indicators will go LOW (0 volts dc) ?

AM Radio discussion. Directional arrays are FUN!
Post Reply
User avatar
abwdvm
Posts: 70
Joined: Fri Jul 17, 2015 7:32 pm
Location: Tx
Contact:

indicators will go LOW (0 volts dc) ?

Post by abwdvm »

[The power level status indicators will go LOW (0 volts dc) when activated. ]

Does this mean that once the 1000w transmitter is set to 1000w, then the 1000w indicator circuit will no longer produce +5v?
technician at an 1000w AM station in Texas
TPT
Posts: 989
Joined: Mon Dec 03, 2007 3:18 pm
Location: St. Marys, WV

Re: indicators will go LOW (0 volts dc) ?

Post by TPT »

I would assume this means that when the transmitter is on high power the status indicator would go low. Usually this indicator circuit uses a small transistor or integrated circuit for this indication, the logic in the transmitter causing the transistor to conduct & going to ground to indicate a change in status. E.G., when the status indication is "no" (not on high power, not on, etc.) you would read 5 volts on this status line, referenced to ground. When the logic indicates "yes," the transistor is turned on, and the status line would indicate zero volts ( or show a short to ground, using an analog volt-ohm meter).

Any external device or indicator attached to this circuit would then pull current through the transistor or IC to the ground reference. So you need to watch what you attach so as not to damage the circuit. Typically such status lines would connect to a remote control, which would interpret the 0 volt or ground switched condition as an indication that the particular metering channel is "on." At a combined site, this could also be used for a remote indication for the studio operator.

For example, you probably could connect a negative lead of an LED indicator to this status indicator (with the positive lead to a 5 volt supply through a resistor, --try 620 ohms or so). When activated the LED would light. ( you may need to connect the supply's negative lead to the transmitter monitoring circuits ground reference). However, a small light bulb may draw too much current and damage the circuit.

Similarly, a small relay could be used to turn on some other indicator, but a bigger relay, drawing too much current, could also cause damage. Your transmitter manual should indicate the current limits.
Last edited by TPT on Mon Jul 20, 2015 7:09 pm, edited 1 time in total.
User avatar
abwdvm
Posts: 70
Joined: Fri Jul 17, 2015 7:32 pm
Location: Tx
Contact:

Re: indicators will go LOW (0 volts dc) ?

Post by abwdvm »

[ The remote indication functions: 1) require current limiting resistors and 2) provide up to 100 mA
for indicators. A +5 volt supply is provided on the ECU rear-panel for remote control
operations. The remote control connections are located on the rear-panel of the ECU. ]


Sound right?
technician at an 1000w AM station in Texas
TPT
Posts: 989
Joined: Mon Dec 03, 2007 3:18 pm
Location: St. Marys, WV

Re: indicators will go LOW (0 volts dc) ?

Post by TPT »

Yes, typical LED's take about 40 milliamperes, so 100 MA would be adequate. A 1/2 watt resistor in the range of 270 to 620 ohms should be adequate to limit the current going through the LED--try different values to see how bright the LED is.

If you are connecting to a remote co0ntrol sample line, there would not need to be any current limiting since you are talking to another IC & the current draw in minimal.
User avatar
PID_Stop
Posts: 690
Joined: Thu Apr 01, 2010 11:58 am
Location: Syracuse, New York
Contact:

Re: indicators will go LOW (0 volts dc) ?

Post by PID_Stop »

Generally speaking, 20ma is on the high side for most LEDs; I design closer to 10 or 15 for most reasonably efficient versions.

Suppose you have a 5-volt power source; take a look at the LED's forward voltage drop, and subtract that from 5 volts. If you have an LED with a Vf of 2.1 volts (pretty common), that means the resistor will have 2.9 volts across it. Apply Ohm's law, assuming that we want 15 milliamps of forward current:

Code: Select all

	E = I R
	2.9v = 0.015a R
	2.9v / 0.015a = R
	R = 193.3 ohms
200 ohms is a close convenient value, and will yield a current of 14.5 milliamps. And multiplying that current by the voltage yields 42 milliwatts, so even a very small (1/4 watt) resistor is more than adequate.

Oh... you really don't want to try and drive a relay directly from something like this. The collapsing magnetic field when the relay turns off can generate a very high back voltage that can easily wipe out your logic output. It's best to use a driver transistor with a swamping diode (or something like a ULN-2803A octal driver).

-- Jeff
User avatar
abwdvm
Posts: 70
Joined: Fri Jul 17, 2015 7:32 pm
Location: Tx
Contact:

Re: indicators will go LOW (0 volts dc) ?

Post by abwdvm »

http://pdf.datasheetcatalog.com/datashe ... 828_DS.pdf

So one chip could control up to 8 lights? I could use existing 28v dc power supply, and all the old 28PSB Miniature Incandescent Lamps/Bulbs indicators?
technician at an 1000w AM station in Texas
RodeoJack
Posts: 200
Joined: Fri May 29, 2015 12:54 pm
Location: Port Orchard, WA
Contact:

Re: indicators will go LOW (0 volts dc) ?

Post by RodeoJack »

I guess, in all of this, I've become confused by how wide this has gotten.

The poster isn't clear about the equipment he's talking about. His mention of an "ECM board" is my first clue. He's a 'technician at a 1000 watt station', so I gather he has an AM-1A for a main transmitter.

Feel free to deride unmercifully if I missed that in a previous post. It's been known to happen.

OK. his quote, "The power level status indicators will go LOW (0 volts dc) when activated." comes from page 22 of the AM-XA manual, discussing the power level buttons related indicator pinouts on the back.

He asked, "Does this mean that once the 1000w transmitter is set to 1000w, then the 1000w indicator circuit will no longer produce +5v?"

Sort of. It can be a misleading way to put it. As I read the manual and the associated pictorial on page 23, the indication is the circuit will sink up to 100ma of current, based on power YOU provide through the indicator circuit (OK. call it "zero volts"). For convenience, the transmitter provides 5 volts for control and indicator usage on pins 18 and 19 of TB1, but you're welcome to provide your own power from another source. The book specifies +5 to +15 volts DC . Your specific question about whether the 1000 watt power level vs 1000 watt indicator is correct, though keep in mind what they're really saying relates to the numbered power button on the front of the transmitter vs the related pin on the back. You can set those controls to most any power level you need.

If I recall, a Broadcast Tools remote control has pullup resistors in its status indicators. Conceivably, you could connect a status number directly to a power indicator pin on the transmitter and it should work. For "non-logic" things, like LEDs and relays from other types of equiment, you'd need the resistor.

The poster now says: (The remote indication functions) "require current limiting resistors and 2) provide up to 100 mA for indicators. A +5 volt supply is provided on the ECU rear-panel for remote control operations. The remote control connections are located on the rear-panel of the ECU.", and he then asks, "Sound right"?

Yep. In "Broadcast Electronics" speak, all of this is what they're trying to say. The pictorial seems to support that power is not provided at the indicator pin.
User avatar
PID_Stop
Posts: 690
Joined: Thu Apr 01, 2010 11:58 am
Location: Syracuse, New York
Contact:

Re: indicators will go LOW (0 volts dc) ?

Post by PID_Stop »

abwdvm wrote:http://pdf.datasheetcatalog.com/datashe ... 828_DS.pdf

So one chip could control up to 8 lights? I could use existing 28v dc power supply, and all the old 28PSB Miniature Incandescent Lamps/Bulbs indicators?
Basically, yes... the 28PSB is well within the current drive ability of a 2803, and it can handle the voltage easily.

The 2803 is a high-current switch to ground, so you would have one side of all of your lamps tied to the +24v supply; the switched side of the lamps go to the driver outputs (pins 11-18), and the supply ground goes to pin 9 of the 2803.

Here's the thing: the ULN assumes active-high logic, meaning that a high input (+5v) turns the load on, and a low input (0) turns the load off. Your transmitter, on the other hand, is active low... so you need to invert the logic levels first before hitting the ULN. Something like an SN74HC540N (an 8-channel inverter) should do the job nicely... you might need pullup resistors on the inputs, on the order of 1K or so. This is more easily done than it might seem at first blush: if you look at the pinouts for the ULN2803A and the SN74HC540N, you'll see that the 540's outputs can line right up with the 2803's inputs, and a SIP resistor pack will line up nicely with the 540's inputs. Just power the 540 from the +5 supply, and it should work just fine.

-- Jeff
Post Reply