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Any issues with 15.5VDC powering 12V amplifiers?

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15.3 VDC system for amplifier - install notes to come

Hi guys,

I will be doing a full writeup on my sub and amp install and Hanshow speaker upgrade in my SR+ M3 - lots of bits of info have maybe been missed in other write-ups (no one is perfect and some things have obviously changed between models and years (and likely country of build I suspect)).

Having said that, one thing that I didn't account for is that the LV power system is 15+ VDC, not 12. Now I realize even 12V systems are usually above this and why you see amplifier dyno testing above 14V but my measured voltage at the LV point (under the back right seat) is 15.3V.

I measured 15.3 VDC at my amplifier input and my Pioneer amp is rated 10.8 to 15.1. It's a 500W monoblock amp powering a 500W ported subwoofer. I had an issue with the sub and don't want to continue without addressing this concern. It's only 0.2V above specification, but I don't know if it routinely goes above 15.3 when driving around.

Has anyone else encountered this issue? I called a local stereo shop and tried to be respectful of their time by not just asking for free advice.

They didnt seem overly knowledgeable about Teslas, but said its not crazy to think the LV goes above this measured rating. I asked if I needed a 15VDC to 12VDC converter , but we both agreed that a 500W converter would be expensive. I asked about 500W resistors to drop the voltage but we both agreed that would be hot and not ideal. Their suggestion was to maybe use a diode which would drop the voltage 0.4 to 0.7 depending on type. I kind of like this idea but don't want to rush into this too much - I feel this must come up but I haven't found much conversation online.

I could get into more details on this topic but I'm short for time. For now, I only want to ask if this has come up before for anyone else. I believe only newer Teslas have 15 VDC systems and this usually isn't a problem. But I think for my amp, it might be an issue.
 
One quick fix would be to use a forward biased power diode in a power lead feeding the amp. It will cause a voltage drop of roughly 0.7 volts. You should make sure it can dissipate 50 watts (which may be overkill). The exact requirements will depend on the efficiency of your amplifier the max average power.
 
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Yes, that was the suggestion I discussed. I have been looking at 100amp battery sharing devices on Amazon for roughly 80 Canadian (50US) dollars. If I only use one input, I am essentially doing the same thing, but it's got the heatsink etc already installed.


There might be more elegant solutions. I thought it was a decent one that has one day shipping and good reviews. Might be useful if I get a 2nd battery too.
 
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After some thought, I'm leaning towards a voltage regulator. I'll keep y'all updated. Hope to hear some feedback on what worked for you guys.


Edit: Bought this. I feel I might be able to use 2 x diodes to maybe get a larger drop if necessary. And if I'm wrong then we can all laugh at me :)

 
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Edit: Bought this. I feel I might be able to use 2 x diodes to maybe get a larger drop if necessary. And if I'm wrong then we can all laugh at me :)

https://www.google.ca/url?sa=t&sour...wQFnoECA4QAQ&usg=AOvVaw2KZuIXEL5o8bqsHQchVEOb
Good choice! I was going to recommend a rectifier over a voltage regulator. You can use all 4 diodes. Or use only 1 or 2 if you want less voltage drop. Of course, measure the voltage drop. They claim "VTM = 1.5V". I don't know what VTM means.

It's generally not recommended to use diodes in parallel when designing with discrete parts but I don't think it will harm anything. One diode (or pair of diodes) will handle most or all of the current because the voltage drop decreases as the temperature increases. Oh well. The voltage drop will only change by about 0.1 volts over a 60C temperature swing so your amp should be fine.

Also, in this case if you want two voltage drops then you can't avoid using diodes in parallel. If you use all four diodes, I recommend you keep the AC input terminals open.

Have fun!
 
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Good choice! I was going to recommend a rectifier over a voltage regulator. You can use all 4 diodes. Or use only 1 or 2 if you want less voltage drop. Of course, measure the voltage drop. They claim "VTM = 1.5V". I don't know what VTM means.

It's generally not recommended to use diodes in parallel when designing with discrete parts but I don't think it will harm anything. One diode (or pair of diodes) will handle most or all of the current because the voltage drop decreases as the temperature increases. Oh well. The voltage drop will only change by about 0.1 volts over a 60C temperature swing so your amp should be fine.

Also, in this case if you want two voltage drops then you can't avoid using diodes in parallel. If you use all four diodes, I recommend you keep the AC input terminals open.

Have fun!
Not sure if it was clear but I meant 2 x diodes in series. So if it's 0.4V then I'd get 0.8V.
 
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Not sure if it was clear but I meant 2 x diodes in series. So if it's 0.4V then I'd get 0.8V.
Yes, I understood. It's a bridge rectifier. You don't have access to each diode independently. There are only four external terminals. Due to the way it's wired internally, if you want two voltage drops then the diodes will be in parallel as well as in series. Two sets of "two diodes in series" in parallel.

Code:
                     o AC
                     |
          -----|>----------|>|------
          |                        |
 - o------|                        |------o +
          |                        |
          -----|>----------|>|------
                     |
                     o AC

You can connect the (-) terminal to your positive power supply and connect the (+) terminal to the positive power input on the amp. Leave the AC terminals open.
 
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Yes, I understood. It's a bridge rectifier. You don't have access to each diode independently. There are only four external terminals. Due to the way it's wired internally, if you want two voltage drops then the diodes will be in parallel as well as in series. Two sets of "two diodes in series" in parallel.

Code:
                     o AC
                     |
          -----|>----------|>|------
          |                        |
 - o------|                        |------o +
          |                        |
          -----|>----------|>|------
                     |
                     o AC

You can connect the (-) terminal to your positive power supply and connect the (+) terminal to the positive power input on the amp. Leave the AC terminals open.
Hi,
Is it possible to get 1 voltage drop? What other configurations are possible? Thanks.
 
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Hi,
Is it possible to get 1 voltage drop? What other configurations are possible? Thanks.
Not as far as I know but maybe there is a clever solution. Germanium diodes have a smaller voltage drop but they tend to be more fragile and aren't available for high power applications AFAIK.

If you want the voltage drop of a single silicon diode then it is trivial to wire up the bridge rectifier for that.
 
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not sure why you'd want a bridge vs just 1 or 2 power diodes. Regardless you still have to arrange for adequate power dissipation (i.e. heat sinks). I would have just gotten 1 or 2 of the bolt-style power diodes (rectifiers) as I am used to bolting them to heatsinks. Not sure how close to 500W you will ever get to (would likely cause hearing damage), but 500W/15V == 33A, so that is about 25W per diode that you need to arrange to dissipate. I suppose that bridge can also be bolted to a good heat sink, so you should do that...

BTW, the original resistor idea wouldn't have needed a 500W resistor, its the same 33A or 25W resistor. The problem with the resistor is that it wouldn't have a constant voltage drop. For example, if you are drawing 500W, its the same 33A so R = 1/33 ohms. However, when the amp is drawing less, say idling at 10W, then 10/15 == 0.7A, V == 0.7 x 1/33 == 0.02V, i.e. practically no voltage drop at all, which was your original purpose. So the diode idea is much, much better.

If you really want a 1V voltage drop, using a power transistor in a simple emitter follower circuit. You could dial up most any reasonable voltage drop you want (1V, 2V, 3V, etc). But the 1 or 2 diode is the simplest best idea.
 
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Sure:
Battery to -, amp to AC (single or both connections)
Or
Battery to AC (one or both), Amp to +
not sure I see how your suggestions gives a 1V voltage drop... if I understand your proposal, you are engaging only 1 diode's worth of voltage drop, which would not be 1V. Each diodes gives you abouit 0.6 - 0.7V so you have to deal in multiples of those values. The use of a germanium diode is more on point, as they have about 0.3V voltage drop, but they really aren't that easy to find, and probably of little point. Again, use a power transistor if you really want more flexibility or better voltage regulation -- let me know if you want a sample circuit.
 
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not sure I see how your suggestions gives a 1V voltage drop... if I understand your proposal, you are engaging only 1 diode's worth of voltage drop, which would not be 1V. Each diodes gives you abouit 0.6 - 0.7V so you have to deal in multiples of those values. The use of a germanium diode is more on point, as they have about 0.3V voltage drop, but they really aren't that easy to find, and probably of little point. Again, use a power transistor if you really want more flexibility or better voltage regulation -- let me know if you want a sample circuit.
My response was for the situation of "1 (diode) voltage drop" with a bridge rectifier, not "1 volt" of drop.

Can also use one silicon diode and one Schottky or two Schottky, but that still varies based on current.
If a situation really needs 1.0V vs 0.7V or 1.4V then yeah, a regulated supply is probably called for, then come the tradeoffs of linear (simple but dissipation) vs switching (complexity)
 
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Can also use one silicon diode and one Schottky or two Schottky, but that still varies based on current.
If a situation really needs 1.0V vs 0.7V or 1.4V then yeah, a regulated supply is probably called for, then come the tradeoffs of linear (simple but dissipation) vs switching (complexity)
yeah, nearly everything discussed here has been dissipative (diodes, resistor, series pass transistor). Given the description of the problem, the 2 power diode (rectifiers) make the most sense (KISS), as long as a decent heatsink is used. The precise voltage drop isn't going to be critical (1.2-1.4V sounds fine), and with any reasonable sound level (probably less than 100W), the dissipative loss will be modest (e.g. 5-10 watts).
 
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