Guest Mick Berg Posted January 18, 2007 Report Share Posted January 18, 2007 Hi.Sorry for starting a new topic but I want to be noticed! I'm adding MIDIbox to a pipe organ console that already has a Devtronix interface. The Devtronix is quite happy accepting keying voltage of up to 20 volts, I have it regulated to 15. What is the maximum the MIDIbox DIN module will accept? Is it actually 5 volts? I'm hoping not to have to do voltage dividers, but I will if I have to. If so would 10k and 5k resistors be correct for the divider?Thanks,Mick Berg. Quote Link to comment Share on other sites More sharing options...
doc Posted January 19, 2007 Report Share Posted January 19, 2007 Hi,What is the maximum the MIDIbox DIN module will accept? Is it actually 5 volts?You guess is right .... it is maximum 6 Volt .I'm hoping not to have to do voltage dividers, but I will if I have to. If so would 10k and 5k resistors be correct for the divider?Theoretical yes. The calculation is right without load. I don't know how much load goes through your circuit nor if it is a constant current .Keep in mind, if your circuit is loaded: - The voltage on the first resistor rises- The voltage on the second resistor dropsJust test your resister network with a little load (5 - 10mA) instead of the DIN and look how much voltage passes the load. The 74HC165 works well between 4,5V and 5,5VMore than 6V on the input will destroy the 74HC165.greetsDoc Quote Link to comment Share on other sites More sharing options...
MRE Posted January 19, 2007 Report Share Posted January 19, 2007 The other option would be to use optoisolators. There are several that can handle higher input voltages and always give a clean 5 volt signal out. (cant remember p/n off hand) A lot less worry, but a bit more expensive than a voltage divider.So long as you are confident that your key voltage is stable and doesnt swing over, the voltage divider should be fine. If you question the stability of the PWS though, I would go for optos. Designer choice ;) Quote Link to comment Share on other sites More sharing options...
Guest Mick Berg Posted January 19, 2007 Report Share Posted January 19, 2007 Thanks guys.The keying voltage is a solid regulated +15 volts. Does anyone know how much current the DIN inputs take? it shouldn't be more than a couple of milliamps, surely? Or another solution would be to measure the input resistance of a DIN input, and put that in parallel with the divider to get the proper resistor values, right? Would it damage the DIN to make that measurement with a regular test meter?I plan to make a big regulated +15v PSU for the whole organ soon, using a circuit from "Bowden's Hobby Circuits". It's so complicated having different voltages everywhere. Mick Berg. Quote Link to comment Share on other sites More sharing options...
doc Posted January 19, 2007 Report Share Posted January 19, 2007 Does anyone know how much current the DIN inputs take?Have a close look in the datasheet of the HC series. I read it somewhere but don't know where ...Would it damage the DIN to make that measurement with a regular test meter?No, a digital meter shouldnt damage the device.But I don't think you'll get exact values that way... but give it a try. b.t.w. a HC165 is a very cheap part. So if you damage one ore two to get the right values it isn't sooo bad, or?greetsDoc Quote Link to comment Share on other sites More sharing options...
fluke Posted January 19, 2007 Report Share Posted January 19, 2007 Would it damage the DIN to make that measurement with a regular test meter?That depends on the voltage it uses to do the measurements. Some use a very low voltage so as not to turn on diodes or transistors. Which would give a incorrect reading in this case. You'd be better to put 5V on the DIN and measure the current the input draws.A more expensive, but safer solution is to use a Zener diode to control the voltage. It would look something like this: ^ | 15V | | | -+- | | | | Resistor to limit current | | -+- | | ----- \ / Diode v ----- | +---- to DIN input | /---/ ^ 5.1V Zener /___\ | | ----- --- - Any sort of diode will do, it's just to stop current flowing through the Zener in the reverse direction, you may not need it. The resistor should be (15-5.1)/I, where I is the typical input current for the DIN inputs. Quote Link to comment Share on other sites More sharing options...
MRE Posted January 21, 2007 Report Share Posted January 21, 2007 Another good choice. Parts cost likely between that of a voltage divider and an opto.Dont forget .7 volt drop on the Diode, so thats actually 14.3 Volts in the equation. (wont make a lot of difference anyway. Note that TYPICAL does not mean MAX I ;)I dont know anything about those old organs. Is the key voltage 15 volts when the key is up, and zero when the key is pressed? or the other way around? (doesnt matter to the circuit, only curious) Quote Link to comment Share on other sites More sharing options...
Guest Mick Berg Posted January 21, 2007 Report Share Posted January 21, 2007 This is a pipe organ console, all it is electrically is a ton of switches, literally thousands in a big one. The voltage is 15v when a switch is closed, ie a key is pressed or a stop is drawn. There's no ground switching in this console although there might be in others. The divider components are very cheap, the zeners and diodes are eight cents each from Mouser.I can't find the info on the input current of the HC165. The data sheets I found on the web don't show it. Anybody have a clue? Thanks,Mick Berg. Quote Link to comment Share on other sites More sharing options...
doc Posted January 21, 2007 Report Share Posted January 21, 2007 Hi,I can't find the info on the input current of the HC165take a look in the family specifications. Maybe you'll find it there.For example here:http://www.cl.cam.ac.uk/Teaching/2003/DigElec/part2-data.pdfgreetsDoc Quote Link to comment Share on other sites More sharing options...
Guest Mick Berg Posted January 22, 2007 Report Share Posted January 22, 2007 Well, in the family specs. I found "The basic family of buffered devices, designated asXX74HCXXXXX, will operate at CMOS input logic levels for high noise immunity, negligible typical quiescent supply and input current."Since input current leakage was also specified as being a few microamps, I think that it can all be ignored.Thanks everyone!Mick Berg. Quote Link to comment Share on other sites More sharing options...
doc Posted January 22, 2007 Report Share Posted January 22, 2007 I agree,give it a try.It would be intresting for me, if your divider works. Please post, if you get it working that way. greetsDoc Quote Link to comment Share on other sites More sharing options...
Guest Mick Berg Posted January 22, 2007 Report Share Posted January 22, 2007 I'm eagerly awaiting delivery of zeners and diodes from Mouser (5 cents each!! - shipping cost more than the parts!) and I'll report on my progress soon.Mick. Quote Link to comment Share on other sites More sharing options...
Guest Mick Berg Posted February 14, 2007 Report Share Posted February 14, 2007 I tried the zener diode divider, but could not get it to work. The pull-down resistors got in the way I think. I gave up after frying several zeners and a register chip!Anyway I ended up with a 12k/4.7k divider network which worked on the bench, but for some reason in the organ I am now only getting 3.5 volts as my high voltage. Possibly high resistance wiring in the console. Is 3.5 volts enough to reliably send the input pin high?Thanks,Mick Berg. Quote Link to comment Share on other sites More sharing options...
MRE Posted February 17, 2007 Report Share Posted February 17, 2007 Yeah.. In THEORY, TTL vt is 1.5 volts, with reliable Input High detection anywhere above 2 volts. 1.5 - 2 is sort of a gray zone, but chips generally switch close to 1.5But, at 3 volts, you dont have much room if you experiance any voltage sag. So, if you can manage to get your key voltages closer to 5, then you are less likely to experiance missed keys on chords or something (just a guess here). But hey, if it all works fine at 3, then there is no need to worry.. 3 volts is fine as far as TTL is concerned. Quote Link to comment Share on other sites More sharing options...
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