Need help please with a Voltage alarm with LT1017

Thread Starter

dscha

Joined Nov 19, 2018
13
Hello!

With help from the community, I have redeveloped my voltage alarm circuit. I started off with an 741 Opamp and it did not work as I wanted it to work. So I did some research and found a LT1017 comparator.

Based on the data sheet I assumed that I can drive a LED with the output (it says in the datasheet that it can source 70 mA). But it does not work. The output voltage collapse under load.

Anyway, I switched it around and used a PNP (2N2907) to drive a relay, LED and buzzer (See schematics).

So I still have issues which I don't understand why. The 22N907 has a hfe of around 250 (measured with my multimeter). The Buzzer, relay and LED use about 70 mA. Let's keep it simple and use a hfe of 100. This means I need 0.7mA as ib and with 14V Vcc we are at 14-0.7 at 13.3V/0.7mA, that would be 20k.

My problem: With this calculated Rb nothing happens. Even with Rb = 1k only the relay and LED works, the buzzer does not kick off.

If I play with Rb and measure ib and ic I get ib of around 10mA and a ic of 50mA. So the transistor has a hfe of 5 according to this.

I know that I must make a massive thinking error, but where is it? I learned that stuff 30 years ago and I am going back into it now a little. Of course I forgot many many things from back then, but what do I do wrong?

Thanks in advance for your input.

best D

IMG_0005.jpg
 
Last edited:

AlbertHall

Joined Jun 4, 2014
11,609
Looking at the datasheet, the base current used for measuring the transistor saturation voltage is one tenth of the collector current and that is a good rule of thumb when you want a transistor to be saturated.
upload_2018-11-19_21-18-7.png
 

Thread Starter

dscha

Joined Nov 19, 2018
13
I made a new discovery. the transistor does not seem to be problem. The comparator does not switch between Vcc (14V) and 0 (which is my Vss) but stays on about 3.5V when the two input voltages are very close together. So this influences the transistor basis current and therefore the Ic. I thought that there will be a "clean switch" between Vcc and 0 when the one input becomes higher than the other. Is this not the point of a comparator? As always, looking forward to your thoughts.
 

ebeowulf17

Joined Aug 12, 2014
3,282
I made a new discovery. the transistor does not seem to be problem. The comparator does not switch between Vcc (14V) and 0 (which is my Vss) but stays on about 3.5V when the two input voltages are very close together. So this influences the transistor basis current and therefore the Ic. I thought that there will be a "clean switch" between Vcc and 0 when the one input becomes higher than the other. Is this not the point of a comparator? As always, looking forward to your thoughts.
Could it be oscillating? Maybe that's not a steady 3.5VDC output, but a high speed oscillation that averages out to around 3.5V as read by a meter.

If so, maybe you just need a feedback resistor to add some hysteresis. I didn't see any mention of hysteresis in the datasheet, so I'm not sure what, if any, is built into the device.
 

Thread Starter

dscha

Joined Nov 19, 2018
13
Could it be oscillating? Maybe that's not a steady 3.5VDC output, but a high speed oscillation that averages out to around 3.5V as read by a meter.

If so, maybe you just need a feedback resistor to add some hysteresis. I didn't see any mention of hysteresis in the datasheet, so I'm not sure what, if any, is built into the device.
That might be the case, but i would have thought that a dedicated comparator should switch cleanly. Will try to find out mote about jow to calculate the values for the feedback resistors
 

ebeowulf17

Joined Aug 12, 2014
3,282
That might be the case, but i would have thought that a dedicated comparator should switch cleanly. Will try to find out mote about jow to calculate the values for the feedback resistors
They call it a comparator, but they include linear circuits in the application data example circuits. I think you need external hysteresis.
0A321003-F5A7-43D5-8229-6D9577EBEFE1.jpeg
 

Bordodynov

Joined May 20, 2015
2,939
I will help you correct your circuit, but you need to clarify what you connected to the non-inverting input. Those. What is the source of the signal and how it is connected to the common wire (and specify the potentiometer value). Lushe if you give a more complete (input) scheme.
 

Thread Starter

dscha

Joined Nov 19, 2018
13
I will help you correct your circuit, but you need to clarify what you connected to the non-inverting input. Those. What is the source of the signal and how it is connected to the common wire (and specify the potentiometer value). Lushe if you give a more complete (input) scheme.
Thanks!

The non-inverting input will measure the voltage drop on an oil pressure sensor of a boat diesel engine. With the Zener Diode and the potentionmeter, I adjust the voltage divider in a way that the comparator switches when voltage drop on the oil pressure sender is around 4.8V. Everything is connected to the board power supply, hence I used 14V when the engine is running.

I want to drive the buzzer as the audible alarm, the LED as an internal control light (which i might loose in the final design) and the relay will drive the control light on the engine panel (will pull the control light to ground)

Does this make sense?
 

Thread Starter

dscha

Joined Nov 19, 2018
13
They call it a comparator, but they include linear circuits in the application data example circuits. I think you need external hysteresis.
View attachment 164069

Thanks!

The non-inverting input will measure the voltage drop on an oil pressure sensor of a boat diesel engine. With the Zener Diode and the potentionmeter, I adjust the voltage divider in a way that the comparator switches when voltage drop on the oil pressure sender is around 4.8V. Everything is connected to the board power supply, hence I used 14V when the engine is running.

I want to drive the buzzer as the audible alarm, the LED as an internal control light (which i might loose in the final design) and the relay will drive the control light on the engine panel (will pull the control light to ground)

Does this make sense?
EDIT: It is actually the inverted input that gets the voltage divider and the voltage Ii want to track.
 

Thread Starter

dscha

Joined Nov 19, 2018
13
That is brilliant. So basically the feedback resistors with 300k and 1k are new plus the capacitors C1-C3. My R2 is currently 1K, so this should make no big difference. Just a more stabile zener diode operation as far as i understand it, correct?
How do you calculate the reefback resistors please? is there a "standard formular" ?
Thanks a lot for your help. Will build it later today :)
 
Top