# How can I get base current in this simple transistor switch?

• goodphy
In summary, the transistor has a V-I relationship like a typical P-N junction, so the forward bias voltage is usually around 0.6 V. However, this number may be different depending on the circuit.
goodphy
Hello.

Please look at the attached figure first, which is cut image from transistor section of HyperPhysics.

In this circuit, I only need to know base current or base voltage, IB. In order to know this VB must be known but it is already given as 0.6 V. How this value is obtained? I don't think this is given parameter for each transistor and it seems enough external parameters to fully describe this circuit is already specified.

Could you please figure out What I've missed in circuit analysis?

#### Attachments

• BJT transistor switch.png
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The base-emitter is a silicon P-N junction, so exhibits the V-I relationship of the typical P-N junction. For the range of currents over which a small signal transistor will be operated the B-E voltage is approximately 0.6v. (At the low end of expected currents, VBE will be around 0.5V, while at the upper end it may be 0.7V.) So you can assume that whileever that junction is forward biassed and is conducting base current, the B-E voltage is close to 0.6V.

goodphy and davenn
NascentOxygen said:
The base-emitter is a silicon P-N junction, so exhibits the V-I relationship of the typical P-N junction. For the range of currents over which a small signal transistor will be operated the B-E voltage is approximately 0.6v. (At the low end of expected currents, VBE will be around 0.5V, while at the upper end it may be 0.7V.) So you can assume that whileever that junction is forward biassed and is conducting base current, the B-E voltage is close to 0.6V.

Oh thanks for quick reply! But I still don't get one point. Can I really assume that forward bias voltage for base-emitter is always 0.6 V in whatever circuit is made with this transistor? I mean...every element of circuit has its own voltage drop along it when the current is going through. When current is going high, voltage drop is high. This is passive element example thus it may be not relevant to transistor but my point is that the voltage drop (here voltage base-emitter) should be externally dependent parameter unless it has voltage regulation. How can I have guaranty of such 0.6 V bias?

You can assume 0.6V - 0.7V whenever this transistor is given forward base current.

Do a google search for the V-I curve of a silicon PN junction (diode).

goodphy
Note that in this problem..

Ib = (10-Vbe)/1000

So if

Vbe = 0.5 then Ib = 9.5mA
Vbe = 0.7 then Ib = 9.3mA

So an "error" in the value you use or assume for Vbe only has a small effect on the value of Ib. You might consider that the exact value of Vbe doesn't matter in this circuit.

Many transistor parameters are subject to manufacturing tolerance. For example the gain of a transistor might vary from 80 to 200. Good engineering design ensures that the circuit will work for all values between 80 and 200.

goodphy
All right. It looks like that I need to accept 0.6 V is typical forward biasing voltage in base-emitter junction in at least active mode. However, If it is possible, can I get some mathematical expression or graphical form of relation between IB and VB? As suggested, I've found figure like the attached image but even this figure doesn't explicitly tell this relationship.

In addition, 0.6 V even holds for other mode such as saturation and cut-off mode?

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• Transistor load line and characteristic curve.gif
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Actually for my last reply in this post, I've found that silicon PN junction depletion layer voltage (due to space charge) or built-in voltage is about 0.6 V. That is why this number is given as firstly suggested by NascentOxygen since this voltage must be overcome to current flows. Although I still don't have clear equation to get VB in circuit.

In cut-off there is no base drive, so you mustn't expect the 0.6V figure to still hold.

If you really wanted to theoretically calculate VB then you would use the exponential relation between base current and base voltage, including its temperature dependence. You would have to employ a numerical method to solve the mathematics. For normal conditions, with these small signal BJTs, the result you end up with won't differ much from 0.6V.

goodphy

## 1. How does a transistor switch work?

A transistor switch works by using a small current to control a larger current. When a small current is applied to the base terminal of the transistor, it allows a larger current to flow between the collector and emitter terminals, effectively switching on the transistor.

## 2. Why is base current important in a transistor switch?

Base current is important in a transistor switch because it controls the flow of current between the collector and emitter terminals. Without the proper amount of base current, the transistor may not switch on or off correctly, resulting in incorrect operation of the switch.

## 3. How can I calculate the required base current for my transistor switch?

The required base current for a transistor switch can be calculated using the formula Ib = Ic/hFE, where Ib is the base current, Ic is the collector current, and hFE is the current gain of the transistor. This formula can be found in the datasheet of the transistor or can be determined experimentally.

## 4. What can cause a lack of base current in a transistor switch?

A lack of base current in a transistor switch can be caused by a few different factors. It could be due to a faulty or damaged transistor, incorrect wiring, or insufficient voltage being applied to the base terminal. It is important to check all of these factors when troubleshooting a lack of base current.

## 5. How can I troubleshoot a transistor switch that is not working?

If your transistor switch is not working, the first step is to check for proper base current. If there is no base current, check for any potential issues such as damaged components or incorrect wiring. If the base current is present, but the switch is still not working, it could be due to insufficient voltage being applied, or the transistor may be damaged and need to be replaced.

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