Accumatch

[jrod666999]

I was just looking through an old '96/'97 Car Audio and Electronics magazine and there was a revier for the Accumatch, and it reminded me that I had a similar idea back then and was wondering how well it would work. They weren't very impressed with the Accumatch, but here was my idea. I have a 50 x 4 old school Cruch amp that has 2 bad channels that I never planned on fixing because it was just a small amp I didn't care about using. anyway now I am using it to push 2-10"'s and if I had a step up transformer between the amp and the subs wouldn't it boost the power the subs got? Say if I had a 1:4 they should get about 200 watts apiece instead of the 50, right? Couldn't be that easy?

[ECM]

It WOULD be that easy, IF your amp was 1 ohm stable. You can do that with an orion hcca or other high current amps, but if your crunch amp is only 2 ohm stable, then 100 watts is the most you would get. That would be with a 2:1 transformer.

Also, those accumatch transformers were low power. Each transformer model was rated at a certain power level depending of the frequency range they were designed to be used in. The woofer transformers were HUGE and expensive. Another issue was that some amps would go into protect mode with the transformers at the output.

[Clipped]

although the voltage would increase, the current wouldnt so you wouldnt have a power increase.

[KyferEz]

The Accumatch impedance matching transformers do work. In my use of them, the increase in audio output WAS appreciable. I used them to drop a 4ohm load to 1ohm. They are NOT worth what the manufacturers charged IMO, but if you can get them cheap it's worthwhile.

I still have 2 1200 watt ones and 2 500 watt ones that I keep around. I got them real cheap from a dealer that didn't know what they were for LOL. I hang onto them just in case I need them for my next system.

If I understand these transformers correctly, then Clipped is correct, but only in certain circumstances such as if the impedance of the Accumatch primary is the same as the speaker.

[jrod666999]

I was not really wanting to use an accumatch. I was asking if I had a transformer wound with a 4 ohm inductor, and the output wound 4:1 would it make a differance?

[KyferEz]

Ah, ok. Well, like I said, if the impedance of the primary on the coil is the same as the sub, then Clipped is right, because with a transformer power in = power out minus losses.

It's only useful if you decrease the load at the amp so that it produces more power to transfer through the transformer.

For example, lets say you're using the amp, bridged into an 8-ohm load and the amp is 2 ohm stable.
If you use a transformer with a primary of 4ohm impedance and secondary with twice the number of turns, then it will work well because the amp would see 2 ohms and deliver maximal output.

However if the load you are pushing is 8 ohms, and the primary of the transformer is also 8ohms then it's not going to work well. Depending on the secondary, it can boost voltage but not current, so the power output remains the same.

It's also useful if your speaker load is too low for the amp. Lets say you have a 2ohm woofer and a 2ch amp that's stable at 2ohm. With the amp bridged it would "see" a 1ohm load. Using a transformer with a 4ohm primary and 1/2 of the turns on the secondary, you will match the impedance and can push that load safely.

I wrote this quickly, so if anyone notices mistakes, please correct me.

[ECM]

If the voltage across the speaker terminals goes up, then current goes up as well. Ohms law.

If you use a 4 ohm transformer on a 4 ohm sub, then there will be no increase in power. The trick is to use a transformer to present the amp a 2 ohm load. Using a 1:2 ratio transformer will work in this situation. Another way to look at it is as a "step-up" transformer. The transformer will step-up the amps voltage to make max power into a 4 ohm load.

[KyferEz]

Originally posted by ECM
If the voltage across the speaker terminals goes up, then current goes up as well. Ohms law.

Yes and No. In the case of using a transformer, when voltage goes up, current goes down. Why? Power in=Power out. Conservation of energy law. So if my amp produces 400 Watts into a transformer, no matter what the secondary does, you will get 400 watts out.

If your amp has 400 watts to the transformer and the transformer boosts the voltage 2x on the secondary, and you then use a low ohm load, the load will cause the voltage to drop. Why? Because only 400 watts was available and the low resistance load causes higher currents which causes the voltage to drop because only 400 watts was available in the first place.

P=VI
P=V^2 / R
P=I^2 * R

So, 400Watts = 40V * 10A when fed into a 4ohm load.
If you feed that into a transformer that has a 4ohm primary, and the secondary has twice the number of turns, you double the voltage output but decrease the current output:
400Watts = 80V * 5A, but only into an 16ohm load. If you try to put it into a 4 ohm load, the current increases to 10A which causes the voltage to drop back to 40V.

[Perry Babin]

The voltage may not drop down to 40v with the 4 ohm load. It would depend on the loss through the transformer and the output impedance of the amp.

With a 1:2 transformer, if the input to the transformer remaimed at 40v and transformer was 100% efficient, the 4 ohm load would see 80v and the current through the load would be 20 amps. Of course, the transformer won't be 100% efficient. More information would be needed to determine the actual voltage the 4 ohm load would see.

[KyferEz]

Originally posted by Perry Babin
With a 1:2 transformer, if the input to the transformer remaimed at 40v and transformer was 100% efficient, the 4 ohm load would see 80v and the current through the load would be 20 amps. Of course, the transformer won't be 100% efficient. More information would be needed to determine the actual voltage the 4 ohm load would see.

How is that possible? The only way I can see for it to work that way would be that the amp would have to increase it's power output, which means there would be more than 400W at the primary, keeping with my example... Or am I missing something? I guess it's possible for the amp to produce more current, but how is it going to do that if the primary remains at a 4ohm impedance?