Hi,

I'm currently listening to a podcast on building a computer from fundamentals and they say this:

Imagine in this visual slate that there's a wire running along the top which carries a voltage, and another wire running along the bottom which is the ground.



And this is the way most of these logic diagram schematics are drawn, is you'll have sort of a bus running across the top that has a voltage, which is just a pressure, essentially, created by a power supply. And anchored at the bottom is another wire, sort of a bus running horizontally that is the ground. You then - you interconnect things in between this positive power supply potential at the top and the ground at the bottom.



If we had two resistors - a resistor is a component with two wires coming out of each end which, as the name sounds, resists the flow of current through it. Essentially what it does is you run current through it, and it gets hot. It dissipates current in the form of heat.


So imagine in this circuit diagram that we have two resistors connected, the first one at the top, coming down to the second one, which then connects to the ground at the bottom. So that we have a circuit formed just with two resistors in series. And for the sake of simplicity we'll assume that they have the same amount of resistance. Well, this forms something called a "voltage divider" because essentially, when we make this circuit with just two resistors in a series, voltage will flow through this circuit.



And the direction of voltage flow is sort of controversial. I can't remember now, I was trying to remember which direction I learned in high school. Some people think of voltage flowing from the negative to the positive. Some people think of it from the positive to the negative. It really doesn't matter. Technically one direction is current flow, the other is the flow of the electrons, which sort of goes, being negative, goes in the other direction. So either way, all you have to have is a consistent system, since it's really sort of an arbitrary designation which way the current is flowing.
So we have this what's called a "voltage divider."


So at the very top is our power supply voltage. What happens is the resistors share this voltage drop, as it's called, between the positive power supply voltage and ground, so that the junction where they're connected in the middle will be at half of that power supply voltage because they evenly divide it.



And so that's sort of the first thing to see is you have two resistors connected together. They form what's called a voltage divider. And the voltage in the middle, or the voltage at their junction, where they're connected, is half of the total voltage.


So now we take out the bottom resistor, and we replace it with a switch, just a standard mechanical switch. It's got two wires; and, depending upon whether the switch is open or closed - "open" means they're not connected, "closed" means they are. If we close the switch, then the switch is essentially a short circuit.



So now that resistor that's still on the upper half of this little circuit, its lower lead is connected through the closed switch to ground. So its lower lead is now at zero voltage, at ground, when this switch is closed. If we open the switch, then we've disconnected the circuit, and the lower lead now has the same voltage as the power supply because there's no current flowing through this resistor. There's no voltage drop across the resistor.



So now we go to the next step, and we replace the switch with a transistor. A transistor is a three-lead device, a three-terminal electronic device. We've all heard of transistors, of course.



The way it works is it's like a - it works like an electronic switch. We put this transistor in the circuit. And so the transistor has an input on what's called the base lead of the transistor such that, when we put a positive voltage on that base lead, on the input of the transistor, the switch closes. That is, the transistor sort of works like the switch that we just took out. But it's controlled with the voltage on its base.



Actually voltage and current get complicated here, and I want to keep this sort of simple so we can stay to what's important. But the idea is that, if we put a positive voltage on the base of the transistor, that is, the input of the transistor, some current will flow through the base, which turns the transistor on.



But remember that when the transistor is on, it pulls the lower end of that resistor that's coming down from the supply voltage, it pulls it down to ground, that is, down to zero. So what we have is an inverter because, when we put a positive voltage on the input of the transistor, it turns on, which pulls that junction between the resistor and the transistor down to zero. So a one goes in, and a zero comes out.


And if we move the voltage on the base of this transistor, the input of the transistor down to ground, then the transistor turns off. And with the transistor off, then that junction between the resistor and the transistor goes up to the power supply voltage. In other words, a one.

Now then I think I understand the first two examples and I made some circuit diagrams:

and then with the resistor replaced with a switch

But I confused about the transistor example. I tried to make a diagram of it but I'm not sure about the connections of the legs:

Could anyone explain how replacing the switch with a transistor works and maybe correct my diagram ?

Thanks

 

Connect base trough 10K resistor to 9V power supply.

 

Wouldn't this eventually settle so that the voltage at the collector is at some point between 0V and 9V? or would the diode junction in the transister be enough to stop this small current from flowing?

 

Hi,

Thanks for the reply I have connected the transistor as you suggested

Is this correct ? Is there a reason it's 629mV and not 0v ?

 

There will be some small leakage current through the transistor when there is no current flow in the base, but it's a very small amount of current.

In the podcast, the narrator talks about "voltage flow". This is not correct. Current flows; voltage is pressure. Current flow through resistance causes a voltage "drop" across the resistance.

He mentions conventional flow vs electron flow. Conventional flow was thought up by Benjamin Franklin many years ago. In conventional flow, current goes from more positive to more negative. In electron flow, the electrons go from more negative to more positive.

When you are using a transistor as a switch, the generally accepted standard for transistor saturation is that the base current flow should be 1/10 the desired collector current.

So, Rb =(V-Vbe) / (Ic/10)
where:
Rb = Base current limiting resistor value in Ohms
V = the voltage supply for the base current
Vbe = the voltage at the base measured in respect to the emitter when the desired base current is flowing. You can usually start with 0.7v for this value; it will change if base current is very low or very high.
Ic = collector current in Amperes

So, your collector resistor is 10k Ohms, and Vcc=9v. Since I=E/R, 9v/10k Ohms = 0.9mA.
Back to the formula:
Rb =(V-Vbe) / (Ic/10)
Let's plug in Ic, V and Vbe:
Rb = (9V-0.7V) / (0.9mA/10)
And work the problem:
Rb = 8.3V / 0.000,09 A
Rb = 92.222k Ohms.
That is not a standard value of resistance. A standard chart is here:
http://www.logwell.com/tech/components/resistor_values.html
Bookmark that page.

Looking at the E24 values (green columns) we can see that you could use 91K Ohms, as that is a standard value. In this case, you could also likely use 100k Ohms. That is a little less base current than the standard rule, but it would likely be close enough to work.

[eta]
Here is the circuit with R2 added as the base resistor. You can see that the Vce (voltage from collector to emitter) is very low now.

 

Hi,

Thanks for the reply I have connected the transistor as you suggested

That's not exactly what he was thinking of, but I understand why you thought you could connect it that way.

Is this correct?

No.

Is there a reason it's 629mV and not 0v ?

Yes.
As the current from your 10k resistor turns on the transistor, the collector sinks current. The Vce settles at the Vbe for the given current flow; in this case, 0.619v. Note that the simulation won't be completely accurate; what you get with actual components may vary. The BE junction acts like a diode that is forward biased.

 

Hi,

Thanks for all your replies and help

So is this called a called and inverter as you get ~ 0v out from a positive voltage in ?


Thanks

 

Hi,

So I continued listening now I understand that and he moved on to talk about using two transistors in parallel to create a NOR gate

So that's an inverter. It doesn't, I mean, it's certainly useful by itself. But we can do something, make one additional change to it to begin to create some logic gates.

And that is, we take another transistor and hook it to the same place. That is, we put these two transistors in parallel with each other.

Another transistor hooked to the same place so that either of them are able to be turned on and pull this output down to ground, that is, hook the bottom of the resistor down to ground.

So now look what we have. If we turn either transistor on by putting a one, binary one into either of the inputs, then that transistor will turn on and pull the output down to ground. And they can both be turned on.

We get the same result. So what we have is a, in logical terms, is called a NOR gate. Which NOR stands for "not or." So if either input is a one, the output is a zero. So we have the beginning of logic.

Is this diagram a correct representation of this ?

Thanks

 

Look at example diagrams