What is the forward bias voltage of the transistor?

Temperature t rises -> rises -> IC= IB, so IC rises -> IE rises ->VE rises ->UBE Drop - >IB Drop - > IE Drop - > VE Fall, the fall of VE cancels out the previous rise of VE, so it can stabilize the static operating point.

So why does a decline in UBE lead to a decline in IB?

At first, I thought about it for a long time, but later I found out that it was related to the characteristics of the transistor;

As you can see from the graph below, the UBE decreases, so that the IB also decreases.

At first, I understood that I wanted to list the relationship between IB and , but I found that this did not correspond to the actual results at all.

Because RB2 is the lower bias resistor, VBQ is the voltage on RB2, and the voltage on RB2 is the voltage on RB2 after RB1 and RB2 are connected in series, so VBQ = VG*RB2 RB1+RB2, and VG*RB1 RB1+RB2 is the voltage on RB1.

In your derivation "ube =vcc-(ib*rb+(ib+ ib)re)", it is assumed that is constant. However, the temperature of the transistor usually increases when it changes, so the above equation does not negate the previous temperature stabilization process.

When analyzing a problem, we should pay attention to which is the cause and which is the effect. Cause and effect cannot be reversed.

This is the negative feedback effect of RE: when the temperature is IC IE, the voltage drop across RE increases, resulting in IB IC, which stabilizes the operating point.

First, set the base voltage to ground constantly.

The IB component of the transmit junction current is positively correlated with the voltage UBE at both ends of the transmit junction.

IE increases, VE rises, and UBE is compressed, making IB decrease and IC decrease.

This question needs to be tangled, and the book only asks to remember that except for the UBE falling, all other quantities are rising...

A certain voltage is added to the base and emitter of the triode, and if the collector and the emitter are on, then this base voltage (mainly polarity) is called positive bias voltage, and this state is called triode positive bias, or base positive bias.

If the collector and emitter are cut off (not on) after a certain voltage is added to the base and emitter, then this base voltage is called a reverse bias voltage, and this state is called triode reverse bias, or base reverse bias.

Forward bias and reverse bias are for PN junctions. In the general amplification circuit, the emission junction between E-B must be positively biased to ensure that most of the carriers can leak and diffuse smoothly, and the collector junction between C-E must be reversed to ensure that the transmission junction spreads to the majority of carriers in the base area.

Most of them can move rapidly to the collector in a drifting way to form a collector current, and only a very small part recombines with a few carriers in the base region to form a small base current. In this way, the transistor has an amplification effect.

Summary. Hello has found out for you that you are not referring to the base bias voltage. The base bias voltage is an important guarantee to control the base current of the triode.

Because the base current of the triode cannot be exactly the same in the manufacture of the transistor, it can only be achieved by adjusting the bias voltage. The base current is an important parameter to ensure the amplification characteristics of the transistor, and the only way to make the transistor sufficient in the triangle of the working characteristics is to adjust the bias voltage. So why not talk about bias current?

Because the amplitude which was not expressed by voltage was large, the amplitude of the current was magnitude. So it is expressed in terms of voltage.

What does the bias of a triode mean?

Hello has found out for you that you are not referring to the base bias voltage. The base bias voltage is an important guarantee to control the base current of the triode. Because the base current of the triode cannot be exactly the same in the manufacture of the transistor, it can only be achieved by adjusting the bias voltage.

The base current is an important parameter to ensure the amplification characteristics of the transistor, and the only way to make the transistor sufficient in the triangle of the working characteristics is to adjust the bias voltage. So why not talk about bias current? Because the amplitude which was not expressed by voltage was large, the amplitude of the current was magnitude.

So it is expressed in terms of voltage.

What does bias mean?

Bias refers to a point in the overall loop and measures its voltage relative to a reference point (1 N of the overall loop voltage).

1.Transistor bias circuit Fixed bias circuit.

As shown in the above figure, the fixed bias circuit in the common circuit of the transistor: the function of RB is to control the base circuit of the transistor, IB is called bias current, and RB is called bias current resistance or bias resistance. By changing the value of RB, you can change the size of IB.

In the diagram, the RB is fixed, which is called the fixed bias resistor.

This circuit is simple and uses few components, but due to the poor thermal stability of the transistor, even though the bias resistance RB is fixed, when the temperature rises, the ICIO of the transistor increases dramatically, so that the IE also increases, and the transistor operating point changes. Therefore, it is a voltage negative feedback bias circuit in the common circuit of the transistor: the base bias resistance of the transistor is connected to the collector.

The circuit does not seem to differ much in form from a fixed bias circuit, but it is precisely this that automatically compensates for the drift of the operating point. As can be seen from the figure, when the temperature increases, the IC increases, so that the UCE decreases, and through RB, the IB will inevitably decrease, and the decrease of the IB decreases the IC, thus stabilizing the IC and ensuring that the UCE is basically unchanged.

The process, called the negative feedback process, the circuit is the voltage-dividing current negative feedback bias circuit in the common circuit of the triode: the circuit is connected to the resistor Re and the base return through the emitter loop to fix the base potential of the resistor R1, R2 to stabilize the working point, which is called the voltage-dividing current negative feedback bias circuit. The following is an analysis of the workpoint stabilization process.

As the temperature increases, the IC increases as the ICO increases. IE has also increased. In this case, the voltage drop UE=ie*re across the emitter resistor Re also increases.

Since the base potential UB is fixed, the transistor emission junction UBE=UB-UE decreases.

The process is similar to negative voltage feedback, which serves the purpose of stabilizing the operating point. However, the feedback of the circuit is UE=IE*RE, which depends on the output current and has nothing to do with the output voltage, so in the circuit, the resistance values of the upper and lower base bias resistors R1 and R2 are smaller, so that the base potential UB is mainly determined by their partial voltage values. The greater the feedback resistance re, the deeper the negative feedback, and the better the stability.

However, if the RE is too large, under the condition that the power supply voltage remains unchanged, it will cause the UCE to drop and the impact will be amplified, so.

If an AC signal is input, it will also cause a voltage drop on the RE, reducing the amplification of the amplifier, in order to avoid this, a capacitor CE is connected in parallel at both ends of the RE to act as an AC bypass.

Principle: The transistor needs to be turned on and the transmitting junction voltage is higher than its threshold voltage (around the silicon tube), otherwise the input signal will not pass when it is below the threshold. The auxiliary circuit that provides the threshold voltage is the bias circuit. There are many ways to connect, and resistor divider circuit is the most commonly used method.

I can't understand this as soon as I copy a paragraph, but there are several sections in the book on analog electronic circuits. I suggest you go and read a book.

In a triode amplifier, there is a junction voltage or starting voltage (silicon germanium) between the transistors in EB. If the amplifier (silicon) does not have a bias circuit, then, if the signal is less than than is input between EB, the BE pole cannot be turned on, and if the BE pole cannot be conducted, there will be no IB (IB=0), and there will be no IC without IB (that is, if the starting voltage is not enough, there will be no starting current), and then the amplifier is in the cut-off zone;

If the signal voltage you input is 1V, then only the top one can generate IB, and if there is IB, there is IC, that is, the input of 1V electrical signal at the input end of the amplifier will be amplified only if it is above.

The input signal of the general amplification circuit is very small (MV or even UV level), which cannot make the amplifier amplify the input signal, in order to make the small signal can also produce IB, it is necessary to make VEB reach this circuit assigned to VEB = in advance is the bias circuit When there is a bias circuit Signal voltage + bias voltage = VEB actual voltage (due to the clamping effect of BE VEB or; Signal current + bias current = ibe. This causes the signal to float above the bias current, which can be effectively amplified no matter how weak the voltage amplitude and current amplitude of the signal is

In a small-signal Class A analog amplifier, the bias current of the amplifier must also meet the requirement of subtracting the negative half-wave of the input signal to avoid negative half-wave clipping distortion

As for the bias circuit, the resistor divider type is mostly used.

I didn't specifically calculate the static working point and AC parameters of this circuit, but according to what you said, you didn't consider the voltage drop of the on-tube of the transistor UBE when calculating the current IE of the static working point (the silicon tube is about volt, and the germanium tube is about volt), which causes the calculated collector current IC to be large, so the base current IB is also large. At the same time, due to the relatively large IC value of the collector, the calculated AC equivalent resistance value Rbe between the base and emitter of the triode is small, which will affect the large voltage gain of the circuit. This should be the reason why you calculated different from the answer.