Simple transistor amplifier - Page 1
Forum section: electronics / projects

author: yalect
author_rank: Regular Contributor
timestamp: February 19, 2025, 10:10:46 pm
content: Hello,I would like to ask you I have this simple amplifier circuit and according the shematic attached, if we considered C1, C3 and R3 as bandpass filter with the resisor of transistor Emmiter R2, which one is High-pass filter and other Low-pass filter?- How to calculateC1,C3,R3assuming R2?Thank you.

author: TimFox
author_rank: Super Contributor
timestamp: February 19, 2025, 10:50:59 pm
content: Bad circuit: the DC bias won’t work.C3 and R3 form a high pass filter.However, the output impedance of your emitter follower depends on the bias condition of the transistor, not R2, and the bias is wrong.If you change the DC conditions radically, then the output impedance of the follower plus C1 would form a low pass filter.

author: johnboxall
author_rank: Unknown
timestamp: February 19, 2025, 11:02:37 pm
content: https://www.talkingelectronics.com/projects/TheTransistorAmplifier/TheTransistorAmplifier-P1.html

author: yalect
author_rank: Regular Contributor
timestamp: February 20, 2025, 01:40:17 pm
content: Hi,Bad circuit: the DC bias won’t work?!.- May the example on the url above will work?.www.talkingelectronics.com/projects/TheTransistorAmplifier/TheTrAmp/TheTrAmpP2.htmlRegards.

author: TimFox
author_rank: Super Contributor
timestamp: February 20, 2025, 02:58:50 pm
content: Quote from: yalect on February 20, 2025, 01:40:17 pmThe three circuits shown with only a capacitor connected to the base areabsurd.A bipolar transistor's base will always draw DC current: how will it flow through that capacitor and what base voltage will result?The article only discusses how to couple the emitter to a load.

author: yalect
author_rank: Regular Contributor
timestamp: February 20, 2025, 10:05:30 pm
content: Hi, thank you for your reply and bieng interested.that circuit, it'scalled emitter followeralso.The circuit of anemitter follower, circuit that is shown on attached file.The main difference between an emitter follower and conventional amplifier is the absence of collector load and emitter bypass capacitor. The emitter resistance RE itself acts as the load and the ac output voltage Vout is taken across it. The biasing is provided either by base resistor method or potential divider method.The circuit consists of a resistance of the order of 10 kΩ in the emitter itself and the collector is connected directly to the supply. Stabilization of operating point is exceptionally good because of high value of emitter resistance RE.Operation of Emitter Follower Circuit:When input signal Vs is applied to the base, the resulting emitter current Ie develops an output voltage Vout equal to IeRE across the emitter resistance RE. This voltage opposes the ac signal voltage Vs as it is in phase opposition to Vs. Thus it provides negative current feedback. Moreover, this voltage (Vout) fed back to the input is proportional to the emitter current hence this circuit is called a negative current feedback circuit.the source :(from a website)thank you.

author: TimFox
author_rank: Super Contributor
timestamp: February 20, 2025, 10:24:46 pm
content: That reads like an AI summary of various statements about emitter followers.I contend that all your circuit drawings show incorrect biasing: you need to define the DC collector (and emitter) current when the input AC signal is zero to get a useful circuit.In your last posted circuit, add a resistor to ground from the base, which will establish the base voltage at a DC value between 0 and Vcc, which in turn will set the DC voltage across the emitter resistor, which determines the emitter current.At which point, the machine sentence "Stabilization of operating point is exceptionally good because of high value of emitter resistance RE" will make some sense.Once you have biased the circuit correctly, the output capacitor and load resistor will form a high-pass filter.

author: Zero999
author_rank: Super Contributor
timestamp: February 20, 2025, 10:32:48 pm
content: Quote from: yalect on February 20, 2025, 10:05:30 pmThe problem with relying on the base impedance for biasing a circuit is it's dependant on the load and transistor's hFEwhich is unpredictable. The hFEcan vary from device to device, even of the same type and is very temperature dependant.

author: Konkedout
author_rank: Frequent Contributor
timestamp: February 21, 2025, 03:15:34 am
content: Quote from: yalect on February 20, 2025, 10:05:30 pmYou need a garbage leaky capacitor to go along with the bad circuit design.As for the schematic in the original poster's post, the voltage at the base of the transistor will depend greatly on the transistor base current, as controlled by the transistor beta.

author: Analog Kid
author_rank: Super Contributor
timestamp: February 21, 2025, 03:58:18 am
content: Here's a properly biased emitter follower, from Horowitz & Hill'sThe Art of Electronics:

author: PCB.Wiz
author_rank: Super Contributor
timestamp: February 21, 2025, 08:30:14 am
content: Quote from: yalect on February 19, 2025, 10:10:46 pmRun it in spice, with a frequency sweep.Spice will also show you the signal clipping limits, your circuit has only a 1V supply, so can only manage low amplitudes before it clips.Spice will also show you the effect of R1,C2 that you forgot to ask about.

author: Zero999
author_rank: Super Contributor
timestamp: February 21, 2025, 11:26:13 am
content: The circuit will work, so long as the input signal is low and the hFEof the transistor isn't on the low side.I've simulated it in LTSpice.The default BF (hFE) for the 2N3904 model provided with LTSpice was on the high side (300), so I reduced it to 140, the average of the maximum and minimum figures provided on the data sheet at 1mA. I had to estimate the maximum by normalising to the maximum figure given at 10mA, as it's not given for 1mA.https://www.farnell.com/datasheets/1849425.pdfI also performed a DC operating point analysis with the hFEswept from 30 to 300, which resulted in VEbetween 12mV and 80mV.Emitter follower bad design.asc(1.93 kB - downloaded 20 times.)The effect of hFEon VETransient AnalysisBode PlotEmitter follower bad design.asc(1.93 kB - downloaded 20 times.)

author: TimFox
author_rank: Super Contributor
timestamp: February 21, 2025, 02:51:02 pm
content: I still don't recommend running that circuit with Vcc = +1 V, as drawn in the original drawing and the Spice simulation.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: February 21, 2025, 06:47:52 pm
content: Quote from: Zero999 on February 21, 2025, 11:26:13 amThe OP also mentions a more logical 10k re, which also works.QuoteThe ac coupling allows the base to go above 1V so the emitter voltage will be ~300mV and swing 200+ mV each way before it clips.A triangle drive in spice is a good way to check clipping.

author: PGPG
author_rank: Frequent Contributor
timestamp: February 21, 2025, 07:55:53 pm
content: Quote from: TimFox on February 21, 2025, 02:51:02 pmWhy not?Long, long time ago (I think about 1972) I got my first IC. It was ULY741 (=uA741). As all OpAms those times it was intended to be supplied by +15V/-15V.I had transistor voltmeter I have done previously with 4 germanium transistors supplied by single 4.5V battery that I was not satisfied with its parameters (when certain voltage was measured as one transistor was heat more then the second one pointer little shifted with time).I found that supplying this OpAmp from +4.5V/0V I am able to find about 1.3V of linear output range (I needed 1V as my 100uA pointer meter needed about 1V for full scale deflection.So I just did it and for many years I was using this voltmeter.I remember seeing in some magazine DCDC converter supplied from single 1.5V cell. If remember well little more than 1V was needed to start but than it operated down to 0.7V, I think.If someone wants to do experiments with working at very small volteges than why not?