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Electric circuit problem

electronicsLearner77

Jul 2, 2015
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I am trying to solve the below circuit,
upload_2022-7-4_23-30-16.png

It was created by me, may be it is wrong. These are the equations i have written
if the input is V1 is 5V
Q1 is ON
Ib1 = (5 - 0.7)/1k = 4.3mA
5 - 5Ic1-Vce1 = 0 -> eq1
5 - 5Ic2 -Vce2=0 -> eq2
5 - 5Ic1 - Ib1 -Vbe=0 ->eq3
can i assume Q2 ON in the above case?
What voltage shall i consider for Vce1 and Vce2?
When V1 is 0V
Q1 is OFF
Ic1 = Ib1 = (5 - 0.7)/6 = 0.716mA
5 - 5Ic2-Vce2=0 -> eq4.
Again what value shall I consider for Vce2?
Please help.
 

Alec_t

Jul 7, 2015
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If Q1 is on then Q2 is off.
You are using LTspice's default transistors, so you can assume Vce =0 when a transistor is on. What does LTS show it as?
Eq1,2, 3 and 4 are missing the factor 1000 for resistance.
Eq3 looks wrong.
 

Alec_t

Jul 7, 2015
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I green before R4. I don't think i am getting correctly.
We don't know where nodes 3 and 6 are. I suggest you get into the habit of applying labels to nodes you are interested in, so that it is easier for us to discuss the circuit operation.
I ran a sim of your circuit and it shows Vce=20mV when a transitor is on.
 

Harald Kapp

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Nov 17, 2011
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I suggest you get into the habit of applying labels to nodes
+1. Use the F4 key to do this in LTSPICE.
You are using LTspice's default transistors,
Can be done, bur is imho a bad option. Better select a common transistor from the library, like 2N2222 or BCX70, depending on your preferences.

Also, while technically not wrong, your placement of the power supply V2 is unfortunate and makes reading the schematic a bit difficult. I recommend you also get into the habit of orienting "+" on top and "-" on the bottom. And do not avoid unnecccessary wire crossings like for e.g. the GND connection of V2.
Here's how I recommend to draw the circuit:
upload_2022-7-5_8-29-42.png
Not a big issue here, but as soon as circuits get bigger, it can become an issue.
To reduce the number of wires crossing your schematic, you may even resort to using net names to create a connection like so:
upload_2022-7-5_8-31-5.png
Here both nets named Vcc are connected. although no visual wire is shown on the schematic. But do this only for global signals like power supplies. Otherwise you wil have great difficulty recognizing a circuit with only net names but no visible connections.

What you can expect:
The voltage at the collector of Q1 is inverse to the pulse voltage V1.
The voltage at the collector of Q2 is inverse to the voltage ath the collector of Q1 and therefore is in phase with the pulse voltage V1.
 

electronicsLearner77

Jul 2, 2015
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Thank you very much for suggestions, i updated the schematic accordingly
upload_2022-7-15_22-16-12.png
The calculations also able to understand except one doubt
When V1 = 5V
Q1 is ON, Q2 is OFF
VA = 0v -> voltage at node A
VB = 5V -> Voltage at node B
When V1 = 0V
Q1 is OFF, Q2 is ON The circuit becomes
upload_2022-7-15_22-15-27.png
The voltage at node A
I = 4.3/6K = 0.716mA
VA = 5 - 0.716*5 = 1.41V
VB= 5V.
It matches with simulation result. Only one question is in the last scenario how to confirm if Q2 is ON, since i am assuming right now it is ON?
 

Nanren888

Nov 8, 2015
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how to confirm if Q2 is ON
Not sure what you mean.
It has 0.7 mA of base current. So it is driven at least. For any real transisitor you can look up characteristics to check the effect of this with a 5k load, from a load line on the characteristic curves.
.
Or <Proof by spice> That LTSpice says Vce is 0 means it is on.
 

Alec_t

Jul 7, 2015
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... or if you run a sim and left-click the transistor's emitter or collector you can plot the current into that node.
 

roughshawd

Jul 13, 2020
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This circuit looks like an amp stack.
Problem with that is that no matter how much power you put in, it only travels down the POLR. You have to use the power you make, or it causes everything else to heat up.
 
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