anyone suggest circuitry for CMOS to TTL logic conversion
As shown in attachment below,
what the problem is input to opamp is 6-8v DC which is first case;
now since opamp is used as a comparator, it gives 4-5V output,which is desired;
BUT, even when input to the comparator is 200-300 mV still output is 4-5V.
Now, the more interesting fact is that if I supply the comparator 6-8v DC from a separate voltage source, comparator gives 4-5V output which again is desired.
Now,when input to the comparator is 200-300 mV from a separate voltage source, output is of the order of 200-400 mV,which is absolutely correct.
But, obviously I can't take 6-8V input from a separate voltage source(I have to use the 6-8V DC generated within my circuit.
Is it the current that is playing underground role?
But since I have used 7809 voltage regulator as supply and there is only one IC before this stage;So I don't think current is the cause.
So what should I do?
Any similar experiences?
Regards,
Kushal Khanal
The input voltage divider should be at vcc/2 for cmos input compatibility. Put two resistors of equal value as a voltage divider here, the junction going to pin 2 of the op amp (1IN-). Use 10k, 1meg is too high.
The output should be able to sink 1mA or so at logic=low and look like an open circuit (or several k) when high for the output to look like a TTL gate output. Put a diode cathode (the arrowhead) directly to pin 1 of the LM358 and the anode to Vcc through 5k. The TTL output will be at the anode of this diode. Remove the other two resistors at the output.
Hey there AA1LL
May be i couldn't make you understand.
Or you might have forgot
1.Are you clear that the IC LM358 perfectly works as a separate module even with the circuitry as in the attachment;
But in the circuitry where it is to take the input 6-7 v or 200-300mV from circuit
it doesn't works as wished?@?@?!
What might be the cause ?
2. And since in LM358 biasing voltage is 9V .Did you regarded this as Vcc ?
Or can I use Vcc as any other voltage.I think I can .
Regards,
Kushal Khanal
i think u must use a voltage divider at the input.,
First of all, you don't need the cap on the inverted input of this circuit.
Second, this is a bipolar device, the current at the base (the noninverted input) has to be above 6 microamps. I would check the circuit that feeds the non inverted input.
Note : you have 8.59 microamps on the inverted input of this circuit. this is ok