Hypex UCD180 Gain Change?

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Hi Julf,

Ok thanks, I guess I just want to change the resistor then. I have UCD180ST's in bi-amped speakers.

I am trying to match a woofer with 86dB sensitivity to a tweeter with 91.5dB sensitivity. So I'm trying to reduce the gain on the tweeter UCD180 so that the tweeter and woofer match in level.

I have found the 3 resistors that Hypex mention in the PDF...

R10 (Rg) which seems to have 1201 written on it.
R11 (Rf) has 2201
R12 (Rf) has 1022


These numbers seem to differ to what Hypex say they should be.
The resistor numbers aren't really next to the resistors, so it's a bit confusing which is which.

I'm not sure how I calculate the value of Rg that I need, and what type of resistor I need to replace it with.

Thanks,
Matt
 
Hi Julf,

Thanks for the info regarding the resistor type, much appreciated!
Thanks for the link as well. I have read that document but I have a couple of problems.

The first is it's actually hard to tell which one is R10 because R10 R11 R12 are all close to each other - it's not clear which is which.

Secondly I don't how what value to change the resistor to.

I know that the tweeter is 5.5 dB more sensitive (so needs to be attenuated) than the woofer, but I don't know how to translate this into a resistor value.
The equation in the application note you linked to, seems to only give the amount of amplification (the modules default are set at 20 [26dB] times by default).

Anyone that can offer any insight, would be much appreciated.

Many thanks,
Matt
 
The first is it's actually hard to tell which one is R10 because R10 R11 R12 are all close to each other - it's not clear which is which.

Can't help you there, as I only have nc400's - different layout, but it should be the one with the 1201 on it.

I know that the tweeter is 5.5 dB more sensitive (so needs to be attenuated) than the woofer, but I don't know how to translate this into a resistor value.
Thus you need 5.5 dB less amplification. 5.5 dB is a ratio of 1.9, so you need to scale down the amplification by a factor of 1.9.

The equation in the application note you linked to, seems to only give the amount of amplification (the modules default are set at 20 [26dB] times by default).
Right, so what you want is to change the amplification from 20 to 10.6 (20/1.9). Using Bruno's formula suggest that changing Rg (R10) from 1.2k to 3.3k (nearest E24 value) gives a gain of 10.5, close enough (this is assuming the values of 2.2k and 1.2k for Rf and Rg respectively, as indicated by the 1201 and 2201 numbers).
 
Can't help you there, as I only have nc400's - different layout, but it should be the one with the 1201 on it.
No worries, thanks anyway. That's the thing - my values are different to the Hypex App notes. I've attached a picture - maybe I'm just not reading it right.

An externally hosted image should be here but it was not working when we last tested it.


Thus you need 5.5 dB less amplification. 5.5 dB is a ratio of 1.9, so you need to scale down the amplification by a factor of 1.9.
Thanks! Just for my future reference, how do you get to the ratio 1.9? Is it a power ratio thing, as I thought double the power was equal to 3db?

Right, so what you want is to change the amplification from 20 to 10.6 (20/1.9). Using Bruno's formula suggest that changing Rg (R10) from 1.2k to 3.3k (nearest E24 value) gives a gain of 10.5, close enough (this is assuming the values of 2.2k and 1.2k for Rf and Rg respectively, as indicated by the 1201 and 2201 numbers).
Thanks, that all makes sense now. Much appreciated. I'm just uncertain how you find the ratio in the first place.

Many thanks,
Matt
 
Can't help you there, as I only have nc400's - different layout, but it should be the one with the 1201 on it.
Ok my fault. The smd values are correct, I was reading up side down - quite hard to tell as the 2's look the same both ways.

R10 = 1.2k as it should be.

I understand using the forumla, but still trying to understand where 1.9 comes from.

Just to throw a spanner in the works, what if i decided to do this the other way round? Increasing the gain of the woofer to match the tweeter?

Thanks,
Matt
 
No worries, thanks anyway. That's the thing - my values are different to the Hypex App notes. I've attached a picture - maybe I'm just not reading it right.

As I read it, R12 is on the top, 2201 (2.2k), R10 (Rg) is below that, 1201 (1.2k), R11 below those in a vertical orientation (another 2201 - 2.2k).

Thanks! Just for my future reference, how do you get to the ratio 1.9? Is it a power ratio thing, as I thought double the power was equal to 3db?
For power, 3 dB is a factor of 2, but here we are talking about amplitude ratios, so a factor of 2 corresponds to 6 dB (or 20*log10(x), to be accurate). A doubling of amplitude results in a quadrupling of power (both voltage and current doubles), so the result is 3+3 dB.
 
Just to throw a spanner in the works, what if i decided to do this the other way round? Increasing the gain of the woofer to match the tweeter

Simple answer - don't 🙂

Increasing the gain involves reducing feedback, while reducing gain increases feedback. The more feedback, the more linear (better) the amp is, and reducing feedback might also result in the amp becoming unstable.

If you reel really brave, and want to try increasing the gain, you can just do the calculation with 0.53 (the inverse of 1.9) instead of 1.9.
 
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As I read it, R12 is on the top, 2201 (2.2k), R10 (Rg) is below that, 1201 (1.2k), R11 below those in a vertical orientation (another 2201 - 2.2k)

Yes that's how I see it too now - I was just being stupid and looking at R10 upside down.

For power, 3 dB is a factor of 2, but here we are talking about amplitude ratios, so a factor of 2 corresponds to 6 dB (or 20*log10(x), to be accurate). A doubling of amplitude results in a quadrupling of power (both voltage and current doubles), so the result is 3+3 dB.

Ah my mistake, getting mixed up with power and amplitude - thanks for putting straight.

I'm still not sure how you calculate the ratio though? Where does 1.9 come from?

Thanks,
Matt
 
Simple answer - don't
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Haha okay thanks.

Increasing the gain involves reducing feedback, while reducing gain increases feedback. The more feedback, the more linear (better) the amp is, and reducing feedback might also result in the amp becoming unstable.

If you reel really brave, and want to try increasing the gain, you can just do the calculation with 0.53 (the inverse of 1.9) instead of 1.9.

Ok cool, that all makes sense. Thanks again, much appreciated.
 
Hi,

Sorry, I'm probably just completely missing it, but I still don't understand.

5.5db is a ratio of 1.9 in relation to what?
What is the calculation?

Thus you need 5.5 dB less amplification. 5.5 dB is a ratio of 1.9, so you need to scale down the amplification by a factor of 1.9.
The default amplification is 20 times (26db).

So are you doing some sort of calculation with 5.5db and 26db, or 5.5db and 20 times? 😕

5.5db less amplification would give me 20.5db.

Wouldn't it be 26db/20.5db? which gives me approx 1.27.

I'm probably just being stupid and doing completely the wrong thing.

Many thanks.
 
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5.5db is a ratio of 1.9 in relation to what?

Difference in gain. If a signal is 5.5 dB louder than another, it has an amplitude that is 1.9 times higher.

What is the calculation?
Amplitude ratio = 10 ^ (ratio in dB / 20).

The default amplification is 20 times (26db).
So are you doing some sort of calculation with 5.5db and 26db, or 5.5db and 20 times?
The 1.9 has nothing to do with either 20 times or 26 dB - it is purely 5.5 dB expressed as a pure ratio. You then divide the default amplification (20) with the 1.9 to get the amplification you want - 10.6 (or 20.5 dB). You could also go the other way - start with 26 dB, subtract the 5.5 dB to get 20.5 dB as the desired amplification, and convert that to a pure ratio, yielding
10.6.

So 10.6 is the amplification you want to go for, either way.
 
This is where you go wrong - a decibel ratio is a logarithmic ratio. You can't just divide one with another.

26 dB is a ratio of 20
20.5 dB is a ratio of 10.6
20/10.6 is 1.9
Thanks Julf! 🙂

I knew that dB's are a logarithmic scale - I don't know why I just assumed the calculation would take care of itself! I shall go back and learn about decibels and how to calculate ratios of decibels.

The 1.9 has nothing to do with either 20 times or 26 dB - it is purely 5.5 dB expressed as a pure ratio. You then divide the default amplification (20) with the 1.9 to get the amplification you want - 10.6 (or 20.5 dB). You could also go the other way - start with 26 dB, subtract the 5.5 dB to get 20.5 dB as the desired amplification, and convert that to a pure ratio, yielding
10.6.

So 10.6 is the amplification you want to go for, either way.
It all makes sense now. Thanks for your patience, time, and help over the last day or so. Much appreciated 🙂

Hopefully other people who are less technically inclined like me will come across this thread and it'll help them too.

Now all I have to do is research how to desolder/resolder those tiny little SMD resistors without destroying anything!

Thanks again,
Matt
 
Thanks for your patience, time, and help over the last day or so. Much appreciated 🙂

No prob - my pleasure!

Now all I have to do is research how to desolder/resolder those tiny little SMD resistors without destroying anything!

With my old eyes and not-as-nimble-as-they-used-to hands, I did manage to destroy some copper traces on one of my modules. Definitely worth practising on some simpler and cheaper smd board first.
 
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