Dear members,

http://www.pi4hmd.nl/menu/varicon/v4150/v43/artikel07.htm
Schematicc 1:
Can somebody explain me how these current source is working? Why is the ground symbol connected to the collector?
The zener sets always 5.7V on the base.
The emitter resistor controls the Ic current, in other words, it stays always the same.
Is that right?
Why is that circuit grounded on the collector?

 

Dear members,

http://www.pi4hmd.nl/menu/varicon/v4150/v43/artikel07.htm
Schematicc 1:
Can somebody explain me how these current source is working? Why is the ground symbol connected to the collector?
The zener sets always 5.7V on the base.
The emitter resistor controls the Ic current, in other words, it stays always the same.
Is that right?
Can i redraw the schematic from the link to this one? Is that the same?
http://www.uploadarchief.net/files/download/cursour1.jpg
ps: sorry i forgot to connect the collector resistor to the Vcc.

It did take a while to figure this one out! It looked like the polarity of the transistor was wrong. However, if you look at the current source, they've grounded the positive end...so it will work.

Positive ground circuits are used in a lot of situations where it might be awkward to interface it to an outside source or signal. For example in telephone office equipment, everything has a positive ground. To interface telephone equipment to negative ground circuits can be really tricky. (I know, I've had to do it numerous times!

Anyway..the circuit functions identically, regardless of the ground point.

eric

Hope this helps.

 

Dear members,

http://www.pi4hmd.nl/menu/varicon/v4150/v43/artikel07.htm
Schematicc 1:
Can somebody explain me how these current source is working? Why is the ground symbol connected to the collector?
The zener sets always 5.7V on the base.
The emitter resistor controls the Ic current, in other words, it stays always the same.
Is that right?

I believe your question is concerning the NPN which has its collector referenced to ground. You will note that the 50V power supply has its positive terminal referenced to ground as well. This preserves the requirement that the emitter in an NPN be more negative than the potential at its collector.

hgmjr

 

It did take a while to figure this one out! It looked like the polarity of the transistor was wrong. However, if you look at the current source, they've grounded the positive end...so it will work.

Positive ground circuits are used in a lot of situations where it might be awkward to interface it to an outside source or signal. For example in telephone office equipment, everything has a positive ground. To interface telephone equipment to negative ground circuits can be really tricky. (I know, I've had to do it numerous times!

Anyway..the circuit functions identically, regardless of the ground point.

eric

Hope this helps.

But is that not a short then? The plus is fysically connected with the ground.
Why they don't use my second circuit then? What is then the real difference between the two circuits.

 

Can somebody help me with this. It's very important that ik know exactly what happened in the circuit.
Why is the "ground" connected with the puls of the voltage supply, is that not a short?

 

Can somebody help me with this. It's very important that ik know exactly what happened in the circuit.
Why is the "ground" connected with the puls of the voltage supply, is that not a short?

In your redrawn sketch there is no way to determine which is the emitter and which is the collector. This makes it impossible to comment on the validity of the circuit.

hgmjr

 

Sorry, i mean in the link that i gave in the first post. the colector is grounded, i don't understand that.

 

In the case of the circuit in question, imagine if you will that the ground symbol is moved to the negative side of the battery instead of the positive side. That would not change the functioning of the circuit. It would change the values of voltages measured but the circuit would operate the same. Would the circuit be more digestible to you then?

hgmjr

 

Well yes i think so. the purpose of that ground signal is to make the voltages negative. How can i do that in practice?

one more question, when i take Rc = 500ohm, is it normal that the transistor is not working well anymore .

 

Well yes i think so. the purpose of that ground signal is to make the voltages negative. How can i do that in practice?

one more question, when i take Rc = 500ohm, is it normal that the transistor is not working well anymore .

Since the collector current is 0.1 Amps then V=ir results in 50 volts. That means that the transistor is in saturation. That would prevent it from working.

hgmjr

 

Yes i agree with you, but normally when he is un saturation, you would measure a few volts, because on the graph (Ic versus Vce), it is close to the zero point.
I build up the same circuit, only with other values. Vcc = 12V
Zener= 5.6V, Rb= 670 ohm
Re = 150 ohm, Rc= 500
when i measure with these values, i measure for Vc 8V and for Ve more or less the same, is that not strange?
Normally there have to flow the Ic(sat), but measured vaue is also less. Do you know why?

 

A value of 8 volts for the voltage drop across Rc does sound a bit low. I would expect the conditions you have created using the new values would still result in a saturated state for the transistor.

hgmjr

 

8.8V to be precise. I will build up the circuit again.
Thank you for your help

 

On reflection, the value of 8.8V or so is probably not too far off. Remember that there will be close to a couple of volts dropped across the 150 Ohm emitter resistor.

hgmjr

 

I think that normal effect, I tried it now with 1000 ohm, and measure te current through Rc, but this is also less 8.4mA, thus that means that the voltage drop is also less.

 

Keep in mind that the transistor is in saturation. That means that neglecting the 200 millivolts or so dropped across the transistor's emitter-collector means that the current is 12V/1150 Ohms. That comes to around 10 milliamps.

hgmjr

 

You do realize that once a transistor enters saturation it no longer acts like a constant current source, right?

hgmjr

 

Well yes that's right.

I have checked that, and the currect was decreased!