Discussion in 'Homework Help' started by theertham, Apr 29, 2012.

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can any body tell me the working of transistor as an invertor circuit.figure is attached

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2. ### blah2222Well-Known Member

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Well for AC-small-signal operation (DC = 0) you can use this model of the BJT to figure out what's going to happen:

In your case, you have the input connected to the base via R9 (10k), emitter tied to ground, and R10 (10k) in parallel with the output resistance (rout) of the BJT.

You need to find Vout as a function of Vin by first finding Vbe as a function of Vin and then Vout as a function of Vbe:

$V_{be} = \frac{r_{\pi}}{r_{\pi} + R_{9}}V_{in}$

Knowing this, you can find out the amount of small-signal current running through the current source. Since the emitter is grounded, the current through this source will run through the parallel contribution of rout and R10 from ground to Vout. With this in mind, we can setup the following KCL equation:

$g_{m}V_{be} = \frac{0 - V_{out}}{r_{out}||R_{10}}$

Rearranging and subbing in for Vbe:

$\frac{V_{out}}{V_{in}} = -g_{m}(\frac{r_{\pi}}{r_{\pi} + R_{9}})(r_{out}||R_{10})V_{in}$

Where:

$g_{m} ~= 40I_{C}$

$r_{\pi} = \frac{\beta}{g_{m}}$

$r_{out} = \frac{V_{A} + V_{CE}}{I_{C}}$

The negative sign shows why this is an inverting amplifier. (Math approach)

Last edited: Apr 29, 2012
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3. ### crutschowAAC Fanatic!

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Its fairly simple.

Start out with zero volts on the input. This means the transistor is off and the collector output is at the supply voltage since there is no current through the collector resistor, R10.

Now start increasing the input voltage. That will cause base current to flow through R9 and the transistor will start to turn on with current starting to flow through R10. The causes an IR voltage drop across R10, reducing the collector voltage.

Thus as the input voltage is increasing, the output voltage is being reduced, thus the circuit acts as an inverter.

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hi
i have attached the output from transistor inverter circuit which has 10k at the collector and 1 k at the base with emitter grounded.the output shows there is a slight curve at the top of the pulse.but when i connected 10k at the base and 1k at the collector with emitter ground,the output became perfect pulse.what could be the reason?

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6. ### Jony130Senior Member

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The effect you describe is caused by the parasitic capacity.
This capacitor is connect between collector and the gnd.