Project PITS Problem

Question

I have been trying to build up a Pick-up Indicator Test System (PITS) to detect the overcurrent indicator on protection relays I test, for the project14 test instrumentation.

Once I figured out that there are some photo diodes that respond better to green LEDs than others, I managed to get the circuit working. However, I then decided to add an output indicator off the negated output of the AD8561 comparator chip I was using. After adding this, both outputs of the comparator remain low and the LED is on all of the time.

When I drop out R6, the 10k resistor to the pnp transistor base, the circuit starts to work again.

Anybody have any ideas what I am doing wrong? The only thing I can think of doing is to increase the size of R6, I had originally calculated this as 14k for a 1mA base current, but from my limited selection, 10k was the closest I had.

I also find that when turned on, the MOSFET has a resistance of 780 ohms, which seems a little high to me?

Kind regards

jc2048 · Accepted Answer

Your transistor is upside down - you've got the collector and emitter swapped. A transistor will work like that, but the gain is very much less than if you use it the right way round so you'd only ever do it in a circuit if you specifically wanted the very low saturation voltage it gives [in the old days, people used to build chopper amplifiers with transistors in 'inverse mode' for that reason]. The comparator outputs are designed to drive TTL inputs where the threshold would be around 1.5V and the high output only needs to be (at least) 2.8V. If you've got a high-brightness white, green or blue LED, where the forward voltage would be a little over 3V and it would illuminate on a couple of milliamps, you could consider something like this The comparator is very fast. Make sure you have the decoupling they recommend, close in to the part's supply pins, and be careful with the layout. You might also want some hysteresis - the input will only move slowly when there's no photodiode current [the reverse capacitance of the diode is quite high and there's only 5M to charge it], so when you slowly transition past the switching point there's a real possibility that noise will cause the output to chatter.

Dudley · Answer

better now?

three-phase · Answer

Hello Rod,

Many thanks for your help.

I have read through the data sheet for the AD8561 and it does not say much about the outputs and if they are differential or inverse from one another.

I have just replaced, the 10k pull down resistor with a 300k one and the MOSFET now goes down to 30 ohms when switched on.

The output voltages are as follows with a supply voltage of 4.52V

Comparator off - Pin 7 is at 0.188V and Pin 8 at 3.86V

Comparator on - Pin 7 is at 3.30V and Pin 8 at 0.190V

These were measured with R6 still removed.

I have not tried an NPN transistor. I did not want to run that in parallel with the MOSFET gate in case it affected its operation. I will have a look and see what is in my box of bits and see if I can find a bigger resistor for the gate.

Kind regards

14rhb · Answer

Hi Donald,

[I could be completely wrong...]

I wonder if those outputs are differential rather that just a plain inverse of each other. By taking one to your PNP you are pushing the feed to your MOSFET below any possibility of turning it on (as the difference between them is not changed) ?

The gate resistor to ground on your MOSFET is just to ensure it turns off, I'd make that a bit bigger e.g maybe 100k to 1M. This would make your Vgs higher and make it conduct better (lower Rds).

Have you tried using a NPN off the main output to drive your LED?

Rod

three-phase · Answer

Many thanks for the information, I believe the led I fitted was high intensity so I can look to rework the circuit and get the transistor connected the correct way around. The circuit shouldn't transition slowly. The photodiode will sit enclosed over an LED on the protection relay and conducts when that LED is lit. I haven't noticed any chatter whilst testing the circuit. Unfortunately, I haven't given all the details of how it will be used, so you wouldn't have known any of this. Kind regards.

jc2048 · Answer

If it turned out you did need the hysteresis, the datasheet has a section explaining how to implement it. BTW there's nothing wrong with having the LED drive and MOSFET drive separate, indeed from a design point of view it's probably better - I just thought it was neat that with a 5V rail you could use the 3V drop of the LED to do the level translation up to the transistor and combine the two.

three-phase · Answer

With the helpful answers, I have made some adjustments and have a rudimentary system functioning. I have increased the pull down resistor to a 680k in series with 300k. Testing this, the resistance of the MOSFET dropped to 22.1 Ohms with the comparator turned on. So I think for a more final version, I will get a 1M resistor as suggested by Rod. I then utilised the circuit identified by Jon and can get the LED to light when the comparator is turned on. An added bonus is that the voltage to the gate on the MOSFET has now increased to 4.51V from 3.38V and the resistance of the MOSFET is down to 1.3 Ohms when it is on. Many thanks to jc2048 and 14rhb for your timely assistance, much appreciated. Kind regards.

14rhb · Answer

Hi Donald,

You are more than welcome. Sounds like you are making headway - which is always good to hear.

I'm not sure how you intend to use the output 4mm or how much current you want out; is 30 ohms a low enough resistance? I looked up the MOSFET you are using and the device conducts completely when Vgs is 10v, although the Vgs threshold is 2-4V. You could perhaps find a device that saturates at a lower Vgs ? There seems to be a handful of MOSFETS that have much lower "Rds(on) Test Voltage Vgs" which is a parameter you can select and filter via on the Farnell/Newark website...but most devices seem to be SMT

I think you should be all right taking the transistor and MOSFET off the same output. The MOSFET gate will be high impedance so won't affect the output. The transistor will be fed via its base resistor to derive the base current. [Famous last words ]

Nice Veroboard layout by the way.

Rod

dougw · Answer

I would flip Q2 (emitter to +5). That opamp may not drive high enough to fully turn on the FET. Switching R5 to be a pull-up instead of a pull down should help, but it may need to be adjusted to get the optimum thresholds.

three-phase · Answer

The 4mm output jacks will connect into the timing circuit of an injection test set, very low current but the voltage from the test set can vary depending on its setup or the the of tester used and I wanted to keep the circuit usable on as many testers as possible.

I will try an npn transistor over the weekend and see what happens.

Thanks for the help.

three-phase · Answer

The photo is a jpeg but the schematic was saved as a png - maybe that is the reason? Kind regards

Project PITS Problem

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