National instruments oscilloscope

[Bassalisk]

Hello, I am trying to understand input AC source signal amplification. But since my image is so blurry in my textbook and I cannot find it anywhere on the internet, I am trying to simulate it in National instruments.

How do you make oscilloscope measure current, instead of voltage? Sinusoid of currents, not voltages.

Thanks

 

[Studiot]

Hello, I am trying to understand input AC source signal amplification.

Strange that we have has another question about this today (although the amplification was DC).

https://www.physicsforums.com/showthread.php?t=499463

How do you make oscilloscope measure current, instead of voltage?

You don't. You insert a low value non inductive resistor in series with the current and connect you scope across that. Remember that the input outer of the scope may be grounded so you would normally connect the resistor into the low side of the circuit.

go well

 

[Bassalisk]

Thanks. It works but how did you come up with that? :D Is it because I basically measure voltage drop across the resistor, but since the resistance is low, I am measuring current?

 

[Studiot]

No you are still measuring voltage, but remember Ohm's law?

If you need more help with you load line post again

 

[berkeman]

Hello, I am trying to understand input AC source signal amplification. But since my image is so blurry in my textbook and I cannot find it anywhere on the internet, I am trying to simulate it in National instruments.

How do you make oscilloscope measure current, instead of voltage? Sinusoid of currents, not voltages.

Thanks

Another way to measure current with an oscilloscope is to use a cable clamp current probe.

If the signal is AC, it can just be a current transformer type of probe.

If the signal includes DC, then it needs to be a Hall effect type of probe.

http://www.google.com/search?tbm=is...=current+probe&gbv=2&aq=f&aqi=g1g-m1&aql=&oq=

.

 

[Bassalisk]

Thanks all. Really appreciate it. Will post again if I encounter any more problems.

 

[Bassalisk]

Hello Studiot,

In process of studying these amplifiers one thing confuses me. That is :removing DC and AC components. What are they? Where do they come from?

I attached the image with my circuit.

Can you just give me few pushes about what does this DC/AC components do? And why would you have it when you have a AC source on base. Yes, you have Vcc but, that, in my opinion, doesn't matter because transistor is "oscillating".

Thank you.

 

[Studiot]

When you consider any circuit containing components other than simple resistors you have to perform two analyses. A DC analysis and an AC (signal) analysis. These analyses are made for different purposes and are normally performed separately.

Most circuits are built to process a signal of some sort.

Active components such as transistors require a source of power for them to function at all. they do not draw power from the signal. Instead they draw power from a power source such as a battery or DC supply. Transistors, in fact control the flow of power from the supply to the load. In order to correctly peform their function transistors need to be set up properly. This is called biasing. Some circuit elements are only present for this biasing, some elements perform a dual role in the DC and AC functions of the circuit and some are present in order to separate these functions.

 

[Bassalisk]

So basically, that DC component comes from that Vcc? And because it branches to few wires, you have to filter that out somehow?

You do that by applying a capacitor, capacitor let's AC out, but holds DC in, right?

Does that C1 have 2 purposes? First is to make the amplitude smaller and the second is to block DC signal?

 

[Bassalisk]

ooooh you do the analysis, and then you figure out what is DC and what AC? I get it now. I know how to do that. Just the point of that was vague to me. Thanks for your help! <3

 

[Studiot]

Here is a 'walking through' analysis of your circuit.

First the DC analysis

We ignore the capacitors, they play no part in the DC analysis.

The quiescent voltage at the base of transistor, Q, is set at 3 volts by the voltage divider R2 & R1.

The emitter is 0.7 volts less than this ie 2.3 volts. This is therefore the voltage across the emitter resistor, R_E. So the emitter current is 2.3/1.3 milliamps = 1.77

This is also sensibly the collector current. So the voltage across the collector resistor R_c is 4.3 times 1.77 volts = 7.6 volts.

Thus the collector resides at 12 - 7.6 = 4.4 volts.

Since the base (input) resides at +3 volts and the collector (output) at +4.4 volts we do not want to have these values imposed on any external circuit we connect so we isolate these using 'blocking' capacitors C1 and C3, which can pass the AC signal, but not the DC.

I will come to C3 in part 2 the AC analysis.

Are you with me so far?

 

[Bassalisk]

Here is a 'walking through' analysis of your circuit.

First the DC analysis

We ignore the capacitors, they play no part in the DC analysis.

The quiescent voltage at the base of transistor, Q, is set at 3 volts by the voltage divider R2 & R1.

The emitter is 0.7 volts less than this ie 2.3 volts. This is therefore the voltage across the emitter resistor, R_E. So the emitter current is 2.3/1.3 milliamps = 1.77

This is also sensibly the collector current. So the voltage across the collector resistor R_c is 4.3 times 1.77 volts = 7.6 volts.

Thus the collector resides at 12 - 7.6 = 4.4 volts.

Since the base (input) resides at +3 volts and the collector (output) at +4.4 volts we do not want to have these values imposed on any external circuit we connect so we isolate these using 'blocking' capacitors C1 and C3, which can pass the AC signal, but not the DC.

I will come to C3 in part 2 the AC analysis.

Are you with me so far?

Just this 3V on the base are not in place. How do you get that? Thevenin theorem?

 

[Studiot]

R2 and R1 form a potential divider across the 12 volt supply, V_cc.

Voltage across R2 = R2/(R2+R1) * V_cc = 9

Voltage across R1 = R1/(R2+R1) * V_cc = 3

That is the purpose of this potential divider.

Incidentally do you know what the cc stands for in V_cc?

 

[Bassalisk]

R2 and R1 form a potential divider across the 12 volt supply, V_cc.

Voltage across R2 = R2/(R2+R1) * V_cc = 9

Voltage across R1 = R1/(R2+R1) * V_cc = 3

That is the purpose of this potential divider.

Incidentally do you know what the cc stands for in V_cc?

Potential related to collector? Voltage controlled collector?

 

[Studiot]

That doesn't tell me if you understood where the 3 volts came from.

The cc is short for 'common collector'. The reason I asked is because your circuit actually shows a common emitter configuration.

The 'common' arises because the transistor is actually a three terminal device. Since we require four terminals 2 for the input and 2 for the output it follows that one terminal has to do double duty and appear in both the input and output circuits. This is known as the common terminal since it is common to both input and output.

In your circuit the emitter is the common terminal. The base only appears in the input circuit and the collector only appears in the output.

 

[Bassalisk]

Ok, I understood the DC part fully now. You can move on.

 

[Studiot]

An AC signal is more complicated than a DC one because we also have to consider frequency.

The purpose of C2 is to short the emitter to ground at signal frequencies. We say that it bypasses R_E and is an emitter bypass capacitor.
As such C2 has a value that makes its impedance small (say <1/10) R_E at the lowest signal frequency of interest.

Employing this capacitor allows the signal voltage (collector voltage) to swing down to Earth (zero) volts and up by the same amount about its quiescent value. Without the capacitor it could only swing down to the top of R_E.
This maximises the available gain.

An analysis of the AC operation require the introduction of a model of the transistor, but I think you have some reading to do before you get to that.

Edit: On your simulator, try applying a signal with and without C2.

 

[Bassalisk]

Thank you very much for your help Studiot. I don't know how can I ever repay you.

 

[Studiot]

See my edit about your simulator.

As for repayment, just become good at peaceful electronics.

go well

 

[Studiot]

I've learned something today.

It does not stand for common collector. It is the power source for the collector in a BJT amplifier. It will be positive for an NPN transistor.

Vc is the voltage ON the collector and Vcc is the voltage supply to this pin, usually on the other side of the load resistor.

Similarly, Vdd is the power supply for the drain on a FET. This will be positive for an N channel FET.

There is actually a Wikipedia article about it here:
http://en.wikipedia.org/wiki/Vcc

Thank you Vk6kro.