At 11:14 AM -0500 04/06/01, Paul Schreiber wrote:
>
>No need. Just use the CV inputs!
>
>If UP and DOWN are 0 on the knobs, then the rise/fall are controlled by
>the CV ins. a BYPASS is UP = 0V, DOWN=0V.
>
>Paul S.
>Mr. No-Mods
That would be fine if I wanted to control the _amount_ of portamento (CC#5
"Portamento Time", or even maybe CC#84), but I want to turn it completely
on or off using my controller which spits out CC#65 (which turns portamento
either completely on or off). It's either on or off (0 or 127) depending
on whether I play legato or depress the pedal. That won't work with the
existing CV inputs.
Please notice, I'm not putting down the 820 at all -- it looks like a great
module. I just want it to work as well as possible with my set-up.
At 12:30 PM -0500 04/06/01, J. Larry Hendry wrote:
>
>I agree it seems like you should step on the switch when you want to bring
>ON the effect.
More importantly, so do Korg and Kenton, and there isn't much I can do to
get around it.
>If you plan to do it directly with a foot pedal instead of
>running the foot pedal through the controller, just be sure you use a pedal
>with a normally closed switch. Then you press on the switch, the ground is
>lifted and the portamento effect is ON. I forget, but Roland and Yamaha
>are backwards. One is normally closed. And, 99% of any foot switches I
>have ever used have a SPDT switch in them. So, if your polarity is wrong,
>open the switch and swap the connection from NO to NC.
I have footswitches with polarity switches, but as you have noted, plugging
directly into the 820 won't record my actions into my MIDI sequencer.
>I agree with Paul that using CV control is the best, certainly if
>sequencing. then you can not only turn on and off, but vary the times as
>part of your work.
Yes, I'm glad that those function have CV control, but it's impossible to
have the same UP/DOWN settings to follow legato from a MIDI keyboard. You
would also need twice the data, and twice the MIDI-->CV channels, to go
from 0 to whatever if your UP/DOWN times were different.
>However, (WARNING DIY CONTENT) in the the DIY spirit, I will offer my
>opinion of how the bypass circuit works.
OK :) I'll go over the circuit again. I have time to work on this since I
plan on building the 820 last.
>It seems to me, that under normal
>(not-bypassed condition), the positive voltage applied through R3 causes Q1
>to operate in saturation. The saturation C/E current is limited by R5. B/E
>current is limited by R3 and R6. Because, Q1 is conducting, D2 is
>effectively shorted out and does not light. In this condition, a positive
>voltage is applied to the switch pins (8 and 9) of DG213, the analog
>switch.
>This positive voltage causes the normal closed switches to open and the
>Normal open switches to close. So, pin 6 and 7 are approximately shorted
>(some resistance) sending the lag-output voltage to the output driver
>circuit. Pins 10 and 11 are open so that the input is not directly coupled
>to the output driver.
>
>When the bypass switch is shorted (jack or switch), the voltage applied to
>the base of Q1= ground. Q1 is cutoff because the B/E junction has no
>forward bias. In this condition, D2 is allowed to conduct and light
>indicating a bypass condition. Current is still limited by R5. With pins
>>8 and 9 of the DG213 at ground, the normally closed switch (10 to 11) is
>closed connecting the input directly to the ouput driver. The normally
>>open switch is open (pins 6 and 7) so the lag output circuit is not
>>connected to the output driver. Notice that the input is alwasy
>connected >to the lag input circuit. So, the circuit continues to lag
>whether >bypassed or not. All the bypass does is select the output.
>
>Oh! D1 looks important to protect Q1 and DG213 from any negative voltage
>that might accidentially be applied to the bypass jack by shorting that to
>ground (through R4).
I could put a diode right on the CV input, leaving me with approximately 4.4V.
>Here is an idea to modify the circuit as you suggested. Disclaimer:
>WARNING!!! I have not tried this. This is just what makes sense to me
>looking at the schematic.
I understand.
>Remove R3. That takes the forward bias off of Q1
>and causes DG213 to be in the "normal" condition. The 820 will be in
>>bypass mode and the the LED will be lit. If you apply a positive CV of
>>sufficient magnitude at J1, you will force Q1 back into saturation and
>the >bypass light would turn off. DG213 would switch to cause the lag out
>to be >connected to the output driver and you would have lag.
>
>What is not clear to me (no data sheet)
I think have the application notes for all of the motm IC's would make the
experience more educational. Do you know who makes it?? I shall attempt a
web search.
>is how much positive voltage is
>required at the DG213 switch pins to activate the switch. It might be more
>than 5 volts. I see Paul is applying considerably more than that. So,
>>that research is required.
Pending research, how is this idea?? Add another transistor with 15V at
the collector, have the input from the jack go through a diode (to protect
against negative voltages), through a current limiting resistor at the
base, then connect the emmiter to the switch input. I'm not an EE, so that
could be completely wrong, but off the top of my head that sounds like it
could work, or am I confused??
>To restore the operation of the panel switch, you could disconnect the
>ground side and take that to +15V through 10K (removed R3). But, it would
>not be exactly the same. The override would be opposite. As it is today,
>either in bypass overrides the other. The new way would be either turned
>>on overrides the other.
Which is what I want.
>Disclaimer: This modification description is strickly for the purposed of
>education and discussion. I do not recommend you make this modification to
>your 820. I have not tested it and my logic could be flawed.
I'm going to look everything over several times, and post back to the list
before I try anything. Frankly, I'm surprised this modification has not
been discussed before. Thank you for your input.