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Misc Electronic Questions

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Oh , so the triangles amplitude has to be higher than the sine waveform to get a HIGH ON time

So the Op amp is acting like a comparator? that saturates or clips the output
 
in the original picture you posted it WAS a comparator (you can usually tell by the pull up resistor on the output), and the one I drew might be as well. Opamps and comparators have the same schematic symbol.
 
What kind of comparator is this?

Most comparators i have seen , has one input tied to a trim pot going to a DC voltage level and the other input is the input signal waveform, it compares the input signal with the DC voltage level

What does the pull up resistor do on the output?
 
Most comparators have open collector output, so they need a pullup to produce high output.
Comparator compares two voltages. The fact that you usually need one of them set permannently and sometimes adjusted by a trimmer doesn´t mean all are connected like that. But you can think of the sine wave as the set voltage, only it is slowly changing in time and thus slowly changing the output duty cycle.
What kind of comparator is what? The one I used in that simulation? That was an tl072 opamp because they are essentially the same thing for this purpose and I couldn´t be bothered to find a proper comparator in the simulator, but opamps make very lousy comparators and that is why you should allways use the true ones in real circuits unless you know what you´re doing.
 
Why are op amps not good to use as comparators?

Most comparators have open collector output, so they need a pullup to produce high output.

I still don't understand the pull up resistor, is there was no pull up resistor what will happen?

Why do comparators are open collector output?

Comparator compares two voltages

Yes I understand that part

What I don't get is how you can put 2 different waveforms and the OP amp internally compares the amplitudes together?

The comparators i have seen is they have an external DC voltage tied to one of the input pins

This PWM op amp comparator, has 2 signals going direct to the inputs and the Op amp internally compares the amplitudes?
 
The comparator doesnt care about amplitudes at all. It only cares about the instanteneous voltage at the two inputs. It doesnt know that the two inputs are some waveforms, it just compares the input voltage all the time.

Without the pullup, the output can only sink current because it has only a transistor between ground and the output pin, so you dont get any voltage output unless you apply some voltage to the output. Read something about open collectors before you ask again.
Opamps are meant for amplification with high gain. When you let the opamp output to swing hard to either end of the possible output voltage, the transistors inside saturate and transistors take long time to get back out of saturation, so in effect the comparator made of an opamp is very slow.
 
Op Amp Compares the instanteneous voltage at the two inputs

If one input has a higher amplitude or instantenous voltage? the output will be a HIGH state?

I thought it was comparing the amplitudes of the two inputs , if one waveform is Higher amplitude than the output will be a HIGH state
 
An opamp or comparator have a very high voltage gain of maybe 1 million times. The very high gain causes them to have a phase shift at higher frequencies.
When an opamp (that has a phase shift) uses negative feedback to reduce the extremely high gain and reduce distortion then at some high frequency the phase shift causes the negative feedback to become positive feedback which causes oscillation. So opamps have a frequency compensation capacitor inside to reduce the gain at high frequencies.

Haven't you looked at the datasheet of an opamp?? The voltage gain is almost 1 million at DC and very low frequencies then drops -6dB (half) per octave. Then at a couple of MHz the gain is less than 1.

A comparator is almost NEVER used with negative feedback so it does not have a frequency compensation capacitor. Then they can switch much faster than an opamp.
 
So All TTL chips only Sink current?

CMOS chips only sink current too?

Adding a Pull up resistor makes the output source current?
 
NO: Here is a Hex inverter (7405) with o/c ouputs. https://www.electro-tech-online.com/custompdfs/2013/04/sn7405.pdf

Here is a Hex inverter (7404) with TTL (Totem-pole) outputs: https://www.electro-tech-online.com/custompdfs/2013/04/sn7404.pdf

The later can source and sink current, but asymmetrically. It can sink 16 and only source -0.4 mA. Most TTL chips are this way.

TTL is probably a bad example to use now, but you have to understand the limitations of each logic family. TTL, LSTTL, CMOS, HCT etc. Early logic families such as RTL, DTL, and ECL you just don't see anymore.

the OP said:
Adding a Pull up resistor makes the output source current?

yes, but let's change makes to lets.
 
so some open collector output only sink current not source current

Do totem pole outputs only sink current not source current

Totem pole output need a pull up resistor?
 
the OP said:
so some open collector output only sink current not source current

All open collector devices only sink current unless provided with an external pull-up.

the OP said:
Do totem pole outputs only sink current not source current?

Totem-Pole devices source and sink current.

the OP said:
Totem pole output need a pull up resistor?

No
 
ok thanks for the info

When an IC chips VCC is slower, the current is Higher? the IC chip will get HOT because the VCC is low and the current gets raised?

How can the voltage goes very low, but the current goes very HIGH

I would think the voltage would have to be High in order to have a high current

I don't get how very low voltage can have a Very high current?

How can the current be so high when the voltage is so low
 
water analogy with voltage and current is

Voltage = How high or water level

Current = The flow rate of the water

SHORTED IC chip when it gets HOT = Is you have a low water level , you can't get a HIGH flow rate , because the water level is so low
 
When the chip is broken, it acts more like a wire, not a semiconductor.

I've seen systems designed for 3000 Amps at 6 VAC and 0.1A at 100 kV DC. Most 3000 A shorts would act as a fuse for the 3000 A system.
 
When the chip is broken, it acts more like a wire, not a semiconductor.

Yes I understand that part

A short seems to "amplifies" the current, where does it get the "extra current" from when the voltage is so low?
 
A short does not amplify anything. It simply passes whatever current is available.
 
A short does not amplify anything. It simply passes whatever current is available.

But if VCC is 5 volts normal, but when the IC chip has a short the VCC is at 2 volts, how does it has Current in the Amps? not milliamps
 
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