yes, that's why when i use the A100 with other modular devices i, very often, use buffer modules even for gates clocking various other sequencers, etc...


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--- On Mon, 11/16/09, yahoo@... < yahoo@... > wrote:

From: yahoo@... < yahoo@... >
Subject: AW: 1 A-180
To: Doepfer_a100@yahoogroups.com
Date: Monday, November 16, 2009, 11:58 AM







 









> I've noticed recently that the A-180 Multiple really cuts the voltage

> signal down quite substantially. I just got an MFB Dual ADSR and noticed

> that when i mult the Gate Output of my SH-101 by 3x, it drops the gate

> voltage down to where the MFB won't trigger anymore. Basically, i had it

> set up where the gate out went into the A-180, then the first 2 outs

> went to the two Gate Ins on the MFB and the 3rd out went to my A-140 EG.

> If i pulled the cable out that went to the A-140, the MFB would fire,

> but not with it inserted. However, it works if i use my 303, which has a

> hotter Gate signal. I checked the gate out voltages coming of the 101

> direct, then at each output of the A-180. I got the following:

>

> Direct Out from 101: 11.56v

> Into the A-180 w/no other cables inserted: same as above

> with 2 cables inserted: 7v

> with 3 cables inserted: 5.5v

>

> This is quite a bit of voltage drop if you ask me. Is there anyway to

> mod the A-180 so it doesn't eat up all my voltages

>

> -Larry



Some basic information about this issue:



The A-180 is a passive module - nothing but 2x4 connected jack sockets. It

works as if one would simply connect the jack plugs manually of if one would

use an Y-cable. Consequently the reason for the behaviour described is

actually not the A-180 but the inputs and outputs of the modules. Ideally

the output of each module should be able to drive any load without any

losses (technically spoken: zero ohm output impedance or resistance). And

ideally the input of each module should not cause any load (technically

spoken: infinite ohm output impedance or resistance). But in reality the

situation is different: each output has a limited driving capability (e.g. 1

k output resistance) and each input causes a load (e.g. 100k for most VCO CV

inputs). As a rule of thumb the relation between input and output resistance

causes the voltage loss: if a 1k output drives a 100k input one has about

1/100 of voltage drop. Not much for a gate voltage but already too much for

a VCO control input(that's why special buffers like A-185-1/2 are

recommended for VCO control). If more than one input is connected to an

output (e.g. by means of an A-180) the resulting load becomes smaller: as a

rule of thumb the load is divided by two if both inputs have the same load.

Otherwise it's a bit more difficult to calculate. For example: the A-140 has

a 10k load at it's input. If two A-140 inputs are connected together the

resulting load is 5k. If the resistance of the output that drives the two

A-140 is also in the 5k range the output voltage is divided by two. If it

was 12V without load it becomes ~ 6V with two A-140.



In the A-100 we payed attention that (with some exceptions) each source is

able to drive a customary number of units (e.g. 3-4 A-140 from a gate

source). The output resistance of A-100 modules is typ. 1k and the input

resistance 10k or more (e.g. 100k for VCO control inputs). Consequently

within the A-100 no problems with gate signals should occur. For voltages

that drive VCOs buffer modules (A-185-1/2) are recommended. These have a

much lower output impedance. If an external voltage source is used the

situation may be different. One needs to know the output impedance to

calculate the expected voltage drop. To be on the safe side a buffer module

can be used (even for gate signals).



Best wishes

Dieter Doepfer

























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