Voltage divider question in TINA

[sunny1]

Hello. I just started using TINA for simulation and I have this Common Collector circuit. The voltage at the base of the transistor should be a simple voltage divider. (10K*12V/43K = 2.79 V). But, when I hit 'Calculate nodal voltages' in TINA, it is showing 0 V at the base of the transistor. I am stumped as to why this is. Can someone please point out what I am doing wrong?

[watchmaker]

Sunny1,

While I am not experienced enough to offer advice on you ckt (I am waiting to see if my thinking is correct), I do have a question about your TINA output.

What do the little red marks at each junction mean?  Do they mean no connection?

I have been trying out various simulators which include TINA.  I am now down to LTspice (which is really shortcut driven), pSpice for TI and Multisim (which can be found Internet Archive).

Some programs do not connect wires nicely, you may have to make the pins of the elements visible.  And not all make connections and rules checking intuitive.

[wasedadoc]

Hello. I just started using TINA for simulation and I have this Common Collector circuit. The voltage at the base of the transistor should be a simple voltage divider. (10K*12V/43K = 2.79 V). But, when I hit 'Calculate nodal voltages' in TINA, it is showing 0 V at the base of the transistor. I am stumped as to why this is. Can someone please point out what I am doing wrong?

Try removing the input waveform source and running the simulation again.

[berke]

Try removing the input waveform source and running the simulation again.

Seconded, most probably an AC source meant to be coupled with a cap.

[sunny1]

Thank You so much! I coupled the AC signal with a cap and I now see the correct voltages in the simulation. Can you please help me understand why my simulation did not work without the cap?

[Slh]

The AC voltage has a DC value of 0V. The AC value only matters for transient simulation.

[sunny1]

Thank You for your response. I am not sure that I follow your comment. I thought that the purpose of a coupling cap is to block DC. Why did I have to add coupling Cap C1 to this circuit to get the correct bias voltage at the base of the transistor in my simulation?

[wasedadoc]

An ideal source has zero resistance.

[tooki]

Sunny1,

While I am not experienced enough to offer advice on you ckt (I am waiting to see if my thinking is correct), I do have a question about your TINA output.

What do the little red marks at each junction mean?  Do they mean no connection?

I have been trying out various simulators which include TINA.  I am now down to LTspice (which is really shortcut driven), pSpice for TI and Multisim (which can be found Internet Archive).

The red marks are the connection handles for components. (The big black dots are only for junctions between traces.)

In school, we used both Tina and LTspice. (One teacher preferred one, the other teacher the other.) They both have their strengths and weaknesses. Setting up simulation parameters is much easier in TINA, and its schematic editor is less loony than LTspice’s*, but it has far fewer components to work with. TINA has an interactive digital logic simulator, which LTspice does not have at all, AFAICT.

*I find drawing a new schematic to be much faster in LTspice than TINA, but editing an existing schematic is much easier in TINA.


You managed to get pSpice working? I tried, but couldn’t make heads or tails of it…

[Solder_Junkie]

Thank You for your response. I am not sure that I follow your comment. I thought that the purpose of a coupling cap is to block DC. Why did I have to add coupling Cap C1 to this circuit to get the correct bias voltage at the base of the transistor in my simulation?

I use Tina a lot (the lowest cost student version). The Voltage generator also has a DC component, right click on it and check the properties, you will see the DC Level defaults to 0 Volts. Hence why you need a series capacitor on the base of that transistor.

SJ

[MarkT]

The point is not really that the DC offset is zero, its that the voltage source will define the voltage at that node as it has zero output impedance, so the resistors don't get a look-in and thus have no effect on anything.