Hi All I have a TEAC A-H500 amp, a TEAC PD-H500i CD player and music stored on a WDTVLiveHub. CD and Hub are connected to the amp using Cambridge Audio Pacific interconnectors and....well....the volume is so darn loud. Even at this level (see below) I could not have it this loud with the wife and kids in the house……..and there is a long way round to control the volume (0 is loudest and Infinity is no sound). I also have the TV hooked up to the amp (With a YouView box connected to the TV) and as you can turn the volume down on the YouView box, I am then able to control the volume when playing through the amp really well. i.e. a good listening volume might be 23 on the amp when using YouView through the TV. My concern is that someone watches YouView with volume on 23 on the amp, then switches to CD and blows themselves apart in a sonic wave (ok, so an over exaggeration) or this unduly knackers my new speakers (MA Bronze 5) Given the volume is this loud across multiple inputs, but not all of them.....what's the best way to deal with this without it costing a small fortune (to me)? I have seen the Rothwell inline attenuators at ~£40 a pair, are these the only (cheapest?) option? And how do you work out what level of attenuation is required? -10dB, -15, -20? Any assistance gratefully received! Many thanks Jon
Have you tried plugging the CD into the CD input and using the CD direct button, the manual seems to suggest that this is slightly less sensitive (95db vs 100db) and so should be more than half as quiet (sound halves in volume with every 3db) ? Obvious question (sorry if I am teaching granny to suck eggs) - you don't have the CD plugged into the Phono input ? The Phono input is significantly more sensitive than the other line level inputs and would cause your issues. Ai Direct Play Function - Teac A-H500 Owner's Manual [Page 9]
No, they are fair enough questions I have tried CD direct and it is not noticeably different in volume at all. Also as per below, nothing in phono....
I experienced a similar problem and YES... attenuators were the answer. In my case, -12db was the perfect balance. You can read about my testing process here - For What It's Worth - LP vs CD SPL Generally speaking, I normally listen at about 10 o'clock to 11 o'clock on the Volume Dial with my turntable. But the CD Player was as loud or louder at 9 o'clock on the dial. Also, the CD sounded very harsh and grating, not pleasant at all. I determined I needed about 10db of attenuation, but -12db was the closest I could come, and that worked out fine. The CD Player is balanced and sounds very good now. Rather than hang the Attenuators off the back of my amp or CD Player, which I perceived would put stress on the connections, I put the Attenuators in-line. I was able to buy some short 18" RCA cables and some Couplers and that allowed me to connect a standard 1m RCA cable to the between the attenuators and the amp. (Black = 18", Silver = 1m) Steve/bluewizard
There is a slight error in your interpretation of the spec sheet... The figures you quoted were for signal noise ratio not power gain. What that was saying is that using the CD direct connection some internal switching is avoided and the s n ratio is a phenomenal 100db. If the poster is capable of soldering and is prepared to spend 2 € or 2£ he can roll his own voltage divider for peanuts.. cut break the rca cable and you will have two sets of cables each with a braided outside and probably red and black coated cable inside. solder a 1kohm resistor to mend the break in each case and solder a 220 ohm resistor between the braid and the other resistor. Cover the solder joins with insulating tape. Then reconnect the cables with the 220 ohm resistor closer to the amplifier. This combination will reduce your signal voltage amplitude to 20% if the original level. You could also use 10k and 2.2k resistors same effect.. Metal film resistors from maplin are fine...
I found this Russ Andrews link, which has a nice table of source voltage and amplifier sensitivity. According to this, and the manuals of my kit (CD 2V RMS, Amp 180mV) I would need <-16dB and > -11dB.....probably around -15dB really.
Finding out the output impedance of a cd player seems difficult, so I am assuming that the about 400 ohms, typical of an IC non power stage... So the 1k resistance is about the minimum I would use and 10k and 2k2 would be better...
That's the kind of comment that could almost come out of the product marketing material for most attenuators I use these when I need fixed attenuation: https://www.amazon.co.uk/s/ref=nb_sb_ss_i_1_14?url=search-alias=electronics&field-keywords=rothwell+attenuators&sprefix=Rothwell+atten,electronics,145&crid=3J8UURQJ05GXB&rh=n:560798,k:rothwell+attenuators
Thanks all for the responses. I remembered I had some Cambridge Audio Arctic cables too and found that these unscrew really easily. Might give the diy method a go with my dad who has always been into tinkering with electronics a lot more than myself.
You don't want to use a SINGLE fixed Resistor, rather you want what is known as an L-Pad, which is two resistors in the shale of an "L", one in series and one is parallel. This will create an impedance matched attenuator. That is, with the attenuator in the circuit, the driving source will still see the same impedance at the load. Let me explain this using a speaker driver attenuator. If we have an 8 ohm Tweeter and we want to reduce the output by -6dB then we would use a 4 ohm resistor in Series, and an 8 ohm resistor in Parallel with the Tweeter. The 8 ohm resistor in Parallel with an 8 ohm Tweeter yield a combined impedance of 4 ohms, then we add the 4 ohms in Series and the combined impedance of the complete network is 8 ohms. This is necessary in a Tweeter because the Tweeter is connected to a Crossover, and for the crossover to work, it must feed a relatively constant impedance, at least at the crossover frequency. Most amp inputs are in the range of 10k ohms to 47k ohms. If you put a single resistor in series, then you are creating a voltage divider with the Input Impedance of the amp. So, you have to ask, what series resistance would you need to cut the signal by 50% which is about a -6db drop? If we assume the input is 47k ohms, which is very common. Then, with a simple series resistor, you would need about a 47k ohm resistor. If we go back to the speaker example, we seen that half the signal is lost in the 4 ohm Series Resistor, and the other half is made available to the tweeter. That's for a -6db attenuation. Here is a LINK to an L-Pad (2 resistor, impedance controlled) Attenuator, down at the very bottom of the page - Crossover Design Chart and Inductance vs. Frequency Calculator(Low-pass) Using that Calculator, to get a -6db reduction and assuming a 47k ohm input impedance on the amp, for an Impedance Stable L-Pad we would need - 23,444 ohms in Series 47,224 ohms in Parallel Just like the Tweeter, in this case, we have 47k ohms in Parallel with 47k ohms input on the amp, which is a combined impedance of 23.5k ohms Then we have roughly 23.5k ohms in Series with that Parallel combination, which brings the Impedance back up to 47k ohms. However, if you simply use a 47k ohm resistor in Series, to get the an 50% reduction in Signal, then the combined impedance as seen by the source becomes 94k ohms. That may be OK .... or not. If there are any frequency related components in the signal chain, then their active frequencies will shift with the load impedance. Going back to the Tweeter example, if you substitute a 16 ohm tweeter for an 8 ohm tweeter, the Crossover will cross at a different frequency. If we assume the Crossover is designed for 8 ohms and a 3000hz crossover, then if you substitute a 16 ohm tweeter the crossover shifts to 1500 hz. The saving grace is that the power in a Source to Amp circuit is very very small, so small size resistors can be used, likely 1/4 watt would be more than fine. And if you have sufficient skills, and you have RCA cables/connectors that can be opened, then if you are careful, you can solder both resistors into the connector. Going back to the Calculator Link I provided, if you want -12b of attenuation, then you would need - 35,194 ohm Resistor in Series with the Amp 15,776 ohm Resistor in Parallel with the Amp -9db Attenuation assuming a 47k ohm input impedance, then you would need - 30,323 ohm Resistor in Series with the Amp 25,847 ohm Resistor in Parallel with the Amp Of course, these are calculated values, you can simply shift to the nearest standard size resistor, and if you can get the resistors in 1% tolerances, so much the better. Steve/bluewizard
Completely off topic but a question for the OP. Did you find the plugs on those yellow CA cables to be an incredibly tight fit and if so did you find a way to solve the problem? I have 3 sets, tried them on multiple units over time and found them all to be almost unusable. In one instance they even tore out a socket when trying to unplug them!
This explanation will help the posters Dad.. With respect, I think this is over eleborate explanation. The output impedance of a cd player is probably in the range of a few hundred ohms. The input impedance of the amplifier is listed as 47k. All that is needed is to reduce the signal level to about a quarter or a fifth of its initial value,so that the amplifiers volume control knob is closer to maximum when listening. A single 47k resistance in series ( in the path) of the signal would reduce the voltage by 50% or half. A 100k one would reduce it to 33% , a 150k to 25% etc. However such sizes will also introduce noise. A better solution is to use a resistance in series R1, of modest size and another one in parallel, R2 to ground. The smaller the resistance of the parallel one, the less voltage gets into the amplifier, and the size of the series resistance determines the load as seen by the cd player. In maths terms the voltage into the amplifier is approx signal voltage from cd* ( R2/( R1+R2 )). More precisely, if the cd player has an output resistance of X ohms and the amplifier has a resistance on input it Y ohms then in the formula above R1 is increased to( R1+ X) and R2 is reduced to Rnew as a result the result of the calculation 1/Rnew =(1/R2+ 1/Y) With reasonable values for R1 and R2, the effects of X and Y can be ignored. So I am suggesting that R1 be above 1000 ohms and less than 22 k and R2 be a quarter that value... The parallel resistor must be electrically closer to the amplifier..
Does anybody else find it strange that a TEAC CD player generates too much output for a TEAC integrated amp that appears (on the surface) to be from the same or similar product range?
Yes . I do find it odd. There are or were standards for these signal levels back in the day. My recollection was a peak value of 0.775v RMS into 47k for full amplitude output.
I believe the formula for dB relative to voltage is - dB = 20 Log(Vo/Vi) Vi is the applied Voltage. Vo is the Voltage output of the circuit. Instead of Voltage you can use the Ratio of Output to Input. For example, if you want to reduce the signal by 50%, then Vo/Vi = 50%/100% or 50/100 or 0.5 dB = 20 Log(50/100) = -6.0206 dB or simply -6dB. In my case, I used an SPL (loudness) Meter and measured both CD and Vinyl Album in order to determine how much louder the CD was than Vinyl Albums. In my case, it was roughly 10db. So, the nearest attenuator was 12dB, and that worked out perfect for me. By the way, if you are a bit Math-Phobic, finding the Log is easy enough. If you have a calculator, it is as simply as pushing the [LOG] button. 0.5 [LOG] x 20 Steve/bluewizard
This is one of those "but " questions. The Bel scale is the log to base 10 of a number. So a 1000 is 10 to power 3, so it's Bel rating is 3. Because the Bel is pretty large the deciBel was introduced there are 10 deciBel in a Bel.. so the example quoted is 3 Bel or 30deciBel. Voltage and voltage ratios should be expressed as 10log( V) in dBs Now the voltage ratio is still 10log(V2/V1) in dBs . .. but if one wants to express a power gain or power ratio and the power in a voltage scales as the square of the voltage then the power in dBs is 20log(V2/V1). .. it's a little more convoluted than that, if one is measuring the power gain in two parts of a circuit as a voltage signal developed across a resistance load of R in ohms... Take for instance our example of the cd player where the signal voltage going into the amplifier is 2 volts and developed accross the 47k input resistance. Then the input signal power is 2*2/47,000 watts or 85 microwatts .P=Vsquared/R. The output power into the loudspeaker might typically be 30v and developed over 8 ohm speaker or 30*30/ 8 watts or 110 watts .The signal voltage gain of the power amplifier is a measly 15 or 11 dB , but the signal power gain is around (110/ 85*10^-6.).. Or 10^8. Which could be expressed as 80deciBel The exact formula for signal power gain in dBs for a signal of V1 developed across R1 as input and then output as V2 accross R2 is 20log(V2/V1)+10log(R1/R2)..
dB for Relative Power is ... dB = 10 Log (Po/Pi) dB Relative to Voltage is ... dB = 20 Log(Vo/Vi) Steve/bluewizard
Not quite . The relative power formula is absolutely correct, The resistance of the respective loads needs to be included as per my formula in the version with voltage. The simpler formula is applicable only when the load and input resistance are the same. ... A situation which often occurs inside a Comms system
Well here is a prototype knocked up with a block and cheap resistors (saves faffing with soldering for the time being) These are 3.3K and 70K resistors I believe, my Dad sent the prototype over and I just split and stripped the cable to add to it. It certainly proves the theory at any rate as previously the same loudness (According to the Mrs) would have been at 37: So looks like we can tweak the resistors somewhat in order to get more control. Just got to determine which values to try next. Thanks for the input all.