Never biased a transistor before, don't understand the method - Page 1
Forum section: electronics / beginners

author: Kyle_from_somewhere
author_rank: Regular Contributor
timestamp: June 14, 2022, 02:58:12 am
content: Hello there. I need some help but before I ask my question I'll explaina bit about myselfand what I know/don't know. Then what I'm doing.I've been futzing around with electronics for a for a few years, mostly specializing in repairing and modifying old tube TVs. My most recent project was to add S-Video to a Composite/RGB only trinitron. This uses opamps, comparators, and a CD4081 to detect what kind of signal is present and then alter it's path using a small relay. I drew the board layout by hand and etched it myself. In the past I've built stereo amplifiers using opamps, baxandall tone circuits etc. Those amplifier projects used linear PSUs I made too. From TV experience I have basic RF knowledge like the layout of a superhet receiver.One area I've never trod before is BJT transistors. Any amplification I've done before has been opamps. The only transistors I've used in projects before now were simple 2N7000 to act as a switch.What I'm doingMy new project is to build a television modulator. The basic idea is simple enough: generate sine wave, multiply frequency a bunch of times, amplify to useful amount, filter harmonics, then do AM modulation with video signal.So the first step is a sinewave oscillator with good accuracy. Such an oscillator is nothing but an amplifier going through a feedback loop, moderated by a quartz crystal. Since opamps don't like to go 20Mhz or higher (at least nothing I can get hold of), I have to use a BJT transistor. Oh dear.So here is my circuit scheme of a 20.25Mhz pierce oscillator with 2N3904:I understand that the transistor must be biased to work, and that this is done using voltage dividers. I know how to set up a voltage divider to get the right current and voltage.What I don't understand is how to use the information provided in the datasheet to find the right voltages and current to apply to Collector, Base, and Emitter. Can someone instruct me, step by step?2N3904 datasheetquartz crystal datasheet

author: Doctorandus_P
author_rank: Super Contributor
timestamp: June 14, 2022, 03:53:51 am
content: The transistor plus the 4 resistors is the classic amplifier.Dimensioning goes sort of like this:You want a relatively big output voltage, and the voltage over R2 limits the output voltage, so you want it to be relatively small.For ease of explanation I simply choose 20% for no other reason then that it gives nice round numbers.So you have 12 * 0.2 = 2.4V over R2. (It's value and current are yet unknown)The Base-Emitter junction acts as a diode, so the Base has to be 2.4V + 600mV = 3V.So there is 3V over R4, and thus 9V over R3Right now we just assume a high Hfe, and forget about the base current for a moment.Next step is to choose an input impedance for the amplifier. You can't choose R3 and R4 too high because R3 does have to source some base current for the transistor. The input impedance will be near R3 parallel to R4.If we design for an input impedance of 1M Ohm, then R4 will be around 1.5Meg and R3 will be about 4.5Meg (3 times larger).Next step is to choose the current you want through R2. I simply choose 10mA, but in real life the transistor parameters shift with this base current, so you have to think a bit deeper about this.We already know R2 has 2.4V over it, so if we want a current of 10mA, then it's value is 2.4V / 10mA = 2.4kOhm.Next the dimensioning of R1Normally you want to be able to get some relatively high output voltage, so it has to be able to swing upward and downward.That means that in "rest" you want the voltage over R1 to be chosen in such a way that the output voltage can swing up and down by approximately the same amount. The voltage should therefore be about (12V -2.4V)/2 = 4.7VThe current through R1 is nearly the same as the current through R2.R1 should therefore have a value of 4.7V / 10mA = 4700 Ohm.The total amplification of the circuit, is the ratio between R1 and R2, so in this case it's 4700/2200 = 2.13You can increase the amplification by putting a (relatively big) capacitor over R2.Note that I did not use any parameters of the transistor itself in these (rough) calculations. It's one of the advantages of this circuit. It's a simple and stable circuit that works with widely varying transistors. In reality all values will shift a bit due to the currents through the resistor.However, because we have chosen a relatively high input impedance, the biasing will shift significantly, so we do a simple approximation as a check:I assume your 2N3904 has an Hfe of 500. With 10mA though R2, the base current will be 10mA/500 = 20uA.We've chosen R3 to be 4.5Meg, so the extra voltage drop over R3 due to the current though the resistor is: V= I*R = 20uA * 4.5M = 20*4.5 = 90V. This clearly will not work as expected, because the supply voltage is only 12V. (It would have worked quite good if the input impedance was 10kOhm though)So the Base current for the transistor sets an upper limit for the input impedance.The quite high chosen current of 10mA though R2 also increases the needed base current and thus also decreases the maximum input impedance.In the next approximation, you can choose R3 so that it is able to deliver enough base current for the transistor, and then pull the base voltage lower with R4 (Spice simulation or a potentiometer on a breadboard) until you have the voltage over R2 that you want.=======================I have not used this amplifier in a piece oscillator, so I don't really know how to dimension it around the crystal. I guess you want to block DC voltage over the crystal, and want a quite low amplification, but that is a guess.

author: ledtester
author_rank: Super Contributor
timestamp: June 14, 2022, 04:05:09 am
content: Learning how BJT audio amplifiers work would be a good way to learn about transistor biasing as well as how to analyze BJT circuits in general.A book that helped me with the subject and that I would recommend is Malvino's "Transistor Circuit Approximations". I found it very easy to follow and it doesn't assume a lot beyond basic circuit analysis.Update: The book is available for checkout at the Internet Archive (archive.org).

author: ebclr
author_rank: Super Contributor
timestamp: June 14, 2022, 04:20:39 am
content: https://sites.pitt.edu/~qiw4/Academic/ME2082/Transistor%20Basics.pdfhttps://www.infineon.com/dgdl/Infineon-ApplicationNote_Linear_Mode_Operation_Safe_Operation_Diagram_MOSFETs-AN-v01_00-EN.pdf?fileId=db3a30433e30e4bf013e3646e9381200

author: Kyle_from_somewhere
author_rank: Regular Contributor
timestamp: June 14, 2022, 05:48:44 am
content: thank you Doctorandus_P. You've cleared some things up but also given me more things to question.The 2n3904 has an Hfe of between 100 and 300 when the collector current is 10mA.I got about 10kohm input impedance by making R3 41k and R4 14k (ignore for moment these aren't common values)put it through the ringer: R2's 10mA/100 = 0.0001 amps0.0001 * 41000 = 4.1V voltage drop over R3?I don't know what an appropriate current on the base should be. There's a graph in the datasheet (fig 16) that I don't understand, beyond that, no mention of it at all. With a 41K R3 we're at 0.29mA.You suggest to put a "relatively big" capacitor across R2 to increase gain. How big is big? pF, nF, uF?

author: Zero999
author_rank: Super Contributor
timestamp: June 14, 2022, 07:02:11 am
content: The current through the R3 & R4 should be around 10 times higher than the base current, so its effect is minimal.The value of the capacitor bypassing R2 depends on the lowest frequency of interest. It needs to have a lower impedance than R2.

author: Kyle_from_somewhere
author_rank: Regular Contributor
timestamp: June 14, 2022, 07:25:08 am
content: I thought that the current passing through R3 and R4 is the base current, because R3 and R4 are connected to base.

author: Zero999
author_rank: Super Contributor
timestamp: June 14, 2022, 07:39:05 am
content: Quote from: Kyle_from_somewhere on June 14, 2022, 07:25:08 amOnly a small current flows through the base. Most of it flows through R4.

author: Kyle_from_somewhere
author_rank: Regular Contributor
timestamp: June 14, 2022, 08:28:44 am
content: I'm trying to simulate this circuit. I'm running into the problem that the information a simulator wants about a crystal (series inductance and capacitance) aren't ever written on a datasheet.Is there a way to calculate those values from frequency, ESR and shunt capacitance?

author: Terry Bites
author_rank: Super Contributor
timestamp: June 14, 2022, 10:50:53 am
content: Start with deciding on an emitter current Ie. Chose R2 to set 1 Volt at the emitter. R2=Ve/Ie. So the voltage at the base needs to be 1V +Vbe, 1.7V. For small signal transistors Ie~Ie. Knowing Vcc and Ic lets you work out R1. Ic*R1, this should normally be equal Vcc/2 so that the collector sits mid way between the rails- this gets you maximum voltage swing at the collector. Gain =R1/R2. All you need now is to work out the values for the base potential divider to get you 1.7V. The choice of components for this divider will depend on the input resistance you want. ie the R3 in paralell with R4. I also depends on the hfe of the transistor, Ie/Ib so the divider must provide Ib without too much change in Vb. Put another way, VCC/(R3+R4) must be>> than Ib.The gain is R1/R2. Experiment. You need to capacitively couple the inputs and outputs to prevent external circuits from upsetting the DC operating points.There are other methods, but this is by far the simplest. Note that this circuit uses negative feedback to stabilise Ie. As temperature increases Ie alo increases. As Ie increases the voltage across R2 increases. This voltage is opposes the base voltage. Happy days.

author: Zero999
author_rank: Super Contributor
timestamp: June 14, 2022, 11:23:55 am
content: Quote from: Kyle_from_somewhere on June 14, 2022, 08:28:44 amhttps://www.pspice.com/resources/application-notes/modelling-quartz-crystalshttps://www.maximintegrated.com/en/design/technical-documents/app-notes/2/2127.html

author: CaptDon
author_rank: Super Contributor
timestamp: June 14, 2022, 06:50:01 pm
content: Usually that type of oscillator has an inductor included in the collector circuit which helps give some phase shift and reliably start and run the oscillator. It can also help clean up the crystal fundamental waveshape. I may also tend to run the crystal at a third overtone and not the frequency marked on the crystal (Unless it is a third overtone crystal like was used in the old C.B. sets.

author: Kyle_from_somewhere
author_rank: Regular Contributor
timestamp: June 15, 2022, 03:09:29 pm
content: Quote from: Zero999 on June 14, 2022, 11:23:55 amThat sort of heavy physics equations is beyond my pay grade. thanks anyway.