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LT1011 Comparator

Thread starter #1
So I'm working on a circuit that has a PWM input signal and my goal is to control a 24V relay coil.
So when the filtered voltage on the PWM drops below 18V (Vref) I want to energize the coil. But the issue is when I add the coil in the ouput of U1 remains hi.

Also I'm not familiar with the strobe and offset signal, so I left them open, what are those really used for?
Comparator.PNG
 

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alec_t

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#2
Your coil has zero resistance. The LT1011 can't provide enough output current. Check its datasheet to see its maximum load and all the pin functions.
 

dr pepper

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#3
Edit the coils properties and give it some series resistance 400 ohms would be a reasonable ball park.
 

audioguru

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#4
The LT1011 comparator has a maximum allowed output current of 50mA so with your 24V supply the circuit has a minimum relay coil resistance of 24V/50mA= 480 ohms when its output voltage will be a maximum of 2.5V.
47 years ago the 24V relay coils were 700 ohms.
 
Thread starter #5
Ok thanks. The relay I'm using is specified to have a resistance of 1440 ohms, I entered that into the simulation and works as expected.
Also does anyone think it might be a better design/practice to have a mosfet switching the load instead of being directly from the op amp?
 

audioguru

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#6
The datasheet for the comparator shows its output voltage loss with load current but the simulation shows only "typical", not the worst case. It is probably acceptable. An un-needed Mosfet will have a cost but have much less voltage loss.
 

dr pepper

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#7
Yes it would be a good idea to buffer the op amp with a transistor.
A 2n2222 transistor and a 10k ought to do the job, the relay will only draw about 20ma.
Check the datasheet for info on the strobe pin, you may need to tie it high or low for stable operation.
 
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audioguru

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#8
You do not need to buffer it with a transistor.
The comparator has a maximum allowed output current of 50mA and the 1440 ohm relay coil draws a little less than only 24/1440= 16.7mA.
The 1uF capacitor at the output of the comparator should be removed because it might destroy the comparator when it tries to discharge it with unlimited current.
 

Tony Stewart

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#9
The Capacitor draws excessive current and this prevents the PWM that you desire by integrating the coil current to produce an average Voltage. Remove it. The Open Collector output will integrate when active low and then flyback for a period of τ=L /R (R=1440 ohms)

Verify L and ensure your PWM On time T << τ in order to turn it off. Usually the release current is 20 to 50% of the rated current @ 24V

With the cap removed the pull down slew rate should be τ=L/ 12 Ohms approx. ( using Vol/Iol slope from the datasheet.)


and make sure your diode is in reverse bias across the coil.

Otherwise you wont be able to pull it down.
 
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ChrisP58

Well-Known Member
#10
The Capacitor draws excessive current and this prevents the PWM that you desire by integrating the coil current to produce an average Voltage. Remove it. The Open Collector output will integrate when active low and then flyback for a period of τ=L /R (R=1440 ohms)

Verify L and ensure your PWM On time T << τ in order to turn it off. Usually the release current is 20 to 50% of the rated current @ 24V

With the cap removed the pull down slew rate should be τ=L/ 12 Ohms approx. ( using Vol/Iol slope from the datasheet.)
Are you talking about C1 or C2?

C1, along with R4, integrate the PWM signal and deliver a DC voltage to the input of the comparator. I don't believe that there is any (intended) PWM activity on the output of the comparator, or on the coil.

Note *
Due to the lack of hysteresis around the comparator, there will be some unitentional PWM like chatter on the coil near the transition point with C2 removed.

See next post.
 

ChrisP58

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#11
Electronic comparator circuits have very sharp, precise transition points. While generally this is good, it can be very troublesome. In your circuit, even after the RC integration of the PWM signal there is still about a 180 milliVolts of peak-to-peak ripple. This will produce chatter on the output of the comparator when it is near the 18 Volts trip point. I suspect that you may have noticed this and that may have been why you put in C2, as that capacitance would tend to mask the problem.

The proper solution is to ad hysteresis to the comparator. What this does is that it makes the rising and falling transition points different. There is no hanging on the edge, because as soon as you get to the edge, the edge moves and you find you're already passed it.

To impliment hysteresis in your circuit, insert a 10K resistor between the top of C1 and the noninverting input of U1. Add a 1 Meg resitor from the output ot U1 to the noninverting input. These values add about 235miliVolts of hysteresis. More than enough to offset the peak-to-peak ripple on your input signal.
 

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Thread starter #12
Electronic comparator circuits have very sharp, precise transition points. While generally this is good, it can be very troublesome. In your circuit, even after the RC integration of the PWM signal there is still about a 180 milliVolts of peak-to-peak ripple. This will produce chatter on the output of the comparator when it is near the 18 Volts trip point. I suspect that you may have noticed this and that may have been why you put in C2, as that capacitance would tend to mask the problem.

The proper solution is to ad hysteresis to the comparator. What this does is that it makes the rising and falling transition points different. There is no hanging on the edge, because as soon as you get to the edge, the edge moves and you find you're already passed it.

To impliment hysteresis in your circuit, insert a 10K resistor between the top of C1 and the noninverting input of U1. Add a 1 Meg resitor from the output ot U1 to the noninverting input. These values add about 235miliVolts of hysteresis. More than enough to offset the peak-to-peak ripple on your input signal.
You are correct! I noticed the issue on simulation when input was near the trip point and why I added the cap. The datasheet of the comparator also says I can use the balance or balance/strobe signal to add hysteresis how is this method different from adding a positive feedback?
 

ChrisP58

Well-Known Member
#13
The method I gave is the classic way to add hysteresis to most any generic comparators, or even op-amps being used as comparators. I'm not very familiar with the special features of the LT1011. Glancing through the datasheet for the part, I did see that feature mentioned, but not much usage information is given. It can probably provide the necessary hysteresis internally, but I don't have the time right now to chase down the details of how to implement it.
 

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