25	14.14 5.16	7.3 
10	8.94	2	3.16 

Now we can use these voltages to show us the simple ratios we want each voltage divider equation to yield (for instance, for the equation relating to the lowest diode, Vin/Vout=9/2...
P	V(8ohms)	8.94	14.14	20	28.28	34.64	
150	34.64 20.56	0.59
100	28.28 15.15	18.56	0.54
50	20 9.3	13.15 0.47
25	14.14 5.16	7.3 0.37
10	8.94	2	3.16 0.22

So, if we take the entire string and call it Rtot, that's the sum of all resistors (R1 is small enough to be negligible here): Rtot=R2+R3+R4+R5+R6+R7. Rtot becomes the denominator in all our divider equations:
P	V(8ohms)	8.94	14.14	20	28.28	34.64 Rtot=R2+R3+R4+R5+R6+R7
150	34.64 20.56	0.59	0.59=(R3+R4+R5+R6+R7)/rtot
100	28.28 15.15	18.56	0.54	0.54=(R4+R5+R6+R7)/Rtot
50	20 9.3	13.15 0.47	0.47=(R5+R6+R7)/rtot
25	14.14 5.16	7.3 0.37	0.37=(R6+R7)/Rtot
10	8.94	2	3.16 0.22	0.22=R7/Rtot

Now we have 5 equations, but we have 6 variables. Need one more to make this solvable. Truth be told, we may not want to, as we need to choose standard resistor sizes, but since we're reverse engineering, it's easy... R2=5600. Now we have a system of 5 simultaneous equations with 5 variables. That should be easily solvable. Hopefully when you do it, it will yield values close to the original design. I think you can probably scale that appropriately to your desired situation.

That being said, I started to do it, and I see that your lowest power levels yield voltages that won't bias the diodes, so you're going to have an issue doing exactly what you want with this topology.

If the math proves what I've done works, then I can show you how to get around this problem (I think) and at least have a few LEDs that fire at a minimum output voltage, so you can have the long scale you want...


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## glaufman (Nov 25, 2007)

OK. Before someone else points it out, some much needed sleep has helped my thinking. The general idea of building up the equations should be sound, but... once a given LED fires, it no longer is part of the voltage divider per se, it should then regulate the voltage across its section of the divider... hopefully I'll be guessing correctly at 1 V, but I suppose this might depend on the LED. That means even though the lowest LED at 10W should see 2V, it won't. It'll see 1V. So I say it'll fire before it's at 2V, specifically at 1V, or 5W. I'll have more time later to figure out the rest of the levels (minimum input voltages at which each LED will fire in the original design). Sleep has also shown me that were this done around 4ohm speakers, 10W would represent 1 V at the lowest LED, so ...


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## Vido (Jul 21, 2009)

Wow, thank you very much, you wrote so much I'm gonna need few hour to translate and understand this all  but I think I know what you meant and I will try to follow your advices as soon I finish all school bussines for this week.
Thanks for giving me idea to make calculator in Excel cause I totally forgot about that nice program 

If you have something more to write, please do, any help is welcome.


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## glaufman (Nov 25, 2007)

I've refined the equations in the spreadsheet to more accurately reflect how the divider should work (LEDs regulating around 1V). I also added a section that takes the RC into account (sort of) to calculate the power each LED would fire at. It's imperfect for a number of reasons, but I suspect it's close enough for your purposes. For these purposes, I assumeed we'd be working with a pure sine wave at some frequency. That allows me to easily calculate the RMS voltage see at the diode from the peak voltage the divider string sees after the diode. Then, at that frequency, the C has a given Z and R1C can be treated as a voltage divider, giving the voltage seen by the whole circuit. Now that's used to calculate the power, assuming the speaker's impedance is 8ohms at that frequency. Here's where it gets quite arbitrary: What frequency? Since I'm reverse engineering, I played with values until the numbers got close: at 75Hz, the lowest LEDs are a little low, and the highest ones are a little high, but hopefully close enough to illustrate the procedure. I suspect the remaining innacuracy is due to the voltage divider equations not taking into account the current flow through the LED when it's firing, but that may have to wait for the next go-around... Also bare in mind that this circuit will naturally respond less to higher frequencies than lower...
Aw, well, I'm kinda enthused now so I just might take a crack at the next version pretty soon... depends how soon my boss comes to talk to me today...


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## glaufman (Nov 25, 2007)

Well, I tried adding in the LED currents... it seems I may have reached the point where the more I try to refine it the stranger the numbers get... got to let the mind rest for a while.. let me know if you try experimenting and what you come up with.


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