# LTC490 wiring caused short

I'm testing a LTC490 with a hand-turned encoder.

As I only have one IC, I tried testing only one channel (A).

I wired as follows:

Pin 1 to 5v, Pin 2 to pin header (to read output voltage with a voltmeter), Pin 3 to CLK (Encoder A output), Pin 4 to GND,

Pin 5 to Pin 7, Pin 6 to Pin 8

The idea was to use the LTC490 to encode the signal, and then feed it back into the IC for decoding. As the frequency is so low, I saw no need to use a terminating resistor, and instead made a direct connection.

It would seem that the LTC490 is providing a path of no resistance between the Vin and GND pins, which caused my 5v power supply to melt, and emit smoke. Can anybody explain what's causing this? I suspect it down to the way in which I've wired it, although I have no idea why.

UPDATE:

On closer inspection of the wiring, I realised I was feeding 5v into GND, and 0V into VDD, which I'm going to assume is what caused the short circuit.

Slightly unrelated, but the test schematic includes capacitors and a resistor (Figure. 2 in the datasheet). Can anybody tell me what these are needed for? I've wired as shown, minus the capacitors and resistors, and it seems to work just fine (of course, I'm using a low PPR encoder, turned by hand). Are they to remove AC noise?

• I should mention that no capacitors were used, not sure why that would make a difference though. – 19172281 Jul 22 '18 at 1:43
• Do all condition of pins meet datasheet specs incl ESD awareness? – Tony Stewart Sunnyskyguy EE75 Jul 22 '18 at 3:38
• Pin 5 to Pin 7, Pin 6 to Pin 8 ... that is not the way it is wired in the data sheet – jsotola Jul 22 '18 at 5:44
• @jsotola, I'm trying to feed the differential signal back into the LTC490. How do you suggest I do that? – 19172281 Jul 22 '18 at 12:32
• Ahhh, I see Pin 5 should feed into 8, and Pin 6 into 7. Do I need to use those capacitors and resistors to make this work? Could you explain what they actually do, and why they're needed? – 19172281 Jul 22 '18 at 12:34

It's really good you solved the main part of this yourself - not an uncommon problem getting Vcc and ground wires mixed up.

Slightly unrelated, but the test schematic includes capacitors and a resistor (Figure. 2 in the datasheet). Can anybody tell me what these are needed for?

Figure 2 is not a recommended circuit for anything other than testing the chip's timing diagram. If you read the "Switching characteristic" section on page 3 it adds those capacitors to presumably simulate the effect of cable capacitance. 100 pF is usually about 1 or 2 metres of cable.

Figure 5 is a typical setup and you should consider using 120 ohm resistors to match the cable impedances but, if cables are short and datat rates are low you can usually get away without: -

• So, for the purpose of this test, I'm feeding the driver's output back into the receiver's input. As the cables are indeed short, and (currently) the frequency of the signal is low, I've made direct connections, without terminating the lines. Does this seem OK to you? – 19172281 Jul 22 '18 at 19:16
• I would still terminate the line because if you do need to run the encoder faster you may get into problems. Other than that local loopback should not be a problem. – Andy aka Jul 22 '18 at 19:23
• In the final application, I won't know what how long/thick the line will be; the idea being for a variety of encoders to be used, etc. Would a 120ohm resistor suffice? – 19172281 Jul 22 '18 at 21:10
• You have to match the characteristic impedance of the line to an equivalent resistance. If you cannot define the wire you must find a way of doing so else termination reflections can and will ruin data integrity. – Andy aka Jul 22 '18 at 23:35
• Anything from 100 ohm to 150 ohm will probably be good enough but if you have probels then you might need to tweak it. – Andy aka Jul 26 '18 at 16:21