In first image there has to be dc path to ground at input yeah?

"in" is from function generator

In 2nd image, from the 4th to last line: from "then tune an AM"...Please can someone explain from that point
Don't understand

Thanks



// WBahn : "As this is homework, show your attempt"

 

What is the question? What's the problem?

 

What is the question? What's the problem?

lol..bedtime, no?

Thanks

In the first image : should there be a resistor from "in" to ground? or can connect function generator directly

2nd image : don't understand from 4th to last line to the last line. Please explain

 

In first image there has to be dc path to ground at input yeah?

There is through the function generator.

from the 4th to last line: from "then tune an AM"...Please can someone explain from that point

The circuit will operate as a simple AM transmitter that can be detected by an AM radio tuned to 1MHz (or whatever the Vin frequency is).

 

There is through the function generator.

oh really? cause in previous chapter, somewhere it says always provide dc path to ground

The circuit will operate as a simple AM transmitter that can be detected by an AM radio tuned to 1MHz (or whatever the Vin frequency is).

so from that simple circuit, can add mic signal as the v mod then will hear from the AM radio??
for real???
that's what a transmitter is for, no?

Thanks

 

1. the current from the generator must be returned to the generator on the black generator lead. This lead is commonly grounded at the generator. If the black lead is not connected to a ground potential in test circuit, an extra current can appear. Not only will this distort the signal, but it can burn things up. There are several methods the cancel this effect.

2. It means you can use a AM receiver to listen to this transmitter circuit. You can tune or adjust the tuning dial to find it.

We are not suppose to directly answer questions on this particular forum.

 

oh really? cause in previous chapter, somewhere it says always provide dc path to ground

Really.
Assuming, of course, that the generator ground is connected to the circuit ground.
If not, than the generator wouldn't work as a generator for the circuit.

 

Really.
Assuming, of course, that the generator ground is connected to the circuit ground.
If not, than the generator wouldn't work as a generator for the circuit.

why you not sleeping? what you doing ?...no wife?

btw, bs transmitter circuit...
my AM radio, no "quiet place"...the frequency not in use, always "shhhh" sound

Anything to do? what if to connect electret mic as the Vmod (and play sound from phone to the mic)...will hear from radio?

edit :Trying shortly

 

In first image there has to be dc path to ground at input yeah?

"in" is from function generator

For most simulators, every node needs to have a DC path to ground otherwise there is a good change that the simulator will not be able to get an initial operating point solution to converge.

Consider two capacitors in series across a battery. In theory, there could be ANY voltage at the junction of the two capacitors -- imagine that the junction point was previously connected to another battery that was then removed. The simulator has no way of coming up with a single value for the voltage on that node, so it can't even start. But if you put even a very large value resistor across one of the capacitors, now the simulator can get the initial operating point to converge to a single value.

It doesn't matter how that DC path to ground comes about, it just has to be there. If the output of your function generator is at a known output voltage at the start of the simulation, then the voltage on that node is known to the simulator just as if it had been a battery connected there.

In 2nd image, from the 4th to last line: from "then tune an AM"...Please can someone explain from that point
Don't understand

As the description says, the circuit is a simple AM transmitter. The carrier frequency is applied to the v_in node and the signal is applied to the v_mod node. The circuit amplitude modulates the carrier with the signal. The radio, when tuned to the same frequency as the carrier, will demodulate the combined signal and recover a replica of the original modulating signal.

It probably says this above, but unless you have a signal going into v_mod when you try to listen to it on the AM radio you will here little to nothing.

 

why you not sleeping? what you doing ?...no wife?

btw, bs transmitter circuit...
my AM radio, no "quiet place"...the frequency not in use, always "shhhh" sound

Anything to do? what if to connect electret mic as the Vmod (and play sound from phone to the mic)...will hear from radio?

edit :Trying shortly

The "shhhh" sound is background static. A "quiet place" is simply a place where there is no station can be heard. Your signal is going to be very weak, so even a weak signal from an actual broadcast station is likely to overwhelm it.

 

Consider two capacitors in series across a battery. In theory, there could be ANY voltage at the junction of the two capacitors --

Hmm..Seems like theory predicts the voltage 100% unless not understand this statement..

The "shhhh" sound is background static

How can remove the background static...In book, author says the signal about 200mV (guess need increase)..so can use mic then bjt amplifier then output as Vmod?..Anyways, FET is the topic, will learn RF

Just to test though : how to remove background static?

 

why you not sleeping? what you doing ?...no wife?

It was 9:44pm at my house when I wrote that.
I usually don't go to bed until after midnight.
And what's having a wife got to do with it?

 

It was 9:44pm at my house when I wrote that.
I usually don't go to bed until after midnight.
And what's having a wife got to do with it?

....

 

Hi Zeeus,
Mod: Please desist from making personal comments, stay on the technical Topic.
E

 

Hmm..Seems like theory predicts the voltage 100% unless not understand this statement..

Huh????

Put two 1 uF capacitors in series across a 12 V battery. Are you saying that "theory" predicts that the voltage across each capacitor is 12 V?

 

Huh????

Put two 1 uF capacitors in series across a 12 V battery. Are you saying that "theory" predicts that the voltage across each capacitor is 12 V?

was thinking about AC, middle voltage peak as 6V

Thanks

the midpoint between cap has to be 0v ? if not 0 then dc current is flowing through cap

 

hi Z,
Theoretically the junction/ midpoint would be Vsupply/2.
In a practical application is possible that the leakage current through the two capacitors will not be the same value.

What would you consider as a possible solution if that was the case.?

E

 

the midpoint between cap has to be 0v ? if not 0 then dc current is flowing through cap

Yes, current flows.
If you connect two equal sized caps in series to 12Vdc, then current will flow through both caps until there is equal voltage across both caps, or 6V at the midpoint.

 

hi Z,
Theoretically the junction/ midpoint would be Vsupply/2.
In a practical application is possible that the leakage current through the two capacitors will not be the same value.

What would you consider as a possible solution if that was the case.?

E

Add high equal value resistors in parallel with each? add leakage cap?
dunno

 

Yes, current flows.
If you connect two equal sized caps in series to 12Vdc, then current will flow through both caps until there is equal voltage across both caps, or 6V at the midpoint.

I thought so to but checked it out on spice and there was 0v...only for AC then peak is 6v