window comparator question need help

Thread Starter

SHING19HK

Joined May 30, 2012
7


please forgiven my poor english, i will try my best to express.

in question a i find out Vul have to be 3V and Vll = 1V ,but i googled window comparator,
although i know Vmax and Vmin is depend on R1 and R2 , i still dont know how to choose a correct values for these.
also above the R1, what is the +5V??
and what are those diodes for?? i only know when Vin= 4V , D2 break down, D1 forward , produces logic high for voltages...

sorry about my english, please help me...
 

#12

Joined Nov 30, 2010
18,217
+5 is the power supply. R1 and R2 have nothing to do with Vul or Vll. They are, "pull-up" resistors because the 311 chip can not provide a positive output from its internal structure. The diodes make sure the next stage can not push negative current into the outputs.
 

Attachments

Thread Starter

SHING19HK

Joined May 30, 2012
7
+5 is the power supply. R1 and R2 have nothing to do with Vul or Vll. They are, "pull-up" resistors because the 311 chip can not provide a positive output from its internal structure. The diodes make sure the next stage can not push negative current into the outputs.
thanks for helping
sorry i never learn that before, i guess i can find the values of resistors by matching LM311 in the attached files???

should i need to find out the max and min current then calculate the R1 and R2?? sorry i am confused. which part should i use for find the R?
 

WBahn

Joined Mar 31, 2012
26,398
To size the resistors, you want to satisfy the following two conditions simultaneously:

1) When the output is HI, I want the output voltage to be no lower than Vhmin when the circuit is sourcing at most Ihmax to the load.

2) When the output is LO, I want the output voltage to be no higher than Vlmax when the circuit is soucing at most Ilmax from the load.

The specific values for Vhmin, Vlmax, Ihmax, and Ilmax depend on the requirements. Like so many homework assignments, these are not given. Your text may have talked about the needs of various logic families and, if so, that is a good place to pick some from. Otherwise, just pick your own that seem reasonable and state them in your solution.

The maximum value for R is dictated by what the circuit is driving. If the following circuit is pulling any current, then the voltage at the output will drop and the larger R is the more it will drop. At some point, the drop will be too much and the output voltage won't be high enough to be seen as a HI.

The minimum value for R is dictated by the LM311's ability to assert a LO. If R is too small, the current that the LM311 is able to sink won't be enough to lower the voltage sufficiently to be seen as a LO.

The LM311 data sheet will give you a good idea of what the minimum value of R can be. As for the max, you don't have enough information. In that case, pick something reasonable and use that. For instance, you might assume that the output is driving nothing more than an LED that takes 10mA and see what that does.

Note that if the circuit being driven requires an asserted LO, that this circuit won't do it. This could cause problems if there are CMOS logic inputs connected to the output. To solve that, you could place a pulldown resistor on the output after the two diodes. You then have to balance that value with the R for the two pullups.

BTW: Your english is understandable enough and, I assure you, it is far better than my ability to communicate in your native language.
 

Thread Starter

SHING19HK

Joined May 30, 2012
7
i just too stupid:'( dont know how to read the date of LM311,is take the Output current: 50mA??? ohh ..i am sorry :'( i really dont know which part is useful, btw,i can understand what u said. I find the Vi at Low level output voltage is Vi ≤ -10mV, is Vi= V in??
thats much smaller than 1-3V... am i got the wrong date:'(??
 

Thread Starter

SHING19HK

Joined May 30, 2012
7
The minimum value for R is dictated by the LM311's ability to assert a LO. If R is too small, the current that the LM311 is able to sink won't be enough to lower the voltage sufficiently to be seen as a LO.
when output 0V,need to smaller than-ve supply current (Max)

so by using 5mA i got, R1=1k,then when output high,Vout = 5V - 0.7V (diode) - 1.6mA(1 TTL load) * 1k = 2.7V
is this correct?? i still not sure...
 

Thread Starter

SHING19HK

Joined May 30, 2012
7
when output 0V,need to smaller than-ve supply current (Max)

so by using 5mA i got, R1=1k,then when output high,Vout = 5V - 0.7V (diode) - 1.6mA(1 TTL load) * 1k = 2.7V
is this correct?? i still not sure...
 

Thread Starter

SHING19HK

Joined May 30, 2012
7
i am lost:'(

i pretty sure this setup is wrong....
i have no idea how to place Vin and Vout
also why use DC power??
how can i get something in oscilloscope by using DC power...
i am new to use multisim , can anyone correct me?? Thank you
 

crutschow

Joined Mar 14, 2008
27,025
The input signal goes to the wire between the U1 and U2 (going to oscilloscope input A). To see the circuit in operation input a 1kHz sinewave with a peak voltage of 4V.

The output is oscilloscope input B (you should probably add a 100k load resistor from there to ground).

The DC power is for the output resistors and the real world op amps (the virtual op amps you used have power built into the model).
 

Bernard

Joined Aug 7, 2008
5,784
1 k looks fine for R1, R2. Looking at schematic, P9, D1, D2 form a diode OR gate, A H on either or both inputs[ anodes] gives a H out. Now looking at inputs of U1, U2: Make a chart of inputs Lo, M, Hi for Us. U1-Lo, out H. Complete the chart.

.........Lo_ M_ Hi
U1 out L__ H _H
U2 out H__ H_ L

Looks like the gate output is always H. Go back to Fig 2 & re chart, Vul= 3V, Vll = 1 V.
 
Last edited:
Top