On 9/18/2013 7:09 AM, Ethan Dicks wrote: Thanks to all for the replies, I'm going to attempt to condense my further answers / questions into this response to cut down on the amount of mail in all of your mailboxes :). My 8/L came with its W076 directly wired to the terminals on the ASR33 that came with the system. I've since wired up my own set of connectors (using some 9-pin Mate 'n Lock connectors I happened to have on hand). The wiring from the W076 (using the W076's labeling) to the Current Loop->RS232 adapter is: 3 -> T- 4 -> T+ 6 -> R- 7 -> R+ Relay +/- : Not connected currently I've covered this before, but just for reference, It's one of these: http://www.bb-elec.com/Products/Serial-Connectivity/Serial-Converters/Curre… (the 9-pin serial variety). I have it wired up as in Fig. 2 of the manual linked to from that page, using the wiring described above. Tony Duell wrote: I just tested this out with the PC sending a "break" (for the record, I'm using a *real* RS-232 port, not a USB->Serial adapter for this -- I recently discovered that my USB adapter doesn't actually support 110 baud, guess how I found that out? :) ). When sending a break, the current only drops to about 19.5mA. So, I guess there could be a hardware problem here, but this assumes that the way I have the 8/L hooked up to the adapter is actually correct. I had thought that the configuration I have is correct (I had thought that the 8/L would be considered an "active" port) but I'll reiterate that I've never fooled with current loops before and so my assumption may be in error. I probably have a lot to learn here :). Thanks again for the help... - Josh

No wonder you can't send any data, then. That would be interpretted as a constant 'mark'. A wiring error is a hardware problem ;-). More seriously, what I meant was that if you can't get the RS232 port to turn off the loop current, then there is no point invesitgatign the software on either machine. OK, I am hopefully not insulting your intellegence too much here. I will try to start with the basics. Suppose we have 2 devices and we need to send an electrical signal from device A to device B. The obvious wa to do it is for device A to output a suitable voltage and device B to have some kind of voltage detector. That's what a voltage-level interface, like RS232 is. The other way is for deviec B to sense the current flowing through its input cicuit nad deviec A to somehow control that current/ That is a 'current loop' interface. As an aside, there is nothing that implies a current loop has to be a digital intterface. Just as a voltage-level interface can be analogue (a sensor can output a voltage depending onbthe quantity it is snesing), a sensor can also control the current in a loop. In fact many industrial sensors, at least over here, have a '4-20mA' output, they are designed for use in a current loop with a suitable analogue current detector. It is perhaps interesting to realise that if the sending device is perfect (that is, the voltage output in a votlage-level sender is independant of load, the current in the current loop is independany of the loop resistnace), then voltage level interface is unaffected by leakage between the wires, but the received voltage (assuming a non-infinite input resistae) does depend on the wire resistace whereas the current loop current is unaffected by the wire resistance but the current through the detector will be reduced by any leackage in parallel with it. In many cases leakage is a lesser problem than wire resistance, hence the use of current loops OK... Let's get back to digital current loops of the type you're tryign toi get to work. The current loop circuit is quite simple. It is a loop containing 3 items : A 'switch' to control thr current; a 'sensor' to detect the current; and a power supply. For practicla reasons it is sensible to put the power supply in the same case as one of the other devices. An current loop device (wheter sender or detector) with the power supply built-in is called 'active'. One without is called 'passive'. You link an active device to a passive device. That way you have the 3 things you need in the loop. 2 psssive devices linked together can't work, there is no power supply to send current round the loop. 2 active devices linked togethter is a bad idea, there are then 2 power supplies in series. The Teletype Model 33 is passive on both transmit and receive AFAIK, It is certainly passive on transmit. The transmitter is just a complex circuit of switch contacts. So anything designed to connect to that transmitter has to be an active interface (and yes, every DEC terminal interface I've looked at has been active). The fact you are getting loop current means you have a PSU in the loop too. That is, there is an active device in the loop. Since the RS232 interfce you are using, wired as you say you have wired it, is passive, the PDP8 must be active. I think, though, you don't need ot worry too much about 'current loop' if that's what is scaring you. You hve the scheamtic of the RS232 adapter -- it's in the document you pointed us at. I am sure you have schematics of the PDP8 side too. In which case you can analyse it like any other circuit. It's just transistors and simple ICs after all. Now for a dirty trick I learnt years ago. I notice there's an opto-isolator in the RS232 adapater between the RS232 input buffer and the current swich transistors. I don't know if the idle state of this opto-isolator is 'on' or 'off'. So for the dirty tricks : 1) You can see if the isolator should be on by measuring the voltage drop across the LED series resistor and seeing if the LED is carrying any current. See if that changes when you send a break. 2) The really dirty trick. If you want to force an opto-isolator off, short out the LED conenctions. If you want to force it on, short C-E on the output-side transistor. For opto-isolators in 6 pin packages, those are pins 1-2 and 4-5 respectively. See if doing either of those will reduce the loop current to near 0. The idea is to find out what section works and what doesn't. There aren't that many components involved. -tony

OK. About the only thing you could have got wrong is the polarity of the wires in the current loop from itnerface to PDP8/L If you have wired the adpater as fig 2 i nthat doucment you pointed us at, then it is passive. I have looked at the PDP8/L maintenance manual on bitsavers. If the scheamtic of the teletpye interface in Vol 1 is complete and accurate (and I think it is), then it is clearly an active device and is actualyl quite simple. I don't think the problem is there. The docuemant on the RS232 adapter also incudes a schematic. I would work through that. My first check would be on the LEd side of the optoisolator in the PC->PDP8/L circuit -- that is the one where the LED is being driven by the PC (RS232 interface). First check that the LED current changes when you send breaks. If it doesn't you need to troubleshoot the RS232 receiver, LED driver, etc. If it does, then you need to troubleshoot the circuit on the transistor side of that optoisolator. It's only a couple of transistors. -tony

AS I understnad the problem, the PDP8/L does, indeed, have a current loop interface, but the OP has got a current loop to RS232 itnerface that he is trying to use with said PDP8/L -tony

You say, it only drops to 20mA. What is it dropping from? If it is still 40mA you have multiple problem. As has been said, the 33 is suppose to be a passive end but it can be configured as the active end by someone before you. If you have 40mA something is not right. It is not transistors turning on, it is an additional supply or wrong resistors. With the output of the opto shorted, the loop should go to 0 mA ( actually about 60 uA because of the 200K resistor ). If as you say, it only drops to 20mA something on the interface should be getting warm. 12V at 20mA is about 1/4w for small surface mount parts you should be able to feel that with your finger. Dwight

I'm looking at the RS-232 converter schematic: http://www.bb-elec.com/Products/Datasheets/r2_Website- Only_0812DS.pdf and this PDP-8/L schematic: http://bitsavers.trailing-edge.com/pdf/dec/pdp8/pdp8l/DEC-8L-HR2A- D_8Lschem_Feb70.pdf power supply: page 25 W076: page 37 and http://www.fairchildsemi.com/ds/6N/6N136.pdf (So all of the following is derived from examining the schematics, not previous measurements. I had to redraw the W076 diagram so it makes a little more sense.) One thing you might check is whether the 'optional' -30V is present on the W076, it appears it can be supplied internally or externally. The power supply schematic indicates it is produced by the 8/L supply, whether it makes it to the W076 will require examination. With the -30V the 'open-state' loop voltage would be +5-(-30)=35V, which exceeds the max specs for the 6N135 opto-isolator in the converter. Could be sending it into breakdown/conduction or damaged the converter. Try removing the -30 if it was present. Without the -30 it appears the W076 loop voltage should be +5-(-15)=20V (diode switching of supply) which the converter might be happier with. The converter design appears to me to be a little on the weak side for V specs. If the -30V was present, you might try checking the converter disconnected from the PDP-8. - Connect T- to ground, and T+ to a V+=(+5 or +12) through a 1K- or-so resistor. - T+ should show ~ +0.2V/V+ switching with the RS-232 signal-in. Another suggestion could be to confirm the W076 is working properly. If I have things more or less right, without the -30 supply and the loop disconnected: - W076 with pins 3,4 open (no external connections to 3,4, measured relative to the PDP-8 ground): - pin-4 should = ~ +5V - pin-3 should = ~ -15V - pin-E should = ~ +5V - W076 with pins 3,4 shorted (and no other connections to 3,4): - pins-3,4 should = ~ -(15-2*0.6) = ~ -13.8V - pin-E should = ~ -0.6V (one conducting diode drop) It looks like the receive-loop current from the W076 will be around 43 mA (a current meter between pins 3-4 for example) which seems odd, but that's what my calcs say, unless I've misinterpreted something in the schematic. But it matches what you've been seeing doesn't it? On 2013 Sep 20, at 11:35 PM, Josh Dersch wrote:

I'm looking at the converter schematic but it is onpage 37?If it is a 8L, I figure 20 mA.It is -30volts to +5V or 35V.R1 and R2 are 1K and 750 or 1750 ohms.35/1750 = .02 amps or 20mA. Point N is pulled to -10v when there is 20mAflowing.One could clamp point P to -10V with a 10 voltzener to ground. That would still get the 20mAwhen the contacts were close but limit the voltageacross the contacts ( RS232 converter ) to 15Vinstead of 35V.A half watt zener would work here as it is only0.2Watts to deal with.I hope this makes sense.Dwight

On 2013 Sep 21, at 10:40 PM, dwight elvey wrote: (Dwight, just for clarity, I'm referring to the the RS-232->current loop dongle as the converter, the W076 interface card being on page 37 in the 8/L schematic.) Some more of the reasoning from the earlier examination of the schematic: - When the loop (3,4) is open: pin-4 is pulled up to +5 through R2 (750), pin-3 is pulled down to -30 through R1(1K) and D2 *if the -30 supply is connected to the W076*. However if the -30 is not connected, pin 4 is (should be) pulled down to -15 through now- conducting D3 and D7. (D3 and D7 are reverse biased/off if -30 is connected). - When the loop is closed (3,4 shorted), -V (15 or 30) has two paths to flow: to +5 through R2 and to ground through R4 (750) and D4 (D4 now forward biased and clamping the negative V (intended so point E to the TTL logic doesn't go too negative). I get the 43mA from the sum of those two paths (25mA to +5 plus 18mA to GND) when -30 is disconnected. From my calcs, when the -30 is connected the current is actually lower, despite the higher V, because of the higher impedance through R1(1K). - Josh, the measurements you presented make sense for the -30-connected case, but not for the not-connected case. They suggest that the -15 is not present. Check whether -15 is supplied to the card (I expect it is as IIRC you said the 8->PC direction was working, should be point B but note doc also mentions point K). Without the -30, -15 should be making it through D3 and D7 to pin-3 (to test, might need a slight load to conduct when open-loop). You said you replaced a broken diode earlier, might the new diode be installed backwards? If it's D3 or D7 it would explain the lack of -15 at pin-3 and would also (counter-intuitively as per above) explain the reduction in the current (~40mA down to 23). I think you should figure out what's going on with the W076 before thinking about modifying the converter. Should try to get the expected behaviour and measurements out of the W076 without the converter involved. If that resolves as calculated, the lower V of 20V ( =5-(-15) ) may still be pushing it for the 6N135 but it may work.

On 2013 Sep 22, at 8:10 AM, dwight elvey wrote: I know, it's horrible trying to figure things out when they're drawn in a convoluted manner as it is. That's why I redrew it, and do so with lots of things, so (as much as reasonably possible) the E potential is consistently increasing in the vertical direction.

On 2013 Sep 22, at 4:10 PM, dwight elvey wrote: I was thinking along that line too if the converter has to be made to work at a higher voltage, but figure Josh should first confirm whether it's working properly at any voltage; it might function at 20V/40mA if it's OK. I've long had the impression that most manufacturers' process to produce a schematic is the design engineer hands off some scraps of the original to a draftperson whose priority is "how do I squeeze these lines and symbols onto a page of paper". I don't have it around any longer, but IIRC the fold-out APPLE II schematic was horrible, with gates oriented in all directions and signal lines traversing back and forth everywhere. No worse than a lot of stuff though.

On 2013 Sep 22, at 10:31 AM, Josh Dersch wrote: Great, the W076 looks good, and there's logic-level switching at point E. (I presume that's actually -0.6V relative to GND at pin-E that you're measuring.) I'd turn your attention now to the converter and check it in isolation, disconnected from the loop. Wire it up as suggested earlier with a 10-15V supply and make sure it's working properly, although I was a little off with the levels to expect. Here's an amended version: - Connect T- to 0V/GND, and T+ to a V+ = (+10 to +15V) through a 1K-or-so current-limiting resistor, - T+ should show ~ +1.5V and V+ switching with the RS-232 signal-in. - if that works, try boosting the V+ to +20 to 24. It looks like D1 and D2 are there to provide some minimum voltage drop so the 6N135 always has some Vcc. ( ref: http://www.bb-elec.com/Products/Datasheets/r2_Website- Only_0812DS.pdf ) What Dwight was saying earlier about it likely not being damaged would be true for a common 6-pin type optoisolator, unfortunately the 6N135 is an 8-pin type with separate Vcc connection, so it would have seen the excessive 35V directly across it (pins 8 and 5). I don't know why they used such a device, the thing isn't going to be working at really high speeds. When I designed a little current-loop adapter a few months ago, I was targetting 60mA and 20mA loops and had to make sure it would tolerate higher voltages (over 70 or even 100V), but still just used a common-as-dirt type opto-isolator. I don't expect the current level is an issue, I don't see any problem with it dealing with 40 or more mA, the 3904s are rated up to 200mA and the darlington pair should be enough gain to keep them from saturating.

Originaly, it was probably designed to connect to a Teletype Model 33 raehr than 'any old current loop device;. Said Teletype was designed to be used in 20mA or 60mA loops, there's a jumper wire to shift to set the receiver current threshold. But the transmitter is just contacts. It doesn't care waht the current is, within reason. If it works at 20mA nand 50mA, it'll work at 43mA or whatever too. My guess is htat DEC designed this thing to sue the supply rails that were available and stnadard components. And that's all. -tony