What is a forward transistor

Base resistance

This article explains how to calculate a Base resistance, sometimes also called the base series resistor, for the use of a transistor as a switch described.

Basics & transistor selection [edit]

A common use of transistors as switches is particularly in the field of microcontroller technology. Here the transistor, an NPN type, is usually operated in a common emitter circuit. When operating as a switch, you want the collector-emitter path of the transistor either to block as well as possible ("switch" open) or to conduct as well as possible ("switch" closed), and neither the transistor nor the microcontroller Should take damage.

Two values ​​are particularly interesting here: The collector current, that is the current that flows through the load to be switched (e.g. LED), and the base current, the current flowing through the base with which the transistor is switched in the emitter circuit.

The transistor should normally be suitable for at least twice the value of the collector current that occurs. For example, the BC547 data sheet may contain a permissible collector current of 100 mA, but in practice a more powerful type such as BC635 has significantly better parameters beyond around 50 mA.

Blocking the transistor is the simpler of the two cases. In the emitter circuit, the transistor blocks as best it can (a very small, negligible residual collector current always flows) when the base current is as close as possible to 0 A. This is it when the voltage applied to the base and emitter is 0 V. For example, when the output of a microcontroller that controls the base sets the base to low, i.e. to ground.

In order to switch the transistor through, it must be driven with a base current. The base current is limited by a resistor, the base resistance. The resistance must neither be too big nor too small. If it is too big, the transistor does not switch fully through and there is too much power loss on the transistor, which can destroy it. If it is too small, it sucks too much basic current from the output of the microcontroller, which may then be overloaded.

In order to turn the transistor through as well as possible, it can be controlled in a range that saturation is called. In this area, a further increase in the input signal does not cause any further noteworthy change in the output signal. Technically, the collector-emitter voltage is minimal and smaller than the base-emitter voltage. The base-collector diode path of the transistor becomes conductive (the diode is operated in the forward direction).

A disadvantage of the saturation mode is that the transistor only comes out of this state relatively slowly when you want to turn it off again. This is problematic for high frequency signals. When switching a relay, motor or an LED, however, this problem does not arise at all, since "slow" is still more than fast enough for these applications.

The current gain of the transistor is required to calculate the base resistance. This sinks during saturation operation and can often only be read out of the data sheets. Most of the time, the current gains given in the tables are not those for saturation. Instead, with a bit of luck, you will find the required value in the diagrams. Alternatively, it can be estimated from the values ​​for non-saturation.

Basically, it should also be said that the calculations do not have to be carried out with high accuracy. Transistor properties vary widely per unit, and to be on the safe side, it is better to work with a little more (e.g. 10%) base current (corresponds to a lower base resistance) than the calculation shows.

Calculation [edit]

  • Obtain the data sheet of the transistor, if possible from the manufacturer of the transistor used. If you don't know or recognize the manufacturer, you have to be content with a data sheet from any manufacturer.
  • Read data sheet!
  • The current gain in saturation is required (DC current gain)
    • Look for an explicit specification of this value in the data sheet. With a little luck you can find it in the diagrams. The current amplifications given in tables do not normally refer to saturation if this is not explicitly stated (for example [math] \ displaystyle {{h_ {FE}} _ {SAT}} [/ math] or similar (saturation = saturation) .
    • If you do not find any information on the gain in saturation, then the data sheet minimal Look up the current gain of the transistor. (Character: [math] \ displaystyle {h_ {FE}} [/ math]; Description: DC Current Gain; if several values ​​are given, select the one for the lowest collector current). Think about it: the higher the collector current, the lower the current gain. Divide this value by 2 to 10 to estimate the value in saturation. In the case of small-signal transistors, a current gain of 20-50 can usually be expected as an alternative, and for power transistors it is more likely to be 10. With Darlington transistors, the current amplification factors of both transistors are multiplied, so one can assume 400-2500 for small-power transistors.
  • Divide the current [math] \ displaystyle {I_c} [/ math] that the consumer needs by [math] \ displaystyle {h_ {FE}} [/ math] by the base current [math] \ displaystyle {I_b} [/ math ] to obtain. [math] \ displaystyle {I_b = \ frac {I_c} {h_ {FE}}} [/ math].
  • Applying Ohm's law one obtains the basic resistance [math] \ displaystyle {R_b = \ frac {U_E-0,7V} {I_b}} [/ math].
    [Math] \ displaystyle {U_E-0.7 V} [/ math] is used, because of the control voltage [math] \ displaystyle {0.7 V} [/ math] at the base-emitter path of the transistor. Instead of [math] \ displaystyle {0.7 V} [/ math] you can also take the value from the data sheet if it is given in the data sheet. With Darlington transistors, a drop in [math] \ displaystyle {1.4 V} [/ math] should be expected if nothing else can be found in the data sheet.

Examples [edit]

Example 1 [edit]

  • Transistor: BC328
  • The consumer needs a maximum current of [math] \ displaystyle {I_c = 100 \ mathrm {mA}} [/ math].
  • [math] \ displaystyle {h_ {FE}} [/ math] is 100 (data sheet), but the calculation is based on an estimated current gain in saturation of [math] \ displaystyle {\ frac {h_ {FE}} {3.3} = 30 } [/ math] (empirical value or measured).
  • From this it follows: [math] \ displaystyle {I_b = \ frac {100 \ mathrm {mA}} {30} = 3.3 \ mathrm {mA}} [/ math]
  • The microcontroller is of the type ATmega or ATtiny and delivers at least 4.5 volts at 3.3mA with a supply voltage of 5V minus 10% tolerance. A further 0.7 V is dropped on the BE path, so that approx. 3.8 V remains.
  • Rb = [math] \ displaystyle {\ frac {3.8 \, \ mathrm {V}} {3.3 \, \ mathrm {mA}} = 1150 \, \ Omega} [/ math]
  • So one will choose a base resistance of [math] \ displaystyle {1 \ mathrm {k} \ Omega} [/ math].
Notice: Round off with this calculation so that reliable switching behavior is guaranteed even if the transistor varies.

Example transistor BC547B (from NXP) with max. 40 mA load [edit]

NXP data sheet: http://www.nxp.com/documents/data_sheet/BC847_BC547_SER.pdf

[math] \ displaystyle {h_ {FE}} [/ math] is at least 100 (data sheet) at 2 mA collector current. A look at the characteristic field reveals that [math] \ displaystyle {h_ {FE}} [/ math] is still in this range even with 40 mA collector current. But, somewhat hidden in the data and the characteristic curve for [math] \ displaystyle {V_ {CEsat}} [/ math] (the collector-emitter saturation voltage) one finds that [math] \ displaystyle {\ frac {I_C} { I_B} = 20} [/ math] is at saturation, and that is the desired state. So one calculates with a current gain in saturation of 20.

From this it follows: [math] \ displaystyle {I_b = \ frac {40 \, \ mathrm {mA}} {20} = 2 \, \ mathrm {mA}} [/ math]

The microcontroller is of the type ATmega or ATtiny and delivers at least 4.5 volts at 2 mA with a supply voltage of 5 V minus 10% tolerance. About 0.78 V is dropped on the BE path (data sheet of the BC547B, estimated from the characteristic), so about 3.75 V remains.

Rb = [math] \ displaystyle {\ frac {3.75 \, \ mathrm {V}} {2 \, \ mathrm {mA}} = 1875 \, \ Omega} [/ math]

A base resistance of <1.8 kΩ, for example 1.5 kΩ, will be selected here.

Older microcontrollers [edit]

Microcontrollers from older series can often only deliver a little power, much less than, for example, the AVRs. A number of 8051 variants can only deliver a base current of 50 µA because they have open collector outputs. The circuit shown above is then unsuitable; the following circuit is better. But be careful, the transistor is turned on when the microcontroller is in reset!

Open base on reset [edit]

Another issue is an open basis: If a microcontroller is in reset, the outputs are mostly high-resistance. As a result, the base of the transistor, which is only connected to a port pin via a base resistor, hangs in the air with high resistance. This is often not a problem with bipolar low-voltage transistors, but strong interference can cause the transistor to conduct at least partially. You avoid something like this by closing off the base in a defined manner against the mass. This is done with an additional resistor between base and ground, usually in the range between 50K-1M. The greater the resistance, the more sensitive the circuit is to electromagnetic environmental influences. The lower this resistance, the more current the circuit consumes in the switched-on state. A value of 1M has proven itself in uncritical environments, whereas in the automotive sector it is better to use 100 K.

When calculating the base resistance, of course, you have to calculate the current that flows through this terminating resistor ([math] \ displaystyle {\ mathrm {I} = \ frac {\ mathrm {U_ {BE}}} {\ mathrm {{R}}} } [/ math]), take into account. Because this is not effective in the base. The base resistance must therefore be designed to be correspondingly lower.

See also [edit]