Robbie010 Posted January 21, 2020 Share Posted January 21, 2020 Hello all, I have been given some great advice and guidance from forum members on the subject of the Type A network, I’m just struggling to get my head around this.... When looking at the original schematic for the type A network and viewing images of the the modern version, the 13uf and 2uf caps appear to be in series to the tweeter, is this correct or am totally mis-reading the schematic? As I read the schematic, the positive line runs to the 13uf cap, then to pin 5 on the autoformer (0 attenuation) then to the 2uf cap and finally to the positive terminal on the tweeter driver. From my current understanding, this achieves 0 attenuation via the autoformer and gives a total capacitance of 1.733 ohms from the two caps in series. Is this correct? Quote Link to comment Share on other sites More sharing options...
glens Posted January 21, 2020 Share Posted January 21, 2020 Yes. Certainly by design. Quote Link to comment Share on other sites More sharing options...
JohnA Posted January 21, 2020 Share Posted January 21, 2020 The schematic is correct. It is an older design technique, but it works. My understanding is that it increases the crossover attenuation slope at a point below the tweeter's crossover point. Quote Link to comment Share on other sites More sharing options...
MechEngVic Posted January 22, 2020 Share Posted January 22, 2020 It looks like the 13uf cap is feeding both the autoformer and the 2uf cap. I wouldn't call the 13uf and 2uf being in series. Quote Link to comment Share on other sites More sharing options...
glens Posted January 22, 2020 Share Posted January 22, 2020 I would because they're in series... Quote Link to comment Share on other sites More sharing options...
MechEngVic Posted January 22, 2020 Share Posted January 22, 2020 4 minutes ago, glens said: I would because they're in series... Even if another component is drawing current from between them? Quote Link to comment Share on other sites More sharing options...
glens Posted January 22, 2020 Share Posted January 22, 2020 What's the path to the tweeter? One capacitor out directly into another. There's a bit of parallelism going on after the first cap, sure, but it's still 1/( 1/13uF + 1/2uF )uF to the tweeter. Quote Link to comment Share on other sites More sharing options...
JohnA Posted January 23, 2020 Share Posted January 23, 2020 22 hours ago, glens said: What's the path to the tweeter? One capacitor out directly into another. There's a bit of parallelism going on after the first cap, sure, but it's still 1/( 1/13uF + 1/2uF )uF to the tweeter. 😀 Not with a node between the 2 caps. The autoformer makes a node that separates the 2 caps into 2 *filters* in series. 2 Quote Link to comment Share on other sites More sharing options...
glens Posted January 23, 2020 Share Posted January 23, 2020 Five bucks? Quote Link to comment Share on other sites More sharing options...
mboxler Posted January 23, 2020 Share Posted January 23, 2020 Like I said in another thread, I'm not sure how the autoformer changes the circuit, but I would also think that the current flowing across taps 0 and 5 would have an affect. Before I send @Robbie010 the autoformers, I think I'll test this out. I don't have the 13uf/2uf caps to test with, but any two caps should do. I'll rig up the circuit with a 26uf cap to the autoformer and a 12uf cap to the tweeter. I have 14 ohm and 8 ohm resistors to use as loads. I'll pass, say, a 400 hz 2.83 volt signal and measure the voltages across each resistor. I'll then move the input to the 12uf cap and retest. If the caps are truly in series, the voltages across the resistors should not change, right? I've been wanting to test this for a while. Maybe it's about time. Mike 1 Quote Link to comment Share on other sites More sharing options...
glens Posted January 23, 2020 Share Posted January 23, 2020 Please also take a measurement at the 8 ohm resistor through both caps alone (no transformer attachment between them). The difference between the two "tweeter" circuit measurements should answer the speculation on my part. Thanks. Quote Link to comment Share on other sites More sharing options...
mboxler Posted January 23, 2020 Share Posted January 23, 2020 Will do Quote Link to comment Share on other sites More sharing options...
Deang Posted January 23, 2020 Share Posted January 23, 2020 14 hours ago, JohnA said: 😀 Not with a node between the 2 caps. The autoformer makes a node that separates the 2 caps into 2 *filters* in series. That's what I think. Didn't Al test this out here years ago? 1 Quote Link to comment Share on other sites More sharing options...
MechEngVic Posted January 23, 2020 Share Posted January 23, 2020 We all know from our circuits classes that two components in series can have nothing between them, and if they do, they are no longer in series. I think what glens is saying is that the 2uf cap will still see the complete 13uf coming from the 13uf cap, so the tweeter still sees a 13uf and 2uf cap in series. With the autoformer and midrange driver drawing current from between the two caps, that idea is suspect. Inductors have some capacitance of their own, and maybe the speaker drivers do too. But is it enough to change the capacitance seen by the 2uf cap? Certainly the tweeter is receiving less current. And certainly the 13uf cap is feeding both the tweeter and midrange drivers. I guess the question is: If we consider influence of the autoformer and midrange driver, does the 2uf cap see something other than 13uf? 1 Quote Link to comment Share on other sites More sharing options...
mboxler Posted January 23, 2020 Share Posted January 23, 2020 Well, here's my results using a 2.83 volt 400 hz signal. My setup includes a 26.38uf mid cap, a 11.76uf tweeter cap, a 14.2 ohm resistor connected to tap 4 (tap 0 connected to ground), and an 8 ohm resistor. First, the two caps in series (or 8.136 ohms) results in .46 volts across the 8 ohm resistor. Nothing else connected during this test. Next, the classic Type A configuration with the tweeter cap connected to tap 5. .424 volts across the 8 ohm resistor 1.296 volts across the 14.6 ohm resistor Last, the new Type A configuration, with the tweeter cap connected to the amplifier input. .651 volts across the 8 ohm resistor 1.785 volts across the 14.6 ohm resistor The math required to reach the Classic Type A results ended up being for a parallel connection. It bothers me the the voltages the 8 ohm resistor are so close between the first and second tests, so I plan on running another one tomorrow at 800 hz. If anyone is interested in the math involved, I'd be happy to post it tomorrow as well. It's kinda nerdy. Right now my brain is fried. Mike . Quote Link to comment Share on other sites More sharing options...
JohnA Posted January 24, 2020 Share Posted January 24, 2020 Your cap values are too far from the OEM values to see much difference at 400 Hz. You want to run several frequencies, too, like 50, 100, 400, 800, 2000, 4000, 10k. Quote Link to comment Share on other sites More sharing options...
JohnA Posted January 24, 2020 Share Posted January 24, 2020 WOW! this takes me back over 40 years! I had to stew on it a while. This is how I think the circuit should be analyzed. The autoformer/transformer is an odd duck and requires a different approach. I believe it acts like a voltage source on the low voltage side. The high side is then a series resistor (the wire resistance) and inductor (the coils in the transformer) to ground. Seen this way it becomes plain the squawker and tweeter caps are not in series. Current and voltage drops can then be calculated in each of the 3 loops. It must also steepen the slope of the squawker high-pass. this has been a mild curiosity for me over the years. I'm glad you all forced me to figure it out. Ain't lernin' fun!!! 1 Quote Link to comment Share on other sites More sharing options...
JohnA Posted January 24, 2020 Share Posted January 24, 2020 Robbie, There is no reason not to try the Type A circuit, but I believe you will be much happier with the Type AA tweeter section. Quote Link to comment Share on other sites More sharing options...
MechEngVic Posted January 24, 2020 Share Posted January 24, 2020 1 hour ago, JohnA said: WOW! this takes me back over 40 years! I had to stew on it a while. This is how I think the circuit should be analyzed. The autoformer/transformer is an odd duck and requires a different approach. I believe it acts like a voltage source on the low voltage side. The high side is then a series resistor (the wire resistance) and inductor (the coils in the transformer) to ground. Seen this way it becomes plain the squawker and tweeter caps are not in series. Current and voltage drops can then be calculated in each of the 3 loops. It must also steepen the slope of the squawker high-pass. this has been a mild curiosity for me over the years. I'm glad you all forced me to figure it out. Ain't lernin' fun!!! Ha! I didn't notice it until you drew it out. The autoformer, while not offering attenuation in series with the two caps, acts like an inductive roll-off, and a third component in the tweeter circuit. That makes it a third order, 18db slope. Quote Link to comment Share on other sites More sharing options...
glens Posted January 24, 2020 Share Posted January 24, 2020 8 hours ago, mboxler said: It bothers me the the voltages the 8 ohm resistor are so close between the first and second tests, so I plan on running another one tomorrow at 800 hz. If anyone is interested in the math involved, I'd be happy to post it tomorrow as well. It's kinda nerdy. The results seem to meet my expectations... Never too nerdy; post your math. Quote Link to comment Share on other sites More sharing options...
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