Video amplifier with BJT transistors - Page 1
Forum section: electronics / projects

author: the_eagle
author_rank: Contributor
timestamp: February 17, 2017, 01:44:27 am
content: Hello,I would like to build a video amplifier with a gain of -25, an input impedance of 75 Ohm and an output impedance less than 1 kOhm.To create this amplifier I thought to put a common base configuration followed by a common emitter to invert the signal.The two transistors are BC237B and for each stage there is a voltage gain of 5.The input signal is 1Vpp and the output signal should be 25Vpp inverted.Doing basic calculations, I came up with this circuit however LTSpice simulation seems not to work. Where am I wrong?Thank you for your help.I attached the schematic and the PWL file for LTSpice, too.

author: Kleinstein
author_rank: Super Contributor
timestamp: February 17, 2017, 08:00:25 am
content: As shown the common base stage would have a small gain, more like 56 Ohms / 75 Ohms, thus less than 1. Also the signal is relative to the supply at the collector side - not a problem in the simulation with idea voltage sources, but is can be in real life.The second stage does not have a well defined gain - more like much more than 5. Also the DC bias looks like being rather high and it could this very well be in saturation.

author: orolo
author_rank: Frequent Contributor
timestamp: February 17, 2017, 11:50:21 am
content: Your first stage is not strictly common base, since there is a 10K resistor from base to ground. The input impedance is not 75 ohm, but closer to 10k. You should bypass that resistor with cap a big enough to be an adequate impedance at signal frequencies. The usual strategy would be to use a big cap (short at signal freq.) and depend on the input resistance of the emitter to match the 75 ohm signal, but that is unpractical here: to get 75 ohm input impedance, the quiescent collector current must be 0.3mA [r_pi = 26/Ic(mA) = 26/0.3 = 75 approx. ] Your input signal is 1Vpp @ 75 Ohm, that means input currents of 13mA peak, which are not by any means small signal compared to 0.3mA. Expect great distortion if you try this.I'm not sure a common base stage at this signal level is a good idea. To accomodate 13mA current swings, you'll need a quiescent current of at (the very) least five times that, 65mA. If you choose the emitter resistance to be 75 ohm, the minimum for a match, to get a gain of five you need 375 ohms at the collector. That means a quiescent voltage drop of 0.065*375 = 24.4V just at the collector. And 0.065*75 = 4.9V at the emitter, 24.4+4.9 is almost more than your 30V supply. You might try an active load at the collector to avoid the resistor drop problem, but at this point perhaps another different amplifier type would be simpler and more efficient. Anyway, I've not that much experience with common base amps, but I'm sure any of the experts will confirm if this is possible.Also, the 'uic' option in the simulator might confuse things. I think that, by default, ltspice will simulate starting from the DC operating point of your circuit. If you bypass that with the 'uic' option (or 'startup'), your circuit will take time to settle down, and you won't see the real transient behaviour of the stabilized amp.

author: MagicSmoker
author_rank: Super Contributor
timestamp: February 17, 2017, 12:45:40 pm
content: Quote from: the_eagle on February 17, 2017, 01:44:27 amSo far, so good...Quote from: the_eagle on February 17, 2017, 01:44:27 amOops, here's where you start to go off the rails... First: the CB stage should deliver most/all the voltage gain, while the CE stage should only act as a buffer (ie - current gain) and inverter.Second: the voltage gain of a CB stage is approximately Rc/Re, while the output and input impedances are set by each resistor, respectively. Therefore, Re needs to be 75R which means Rc needs to be 1875R to get a voltage gain of 25 at an input impedance of 75R.Third: the base of the CB amplifier needs to be biased properly for the expected emitter current (~6.7mA peak) andgrounded at signal frequencies!That means you need a DC bias network feeding the base which is bypassed to ground with a capacitor (rather than just a 10k resistor from base to ground).Fourth: coupling capacitors should be appropriate in value for the frequency range of interest; with 10uF LTSpice is probably spending the entire simulation time trying to charge them up! Try 100pF instead.Here's a good PDF on designing/biasing BJT amplifiers:http://www.seas.upenn.edu/~ese319/Lecture_Notes/Lec_9_CCandCBDesigns_08.pdf

author: the_eagle
author_rank: Contributor
timestamp: February 17, 2017, 11:08:11 pm
content: Thank you very much for your help.What do you think about this circuit? I tested it on the bench and it seems to be working however on LTSpice it's not.

author: Zero999
author_rank: Super Contributor
timestamp: February 17, 2017, 11:44:38 pm
content: Quote from: the_eagle on February 17, 2017, 11:08:11 pmYou've used a custom symbol for the voltage in that LTSpice file. Why not use the standard default symbol so anyone can open it?

author: the_eagle
author_rank: Contributor
timestamp: February 17, 2017, 11:46:45 pm
content: Quote from: Hero999 on February 17, 2017, 11:44:38 pmYes, sorry I put the PAL TV signal which is the zip in the first post. I am using it to check that relevant frequencies are not cut and there is the required voltage gain.

author: orolo
author_rank: Frequent Contributor
timestamp: February 18, 2017, 10:12:39 am
content: Ok, I see several problems.The input signal V1 in the simulation is not at 75 ohms. You should add 75 ohms of series resistance to V1.The input impedance of the amplifier is not 75 ohms. Since the common base stage has 16mA of quiescent current, the input impedance looking into the emitter is in the range of 1-2 ohms (disregarding parasitics). To match 75 ohms you may need to put a small resistor in series with the base capacitor. The idea is that the input impedance of a common base stage is the parallel combination of the emitter resistor and the input impedance of the transistor. It's not olny about emitter resistor, you can't forget about that second term.The result is that you have almost no input impedance, so your gain goes through the roof. Hence the clipping.Even at 16mA quiescent current, the 1.8k resistor sinks your collector voltage almost to -14V. That resistor is oversized. I don't understand why you need all the gain in the first stage.I have attached a modest gain (about 3x) but, I in principle, correct common base stage. Maybe from there you can try and crank up the gain to the point you need.

author: vk6zgo
author_rank: Super Contributor
timestamp: February 19, 2017, 12:12:43 am
content: Quote from: MagicSmoker on February 17, 2017, 12:45:40 pmThe frequency range of interest is from around 50Hz to 5MHz,so 10uF is quite appropriate.100pf would be useless.To the OP,---I suggest having a look on the 'Net for schematics ofrealvideo amplifiers, to point you in the correct direction.It may be more satisfying to produce a design from scratch, but you can waste a lot of time re-making the same mistakes others have already made.

author: MagicSmoker
author_rank: Super Contributor
timestamp: February 19, 2017, 11:44:45 am
content: Quote from: vk6zgo on February 19, 2017, 12:12:43 amI did not see where the OP specified such a wide frequency range, but that would explain why he persists in using such large coupling capacitor values. Still, if you use a 10uF (much less 100uF) coupling capacitor between amplifier stages in SPICE then expect it to take a really long time before it charges up to the bias point voltage; a 3ms transient simulation, like he specified, ain't gonna cover it.

author: Zero999
author_rank: Super Contributor
timestamp: February 20, 2017, 12:48:17 am
content: You won't be able to get that much power gain, at such a wide bandwidth, using just two general purpose transistors.I'd recommend a current feedback topology. Here's something I threw together in LTSpice. No doubt it can be improved.