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ssm2404 schematic help
ssm2404 schematic help
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Old 8th April 2009, 07:19 PM   #1
feathed is offline feathed  United States
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Default ssm2404 schematic help

Would someone please look at this and tell me what the input impedance is?

I am ignorant and confused. I have a tube preamp with an output impedance of about 3K ohms and I would like to feed it's output to the input of a modified Behringer DCX2496 that has an input impedance of about 8K. I would like some kind of buffer with about 50-100K input impedance between the two. I also need to switch between a total of three inputs so this switch seems to serve two purposed at once if it will load my preamp correctly. Any help with this or directions to another solution would be greatly appreciated.
Thanks, Doug
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Old 8th April 2009, 07:51 PM   #2
Iain McNeill is offline Iain McNeill  United Kingdom
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Well, I'm sure you've seen that this is a mono product.

But to answer your question there are a lot of options for loading. There's a jumper selectable input resistor and a jumper seletable output resistor. When the analog switch is turned on, both input and output resistors are in parallel with the input impedance of the next stage (amplifier). E.g. if you selected none for input (don't want 50ohms) and selected 100K for the output you would present 100K//8K=7.4Kohms to your tube preamp.

If you replaced the resistors, you could dial in anything you wanted and even change the loading depending on the input selected.

BTW, I don't see any buffers. The SSM2404 is just a standard CMOS switch albeit with only 1ohm on resistance.
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Old 8th April 2009, 07:55 PM   #3
pacificblue is offline pacificblue  Germany
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You seem to expect this device to operate like a buffer, which it is not. The input impedance seems to be selectable via jumpers. Be aware that the Buffer Load and Input Impedance resistors will act in parallel to the DCX's input impedance and lower it accordingly.

The Behringer DCX2496 has 20k input impedance according to its manual. If you modified it to 8k that switch will lower it even further.

If you want 50-100k load impedance for the tube preamp, you will have to use a buffer between tube preamp and switch.

On the homepage there is a hint to ACD, who is a forum member here. PM him, he will probably be the best person to explain that product to you.
If you've always done it like that, then it's probably wrong. (Henry Ford)
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Old 8th April 2009, 10:18 PM   #4
feathed is offline feathed  United States
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Thanks Iain.

E.g. if you selected none for input (don't want 50ohms) and selected 100K for the output you would present 100K//8K=7.4Kohms to your tube preamp.

That's what I calc. but I was confused by the statement: "load impedance 10K,100K or none". I wasn't knowledgeable enough to to sure.
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Old 9th April 2009, 04:29 AM   #5
Jan Dupont is offline Jan Dupont  Denmark
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Regarding the Input load selection (50R or none). Choose 50R for impedance match if the source is fitted with 50R line buffer, or none if the source hasn't this buffer fitted.

The Load Impedance 10k, 100k or none is there in case that the input of the following circuit has a very high input impedance >1MOhm. If you look at the datasheet for the SSM2404 (to large to be attached here) you see that the SMM2404 specs varies with the load impedance.

All resistor values can of cource be choosen to suit ones own needs.
Free Schematic and Service Manual downloads www.audio-circuit.dk, Spare time company (just for fun): www.dupont-audio.com
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