What is a BC547 transistor

Transistor for beginners

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  1. The transistor as a current amplifier
  2. The transistor as a switch
  3. Darlington transistors
  4. Bipolar power levels
  5. The transistor as a current regulator
  6. The transistor as an analog amplifier
  7. The transistor as a sine wave oscillator
  8. The transistor as an RF amplifier

1 The transistor as a current amplifier

Transistors are small black boxes with three legs: C, B and E. Or huge shiny metal parts with two connections and the metal part as the third connection.

A small plate made of high-purity silicon sits in the box or metal parts, symbolized by the slightly thicker vertical line. Two parts of the surface of the plate are contaminated with tiny amounts of a trivalent or pentavalent (semi-) metal, a small puddle in between with the other-valued (semi-) metal. These impurities, known as doping among insiders, do not really match the actually tetravalent silicon, but they give the part its amazing semiconductor properties.

Depending on the doping, there are two types of transistors: NPN and PNP. With NPN transistors, as shown in the logo and in the connection diagram, the positive pole of the power supply goes to the collector (C) of the transistor, the negative pole to the emitter (E). If you do that, almost nothing happens, only a tiny reverse current flows through the transistor, the distance from the collector to the emitter is as good as blocked.

The BC547 is an NPN transistor (N = negative = 5-valent metal as doping, P = positive = 3-valent metal as doping). With the PNP (e.g. a BC557) almost everything is the other way around:
  • the arrow points to the silicon plate, not away from it,
  • the positive pole goes to the emitter, and
  • the negative pole to the collector.
But here, too, almost nothing flows through the CE route, as is the case with the NPN.

That only changes when we bend the third connection, the base B, a little and push in some electricity. But be careful, not too much of a good thing, otherwise our transistor will detach itself from being and go into the eternal hunting grounds of the transistors. In practice we do this with the following circuit.

A current now flows through the base of the transistor, slowed down by the large resistance. As a diode, the base-emitter path also has a forward voltage, like any diode, namely approx. 0.65 V (with silicon diodes). According to Mr. Georg Simon Ohm, the base current is therefore:
I = U / R
I [A] = (Ubattery [V] - UBE route [V]) / R [Ohm]
I [A] = (9 - 0.65) [V] / 100,000 [Ohm] = 0,000.083.5 [A]
I [µA] = 1,000,000 * I [A]
I [µA] = 83.5 µA

That is very, very little electricity, which is not even enough to let a light-emitting diode shine even a little.

The ammeter in the collector of the transistor now shows a much higher current, namely around 20 mA or more. That would be enough to make a light-emitting diode shine brightly. The transistor increases the current through the base by z. B. 250 or 300 times. This DC amplification factor (English: gain) is called hFE ("h" stands for small signal, "F" for forward, i.e. forward current, "E" for emitter circuit). In older documents, this DC gain is also often used with the Greek lowercase letter beta β abbreviated.

However, hFE is different for each transistor, depending on the temperature and also somewhat on the collector current. "Bad" transistors with a low hFE get an "A" on the back, "better" transistors get a "B" and the very "steep" ones get a "C". If you don't know anything, the letter is missing. As a rule, you can remember: 100 can all, except power transistors, they can only be 40 or very thick even 25. Tolerate heat output, which they can give off to the outside with their chunky metal surfaces.

At first we don't care about the high or low hFE, we now know how to measure it. I made it with ten copies of the transistor BC547B. The 9V battery had 9.31V, the base resistance measured 99.5kΩ, so the base current was
I [µA] = 1,000,000 * (9.31-0.65) / 99,500 = 87.04 µA.

That's what came out of the 10 measurements. The second column shows the measured current in [mA], the third the resulting DC voltage gain hFE (hFE = ICE / BE) and in the fourth I have the thermal power of the transistor
P [watt] = UCE [V] * ICE [A]
P [mW] = 1,000 * P [W]

#I.CE [mA]hFEPCE [mW]
The hFE values ​​are all between 290 and 390. This means that the small base current allows a collector current to flow that is approx. 300 times higher.

If you do that, you will immediately see another effect: the measured values ​​of the current increase quite rapidly at the beginning of each measurement and you have difficulties to read something reliable because it is still a little more current for a long time. This is due to the fact that the thermal power generated by the transistor (see last column, approx. 300 mW) causes the temperature of the transistor to rise to 50 ° C during the measurement. The increased temperature also increases the collector current and the hFE.

However, the BC547 must not consume more than a maximum of 500 mW, because otherwise it will get too hot and be damaged. This is because the heat that is generated on the silicon plate is only released very slowly to the plastic housing and the connecting wires. This is expressed as thermal resistance in degrees Celsius per watt. With the BC547 this is 250 ° / W. Since it breaks on the silicon at 150 ° C, that's at 25 °
PMax = (125 - 25) [° C] / 250 [° C / W] = 0.4 [W] = 400 [mW]

At some point the heated transistor loses so much heat during the measurement through its connecting wires and through the air around it that its temperature no longer rises so much and you can then read the current a little more precisely. So anyone who relies on a very specific hFE value has already lost when summer or winter arrives. The effect can also be used to measure the outside temperature, but then the current through the transistor must not be as high as in the present case.

How about a PNP? The collector comes to minus via the ammeter, the emitter to plus. The resistor now comes to minus and the voltage at the base is now approx. 0.65 V below the operating voltage.

The same now applies to direct current amplification: the collector-emitter current I.CE is hFE times greater than the base current I.BE. Now the transistor is "on" if you pull the base a little to zero volts. Also the opposite of the NPN.

2 The transistor as a switch

What is all this for? If you are using a small stream, e.g. B. with 8.65 mA from a CMOS digital output, would like to switch on a very large current, for example to switch a DC motor of one ampere on and off, the transistor DC amplifier does that. For this you need a transistor with an hFE of 1,000 [mA] / 8.65 [mA] = 116, which can withstand 1 ampere collector current, which z. B. a BD439 can. Unfortunately, it only has an hFE of approx. 140 and therefore only manages one ampere if you don't catch a Monday transistor.

In this (and in other circuits with motors or relays) a diode is shown in the reverse direction. What for? If you let a current flow through a coil (which is built into the motor), it builds up a magnetic field. The magnetic field contains some of the energy that enters the coil with the current (the part that is not converted into heat). If the current were to become a little lower, this magnetic field would decrease and the stored energy would counteract this decrease: the decrease in the current is slowed down by current flowing "backwards" from the magnetic field into the power source. This until the stored energy corresponds to the new current flow, then the magnetic field is partially discharged.

If the transistor now delivers less current and the magnetic field resists this change, where does the current go? The CE path cannot divert the "reverse current" from the magnetic field to ground, because it acts as a diode and negative voltage at the collector is not allowed to pass through to the emitter. And even more dramatic: Where should the stored energy go when the transistor abruptly and completely switches off the current? Then there is no longer a power source because the CE line doesn't let anything through backwards. Well: Energy that is not let through creates a higher voltage: when switched off, such a motor magnet can get voltages of a few hundred volts and more. And that would destroy the non-conductive CE line because it is not designed for this high voltage (with the BC547 that's just 50 volts).

The diode is therefor: it offers the reverse current a convenient way out. If the voltage on the motor becomes negative, it becomes conductive and just short-circuits the voltage. The diode has to cope with about the same amount of current as was used to charge the magnetic field (here approx. 1 A). A 1N4148 would be overwhelmed and quickly broken. It would have to be a 1N4000 or something thicker.

Why this circuit when you can turn the motor on and off with a very ordinary mechanical switch? Well, mechanical switches age and at some point give up their switch spirit. And because of the high voltage kickback every time the motor is switched off, a few micrograms of iron burn off in the switch and turn into iron oxide, commonly known as rust. At some point the switch is mostly rust, the contacts suffer and their resistance increases. Which then accelerates aging, because every time you switch, heat is also generated (P = U * I, I = U / R, P = U2 / R) and at some point the switch refuses to take its name seriously: it only switches sometimes and sometimes not at all. Lummerland simply burned down.

In contrast, transistors practically do not age at all. And the kickback diode does not really care how often cut-off current has flowed through it. That alone should be a good argument to leave out the mechanics and rather let the transistor switch and take over.

The same also applies to any relay contacts that smart guys use instead of the mechanical switch. You can sometimes even watch them burn down if they have a transparent housing: every time they are switched off, a short flash of lightning at the contact. Nothing flashes with the transistor.

3 Darlington transistors

What to do if the amplification of a transistor is not sufficient to fully control a very large collector current (e.g. 5 amps) with a small base current? A second transistor can help: it additionally amplifies the base current and thus drives the base of the BD439. With a BC547A, this increases the gain to 180 * 140 = 25,200. The circuit is called Darlington and comes with both transistors and the resistor integrated in a single housing, e.g. B. as BD677 or, for even higher currents, as MJ3000 as shown in the adjacent picture.

Since two base-emitter paths are now eating voltage, the Darlington needs approx. 1.3 V for control. But the structure now has so much reinforcement that it also fully controls the high starting current of the motor with 5 A.

The remaining voltage on the CE line at full modulation is approx. 0.2 V and does not affect the motor anymore. Because the power of the CE line in this case is P = U * I = 0.2 [V] * 1 [A] = 0.2 [W], we don't actually need a power transistor for this, because that can also be done a BC547. It only consumes power for the short time it takes for the transistor to go from zero to full. But the BC547 would definitely break at 1 [A] collector current, because it is only made for a maximum of 0.1 [A]. The low thermal output in the switched-through state means that we do not have to cool the power transistor to a large extent, it only has to withstand the 1 [A]. So always keep an eye on the limits specified by the manufacturer (voltages, currents, thermal power) when selecting the transistor type. You can get the data sheets for all transistors from the Internet by searching for "Transistor BC547 datasheet".

4 bipolar driver transistors

What if you want the engine to turn clockwise today and counterclockwise tomorrow. Smart guys simply swap the connections of the motor or use a two-pole switch and you're done. But that can also be solved electronically with transistors. We need two transistors for each connection wire of the motor: one that can pull the connection to plus, and another that can pull the same to minus. Only one of the two transistors is switched on at a time, the other remains switched off so that a messy short circuit does not kill the transistors. Exactly the other way around on the other side of the motor and the forward / reverse control is perfect.

The four transistors, two each BD439 and BD440, now alternately drive the bridge. In the state shown, the left BD439 is switched on because its base is above 1 kΩ and the lower Zener diode is at plus potential. If the left switch were SL. polarized differently, then the base of the upper transistor, a PNP, would be negative via the upper Zener diode and the 1 kΩ resistor and it would be conductive. When the switch is open, no current flows at all, because the two Zener diodes add up to a voltage above the operating voltage, so they are not conductive.

So that it doesn't get boring, the right side is switched differently: here the NPN is on top and the PNP is on the bottom. If the lower left bridge is switched on as shown, the motor at the emitter of the upper right BD439 is at low potential via the lower left CE path. The switch now sets the base of the upper right transistor to positive potential via 1 kΩ, supplies the base with current and the upper right transistor becomes conductive. The red arrows show how the current runs in this case;

If the left switch is in the lower position, then the upper left transistor is conductive and sets the emitter of the lower right transistor to positive potential via the motor. With the right switch SR. the base of the lower right transistor is pulled to zero via the resistance of 1 kΩ, base current flows and the lower right transistor becomes conductive. The fact that the base of the upper transistor is at a negative voltage compared to its emitter doesn’t matter, because the base-emitter path is a diode and blocks the flow of current in the wrong direction. Both base connections can therefore be easily connected.

The construction on the right-hand side is of course easier because two resistors and the two Zener diodes on the left-hand side are omitted. We could have wired the left side like this too. But since both the left and the right switch go to plus in the upper position, we could simply omit it and would only need a single switch. Kind of clever, right?

Since the motor receives current sometimes forwards and sometimes backwards with the bridge circuit, the derivation of the reverse current from the motor magnetic field is of course not so easy with a diode. There have to be four diodes for that.

The principle of the NPN-PNP bridge circuit also applies to LF amplifiers, see the example below.

5 The transistor as a current regulator

In the long run, the only-off or only-on of the transistor switch is a bit boring, transistors can do a lot more, namely regulate.

Let us assume that we wanted a light-emitting diode with a constant current of z. B. light up 20 mA. This is of course easier with a resistor. It just has to be so large that it absorbs the difference between 9 V and the LED voltage in the direction of flow (e.g. 2.0 V) and would amount to
R.V. = (9 - 2.0) / 0.020 = 350 Ω

With an E12 resistor of 330Ω, this would result in 21.2 mA. What if, instead of one light-emitting diode, we now connect two in a row? Then burn up the 2 * 2.0 = 4 volts. And the resistance should now only be 250Ω.

It is more elegant with a transistor. To do this, we set its base to half the operating voltage with a voltage divider made up of two resistors and set the emitter to a high voltage with a resistor. The transistor now adjusts the current through the collector-emitter path so that the emitter voltage is 4.5-0.65 V = 3.85 V.This is the case when the current through the emitter resistor of 180Ω brings its voltage to 3.85 V. So the size of the emitter resistor regulates the current, in this case up
I.CE = (4.5 - 0.65) / 180Ω = 0.021 A = 21 mA

And it always does it that way, regardless of whether one or two LEDs are connected to the collector or whether the light-emitting diode has 2 or 3 volts and eats up voltage: the CE line eats away the excess and the LEDs always have the same flow current.

Of course, our calculation is not entirely correct, because
  1. the base-emitter path also needs current, in this case 21mA / hFE or approx. 100µA, so that the voltage divider has a slightly lower voltage because the base-emitter path is parallel to the lower voltage divider resistor,
  2. Not only the CE current but also the base current flows through the emitter resistor. But since the 100µA base current is hardly significant compared to the 21mA CE current, this is practically not noticeable.

In order to avoid this inaccuracy and also to make the circuit fit for (almost) any operating voltage, from 4 to 40 volts, we only exchange a few components:
  • instead of the lower voltage divider resistor, we take two normal silicon diodes one behind the other in the direction of flow, and
  • we make the emitter resistance 33Ω.
Now the voltage at the base always remains at 1.30V, regardless of which operating voltage we apply. At 4 volts, I = (4 - 1.3) / 10,000 = 0.00027 A = 270µA flow through the upper resistor, at 40 V halt (40 - 1.3) / 10,000 = 0.00387 A = 3.87 mA. And: if the base-emitter path now requires 20 mA / hFE = 100µA, then this only reduces the current through the two diodes by this 100µA. This only no longer works if the base current exceeds 270µA, but then we would simply use half the resistance instead of 10kΩ and everything would be fine again.

Now it doesn't matter how many of the LEDs we switch one behind the other. If that gives the operating voltage, we can z. B. connect 15 in a row, makes 30 V. Together with the two voltages
  • 0.65 V for the emitter resistor, and
  • 0.1 V for the saturation voltage of the CE line at 20 mA,
the operating voltage would have to be above 31 V so that all LEDs light up with the same current.

Please also note that with an operating voltage of 40 V and only a single LED, the transistor has to absorb the entire residual voltage of (40 - 2.0 - 0.65 - 0.1) V and at 0.02 A a heat output of P = U * I = 0.783 watt = 783 mW must be able to handle. A BC547 would fail and quickly give up the ghost and die from heat, because it can only reach up to 500 mW under particularly favorable circumstances. But only really crazy freaks operate a single LED with 40V operating voltage.

The circuit with a constant voltage at the base can do even more: since the transistor always tries to set the voltage at its emitter connection, which is 0.65 V lower, it can also be used as a voltage regulator. To do this, a voltage that is 0.65 V higher is applied to its base, e.g. B. with a Zener diode and a resistor. The circuit to be supplied is connected to the emitter. No matter what current it is consuming: the transistor always keeps the voltage constant.

But be careful: the circuit is not short-circuit-proof: in the event of a short circuit, the transistor supplies everything that the connected power supply unit and the series resistor on the Zener diode (sometimes hFE) can do and may destroy itself in the process.

By the way, the electrolytic capacitor is sorely needed, as Zener diodes noise quite heavily and this noise would then also continue to the output voltage with amplification. And very noisy supply voltages can make CMOS or microcontroller circuits really think about whether this is the right switching time or maybe not yet. The electrolytic capacitor therefore flattens the noise voltage.

This shows with a somewhat more complex circuit, what a bridge circuit can still serve: it makes a constant current for a duo-LED red / green. Depending on whether the Duo-LED is connected to zero volts (red) or to +5 volts (green), either the upper NPN or the lower PNP drives the LED. The current is determined by the two emitter resistors of 33Ω. The two resistors can easily be of different sizes so that the circuit can be set to the same brightness impression of the LEDs.

The design provides for the base voltage of the NPN to be set to just below (2.5 + 0.65) V so that the NPN is just barely conductive when the LED is not activated. For the base voltage of the PNP (2.5-0.65) V was selected so that the quiescent current remains very low when the LED is not connected.

If the LED with its red cathode is now pulled to zero volts, a voltage of +2.0 V is established at the intersection of the two emitter resistors. Since the base voltage of the NPN is 3.14V, current now flows through the NPN until its emitter is at (3.14-0.65) = 2.49V. This is the case with a current of (2.49 - 2.0) / 33 = 15 mA through the red LED.

If the LED is at +5 V, then the crossing point of the two emitter resistors is at +3.0 V. Since the base voltage of the NPN would now have to be at least (3.0 + 0.65) = 3.65 V in order to supply the base current get, but is only at 3.14 V, the NPN is silent. The PNP is now active for this: its base voltage is 1.86 V and the emitter is therefore (1.86 + 0.65) = 2.51 V. The difference to +3.0 V now drives the green LED.

Incidentally, the current through the three resistors is 160µA. This is only a little compared to the base current of 0.015 / hFE = 0.015 / 300 = 49µA, so that the base voltage by driving the LED has only a small effect on the voltage in the resistor divider. Since this leads to a reduction in the LED current, the change caused by the base current is in any case not harmful. If you want to be even more precise, reduce the three resistors roughly proportionally, then the voltage divider current increases and the base currents then have even less influence.

Circuits like this are designed with a spreadsheet. In their cells you write z. B. the size of the three voltage divider resistors, this is used to calculate the base voltages and, in turn, the emitter voltages. This then also results in the current. With the calculation you can play with the values ​​and see whether it could work.

6 The transistor as an analog amplifier

The constant current or constant voltage circuit makes little use of the gain that transistors thus have. This is different with analog amplifier circuits: they are supposed to amplify LF voltages and can only do this if the transistor provides the amplification.

Amplifiers can be designed for small signals of a few millivolts or for large signals of several volts. Here first the small-signal variant.

Simply applying the NF signal to the base would not do anything. The peak of the alternating current does not reach voltages that would impress the base at all (it only starts at 0.65 V), nor do the negative voltage peaks make an impression on the base, as they have the base-emitter diode with completely wrong polarity in the blocking direction, not getting over it.

A base current must therefore be set from the outset in order for it to work. This is what the 1MΩ resistor does in this circuit: 8.4µA already flow from the start and cause a collector current of 1.7 mA with an hFE of the transistor of 200. The collector resistance of 2.7kΩ is dimensioned in such a way that it drops about half the operating voltage.

The LF input voltage now makes its contribution to the base current. For this purpose, it is separated in terms of direct current with a capacitor, because there are + 0.65V at the base, and they should stay that way. The same on the collector: the further signal processing takes place via a capacitor, which lets the alternating voltage through and blocks the direct voltage.

The circuit reverses the polarity of the input signal: if the base current increases due to positive voltage peaks, the transistor draws more collector current and the voltage at the collector drops. Conversely with negative voltage peaks: the collector current decreases, the collector voltage increases. So it is an inverting amplifier.

The structure amplifies the signal approx. 61 times (part of the amplification of the transistor disappears when the output resistance of the circuit is reduced), so it turns a 5 mV microphone voltage into approx. 250 mV amplified signal. If we need more, it can be one more transistor.

The circuit is not very large-signal voltage-proof. Even with a little more than 65 mV input voltage, the output signal is distorted because the transistor draws more than I = 9/2700 = 0.0033 A = 3.3 mA of collector current and the collector voltage tends to zero at the positive voltage peaks of the AC signal. In order to be able to do more, a little more collector current is required, e.g. B. 5 mA, and a little less collector resistance. But we build the amplifier a little differently.

With this amplifier we recognize some elements from the chapter on constant current:
  • the base is brought with a resistance voltage divider from 68kΩ to plus and 22kΩ to minus in terms of direct voltage to a slightly higher level (+1.92 V),
  • an emitter resistor stabilizes the collector current to 5.73 mA in this case, while the emitter resistor of 220Ω sets the voltage at the emitter to 1.26 V.
This means that the operating point of the transistor is approximately the same even when the hFE changes (due to temperature and model variance).

Since the emitter resistance would also cause negative feedback for the alternating voltage change impressed on the base current when the base is activated and thus reduce the gain, the emitter resistance is short-circuited in the middle of the 10µF electrolytic capacitor for alternating voltage. Its impedance is Z at 100 HzC. = 1/2 / Π / 100 / 0.000.01 = 159Ω. Since the two resistors, 220Ω and 159Ω are connected in parallel, the resistance for 100 Hz AC voltage is only R.E (100) = 1 / (1/220 + 1/159) = 92.3Ω and at higher frequencies even much, much less. It therefore does not slow down the gain so badly.

In this case, the voltage gain is around 38 at 1,000 Hz, and the maximum level of the input signal that can be processed is 160 mVpp. V.pp is pronounced as volt-peak-peak and is the voltage difference between the highest and the lowest point of the voltage for alternating voltage. The conversion of Veffectively in Vpp is obtained by multiplying by 2 and with the square root of 2. 160 mVpp therefore correspond to 56.6 mVeff.

At 5 mA, the collector quiescent current is a little more powerful than in the previous example. This level can therefore be added next to the first if more reinforcement is needed. Since both amplifier circuits invert the input signal, the inversion of the first stage is canceled out in the second and we get a non-inverted signal at the end.

If you need even more power, e.g. B. to drive an 8Ω loudspeaker with two watts, it needs a little more than just the 5 mA, because P = U * I and I = U / R and both together are P = U2 / R and U = √ (P * R) = √ (8 * 2) = 4 V. At 4 V, however, I = U / R = 4/8 = 0.5 A, i.e. 100, already flow through an 8Ω loudspeaker times more than in the above amplifier stage. Since the powers, voltages and currents are effective quantities, this stage already needs 4 * 2 * √ (2) = 11.3 Vpp and can no longer be achieved with a 9V battery, so more is needed.

In the example shown, the bridge circuit appears again with an NPN and a PNP, which we already know from the motor driver. In this circuit, however, it works analogously and not as a switch. The LF input voltage is applied to the positive input of an operational amplifier. This compares a tenth of the voltage at the emitter of the NPN and PNP or the loudspeaker with the input voltage. If this is lower than ten times the input signal voltage, the voltage at the output of the operational amplifier is higher, if it is lower it is reduced. The two bases of the NPN and PNP transistors are controlled according to the input voltage and multiplied by ten.

But be careful, the circuit is incomplete. To make the whole thing more stable, some anti-vibration measures have to be built in, because such operational amplifiers are very fast. The operating voltage of the operational amplifier should also be stabilized so that it does not depend on the level of the loudspeaker. If you are looking for functioning audio amplifiers according to this principle, you will find it on the Internet, e. B. here.

7 The transistor as a sine wave oscillator

But transistors can do a lot more than just switching and amplifying: they can even be made to vibrate and then produce very nice sine waves. And that's how it works.

The analog amplifier from the previous chapter has a very useful property for doing this: it reverses the polarity of the analog input signal.

If the voltage at the base is a little higher than before, the emitter current increases and the emitter voltage becomes higher. However, the higher the emitter current, the higher the collector current. Since a larger collector current at the collector resistor leads to the voltage dropping across it becoming larger, the voltage at the collector drops. This is shown in the illustration for different base voltages. The signal at the collector is therefore opposite to the base voltage. In electronics jargon, this is called "phase reversal".

The gain of the transistor in this case is exactly 5.7 because the collector resistance is 5.7 times greater than the emitter resistance.

If we were to simply couple the signal at the collector back to the base, we would get an oscillator: increasing base voltage reduces the collector voltage, decreasing collector voltage lowers the base voltage and thus increases the collector voltage. Since both are done at maximum speed, we get a fairly fast oscillator: so close to the cutoff frequency of the transistor, at which its gain becomes unity. This is around 300 MHz for a BC547B, and a bit too fast for a loudspeaker and the human ear. So there has to be a brake in the feedback that delays it a little.

Such brakes are RC elements made up of a resistor and a capacitor. These delay the signal because the capacitor needs some time to be charged and then discharged again. The complete circuit then looks like this:

The feedback takes place via capacitors with 22 nF and resistors with 10 kΩ. In the third stage, the resistances of 15 kΩ and 47 kΩ in parallel and the base resistance also set the 10 kΩ.

As with the amplifier circuit, a capacitor of 1 µF is connected in parallel to the emitter resistor, which increases the gain in the case of alternating voltage. The capacitor has a capacitive impedance of Z at 598 HzC. = 1/2 / Π / 598/10-6 = 266 Ω, parallel to the 470 Ω both have 170 Ω together. While the DC voltage gain of the transistor V= = 2700/470 = 5.7, its ac voltage gain is equal to V.~ = 2700/266 = 15.9 increased. Enough, but not excessive, reinforcement.

A little more math. The circuit oscillates at 598 Hz. This corresponds to an angular frequency of ω = 2 * Π * 598 = 3,757 (small omega). The angular frequency ω corresponds to ω = 1 / R / C. With f = 1/2 / Π / R / C the oscillator can be designed with any desired frequency. But don't expect miracles from the formula: the resulting frequency always deviates a little from the calculated one, if only because the capacitors deviate by 10% upwards and downwards. It should have been 723 Hz here, but it is only 598. If you want it to be frequency-accurate, you can use a microcontroller and an R / 2R network, as with this one. That then has quartz accuracy. But if you just want to peep around a little, you don't need it and this is better off with this one.

On the left you can see the cute structure of the whole thing on the breadboard. On the right is the beautiful result of the effort on the oszi: a straight sine. You can only get better with a real LC oscillator.

8 The transistor as an RF amplifier

What if it is now a little faster than 600 Hz. Then you can either build a broadband amplifier, without a resonant circuit but with a band filter, or design the circuit specifically for a single frequency.Depending on the frequency range, you can continue to work with a BC547 (whose transit frequency, at which the gain drops to one, is at least 300 MHz) or with special HF transistors. They have a smaller base puddle than doping, so they have a little less hFE, but also at even higher frequencies.

The advantage of a fixed frequency amplifier stage is that you can use an oscillating circuit in the collector that can be tuned to the frequency to be amplified (here: 455 kHz) (here: with a coil of approx. 100µH). The DC resistance of the coil is very low, but the AC resistance at the resonance frequency is tough:
  • At 455 kHz the impedance is ZL. the coil ZL. = 2 * Π * 455,000 * 0,000.1 = 291.5 ohms.
  • The impedance of the capacitor is denoted by ZC. = 1/2 / Π / 455.000 / the same size.
  • However, since the quality of the resonant circuit is 40 or 100, depending on the quality of the coil, the resonant circuit has an impedance of 11.7 to 29.2 kΩ (resonance increase).
This means that the collector resistance is much greater than with the LF circuit (0.82 kΩ), each change in the collector current therefore causes 14 or 36 times as high amplitudes at the collector and thus a much higher gain.

Two such amplifier stages for a reception frequency of 77.5 kHz (time signal transmitter DCF77 on long wave) are described here.

9 Conclusion

Transistors can be used in very different ways: as amplifiers as well as regulators. It is therefore worth learning how such parts work and what they can be used for. Even if you do use a fully integrated IC, it is worth knowing how its innards work.

© 2019/2020 by Gerhard Schmidt