In the amplification circuit, it is known how to find the parameters of the power supply voltage, wh

This requires consideration of the connection method of the transistor.

In general, for commonly used common emitter amplifier circuits, we design them to ensure maximum output dynamics. Therefore, the voltage of the C pole can be determined to be about 1 2 VCC, which ensures that the output is dynamic to the maximum. (Let the power supply voltage be VCC, and the co-generation circuit In the case of the need to control the amplification, the transmitter may be equipped with an AC feedback resistor, or a stable feedback resistance at the DC working point, or both, then the need to determine the C pole voltage is slightly higher than 1 2VCC.)

The theoretical design can be ignored for easy calculation).

The DC point operating voltage of the VC is determined, and the DC load resistance RC of the collector is determined according to the input resistance required (or the load resistance required) of the subsequent stage. This allows us to calculate the DC operating current IC (IC VCC 2RC) at the C pole of the transistor.

According to the value (or estimate) of the triode, you can calculate the base current of the triode ib( ic Then you know the required bias current, and the bias resistance can be easily calculated according to your bias connection.

Can it solve your problem?

This one is complicated. Amplification circuits include voltage amplification circuits and current amplification circuits. There are three types of signal sources: co-emission, co-base, and co-collection. The calculations are all different.

1.First of all, in order to ensure that the conduction condition of the PN junction of the transistor is not in the cut-off state, the potential difference of the PN junction of the silicon material tube must be greater than the volt, and the potential difference of the PN junction of the germanium material tube must be greater than the volt.

2.current direction; If the current is to flow from the emitter to the collector, the collector has a higher potential than the emitter.

3. It is also necessary to consider that the transistor is not in a state of current saturation.

4.overloading of transistor power, etc.

The first floor says everything.

To ensure that the PN junction conduction condition of the transistor is not in the cut-off state, the silicon material tube.

The calculation formula of each amplifier is different, and the basic co-injection amplifier is used as an example, which is briefly discussed.

In the first step, the liquid vertical alarm determines the amper-volt converter RC according to the load resistance RL according to the principle of maximum efficiency, and the specific formula RC=RL;

In the second step, calculate the base bias resistance rb according to rc, rl and rb, the specific formula rb (critical) = rc + rc rl), where rc rl = r'l, which indicates the parallel resistance of RC and RL, and critical indicates the critical.

For example, knowing =100, RC=RL=1K, you can calculate the base bias resistance rbrb(critical)=100 (1k +1k 1k)=150k.

If it is known or measured, no adjustment is required.

I don't know if it meets the taste, if you have different opinions, please correct.

Current parallel negative feedback circuit.

Negative current feedback increases the output resistance and stabilizes the output current; Negative voltage feedback reduces the output resistance and stabilizes the output voltage; Negative feedback in series increases the input resistance, and negative feedback in parallel decreases the input resistance.

Therefore, a negative feedback circuit should be selected for current paralleling.

It should be a negative feedback in parallel with the current. Look at the picture on the PPT.

Uh, the picture can't be sent.

The characteristics of current parallel negative feedback are: small input resistance, large output resistance, and stable output current.

The calculations are also inaccurate because the magnification is uncertain. It is usually a method of adjusting the quiescent operating current with an adjustable resistor. Connect a potentiometer first, from large to small, to the specified quiescent current.

The design and calculation formula of each amplifier is different, taking the basic co-injection amplifier as an example, let's talk about it briefly.

In the first step, the amperage converter RC is determined according to the load resistance RL according to the principle of maximum efficiency, and the specific formula RC=RL;

In the second step, calculate the base according to RC, RL, and .

Bias resistance rb, the specific formula rb(critical) = (rc+rc rl), where rc rl=r'L, which indicates the parallel resistance of RC and RL.

critical means critical.

For example, knowing =100, RC=RL=1K, you can calculate the base bias resistance rbrb(critical)=100 (1k +1k 1k)=150k.

If it is known or measured, no adjustment is required.

I don't know if it meets the taste, if you have different opinions, please correct.

1) Your analysis [indicating that the voltage provided by VCC to the base through the bias resistor RB is sufficient for the transistor to turn on] is correct;

2) And that iceo, which belongs to the interference information (which is useless), depends on whether you will be disturbed;

3) Since the transistor is turned on, it may be in the amplification region or saturation region;

And the condition: VCC>>VBE, just to show that the base current generated will be very large, so that the transistor will inevitably enter the saturation zone;

However, it is obvious that the questioner here is irresponsible, and the condition of only giving one VCC >>VBE is okay;

Usually a specific value of the vcc is given, and the students are asked to calculate and judge;

However, it is necessary to give an ICEO, which will be reminiscent of other parameters of the transistor, such as the VCEO withstand voltage parameter;

What is the value when it is said that VCC is much greater than VBE? 100v? Then the transistor will be burned, and it will be a fart;

<> at this time, the transmitting junction is positively biased, the collector junction is reversed, and the standard amplification state is in a standard state, but the static working point of this circuit is very high, and it is very easy to saturate.

VCC>>VBE--- can ignore VBEICEO when calculating IB0--- IC can ignore IC, so IB* =VCC*80 100=>VCC RC=VCC transistor works in the saturation region.

1. Remove the capacitor, and what remains is the direct flow path.

2. Static value;

ib≈12/300k=

ic=βib=40*

uce=12-rc*ic=12-3*

3. If you want UCE=6V, then IC=(UCC-UCE) RC=6 3=2MA, corresponding to IB=IC=2 80=

ib≈ucc/rb rb≈12/

4. The figure shows saturation distortion, and VCE should be increased in order to eliminate saturation distortion, so rb should be increased.