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We haven't got an integrated version but a 2 chip version for half-duplex RS485.

For RS422 I can do something similar for RS422 (or full duplex RS485) but only if a two-chip solution is acceptable to you.

Let me know,

Thomas

Hezi,

I have attached the pdf version on an isolated RS-232 to RS-485 converter. The article is published in Planet Analog and EE-Times. With regards to the Condensing issue, the only way I can think of protecting the circuit against high humidity is by using a hermetically sealed chassis.

Regards, Thomas

Hi Thomas,

How is the 2ms high time at data rate of 9600bps is calculated in the above article?

"The time constant of the monostable output is defined by an R-C network with C
= 220 nF, and R = 10 kOhm for a 2 ms high-time at a data rate of 9600 bps, and
R = 100 kOhm for 20 ms at 1200 bps."

Thanks

Hi Kepcoate,

an old friend of mine used these delays for certain 485-protocols, which required a specified inter-packet delay, that is the time between one packet or character of data and the next one following. In the baove case I believe the interpacket delay was 0.75 T-char (the character time). A character or packet or whetever they call it nowadays typically consists of 11 bits: 1 start bit, 8 data bots, and 2 stop bits. At 9600 bps a character length is 11 / 9600 = 1.146ms. Dividing the 2ms by 1.146ms gives you a factor of approx.1.75, indicating that the high-time of the mono-flop is 75% longer than the actual character length. In Modbus RTU, another industrial protocol for RS485 data links, the inter-packet delay is even 3.5 times the character length or Tchar.

So if you only want to keep the 485-driver active for the length of the character, simply divide the 11 bits by the baud rate you are using and then add 10% for temperature variation of the mono-flop and whatever percentage you need for covering the tolerances of the external RC.

Regards, Thomas.  

I just checked looked at a few older converter desings using mainly throughhole components. Here the older mono-flops or one-shot circuits had much larger tolerences. Also the resistors and caps used had 20% and 30% tolerances respectively.

Hence as a rule of thumb I suggest make your time constant 1.5 x the character length. (which isn't too far away from the 1.75 above anayway).

So I hope this simplyfies it. Good luck with your design.

Regards, Thomas

Thanks for clarifying this further. That helped a lot.

Thomas,

I was looking into selecting the wattage of 120 Ohm termination resistor. So I did some scope shots under no load and no termination condition. See below. CH 3 shows the RS485 differential output , CH1 and CH2  are A and B outputs with reference to GND. So pk to pk is roughly 10V.  Therefore P=VxV/R= 10x10/120 = 0.833W. This wattage seems high compared to 0.5W recommended by some app note online. Any thoughts or recommendations on this?

Hello Kepcoate,

The Duty cycle (in the current flow) will be never 100%, in your example only 16%.

I think roughly 50% would be enough, so 0.833 x 0.5 = 0.4165W

-Leo

Hi Leo,

If you look at the scope shot , if the bus is idle the 10V pk to pk is present 100% of the time. That is why  my calculations assume 100% duty cycle.

Other thing is, per RS485 spec the receiver common mode voltage can be 19V ( -7V min to 12V max). Going with this number the wattage  is P= 19*19/120=3W. Not sure if I am doing this part correct.

Thanks

Good day,

the worst case condition is if A or B is constantly at 5V while the other output is 0V. Then you have 5V dc across a 120 ohms resistor which yields a power of 5² /120 = 0.2W.

It is the differential voltage that determines the power dissipation in the termination resistor, not the comon-mode or ground referred voltage.

Good luck , Thomas

Hi Thomas,

If you look at my CH3 differential scope shot, it show 10V as differential. With just that number, I am getting 0.833 watt. The 19V is differential if -7V to 12V is common mode by reasoning of the attached scope shot. My scope shot is 5V on each A and B output w.r.t. GND but differential is 10V.

Thanks

Hi Kepcoate,

Channel 3 is the math-function of your scope calculating Va - Vb, it is not a real 10V drop across the resistor. The differential voltage across a 60 ohm resistor is approximately 2V (sometime even 3V), either in one or the other direction. The limited differential output voltage of less than 5V is due to the internal voltage drop across the driver output transistors. In your measurements without termination the only load is the 1M resistance of you scope probe.  Because there is only leakage current floating, no significant votage drop occurs across the output transistors and hence you can measure the open-circuit output voltage. Under loading condition however, with two parallel termination resistors and multiple transceiver nodes, Vod will drop drastically.

Let's assume you have a common-mode output of Vcc/2 = 2.5 and a high output voltage of Vod = 3V under load.

Worst case 1: When A is always high and B is always low then Va = Vcm + Vod/2 = 2.5V + 1.5V = 4V and Vb = Vcm - Vod/2 = 2.5V - 1.5V = 1V. The differential output voltage across a 120 ohm resistor is therefore Vod = Va - Vb = 4V - 1V = +3V.

Worst case 2: When B is always high and A is always low then Va = Vcm - Vod/2 = 2.5V - 1.5V = 1V and Vb = Vcm + Vod/2 = 2.5V + 1.5V = 4V. The differential output voltage across a 120 ohm resistor is therefore Vod = Va - Vb = 1V - 4V = -3V.

In both cases the power dissipation is P = V² / R. Because V² is always a positive number your can simply calculate P = (+/-3V)² / 120 W = 75mW. So during normal operation and assuming no fault condition, you could easily use 1/10W termination resistors.

However when considering fault conditions, the power rating of the resistors can go way up.

Let's assume one end of the termination resistor at the far cable end looses connetion to the B-line but makes contact with the receiver ground instead. Let's further assume that the ground potential difference between driver and receiver is -7V. In the case that the driver output terminal A was high, the output voltage potential at A will drop suddenly to say 0.5V. Now you have a voltage drop of 7.5V across the remote 120 W resistor and the power dissipation rises to P = 7.5V² / 120 W = 0.47W.

Other fault scenarios can include accidental contact to parallel running 24Vdc power lines under full comon-mode, where termination resistors have power ratings of several watts.

I hope the above clarifies the power calculation for your termination resistors.

best regards, Thomas

First of all thanks for doing this effort to give a face to the 120Ohm termination wattage selection process.

I did some measurement using 120||120 = 60 ohm load and I measure 66mW on 60Ohm. The CH3 voltage above was measured with a differential probe. Although the math function gives same waveform. The key is the 60 ohm did load the driver output quite a bit. From the above calculations it seems 1/2W will be in fact suffice for most cases.

Thanks Again.

Glad I could help. - Thomas