Properly understanding BJT gain and bandwidth - Page 1
Forum section: electronics / projects

author: ksk
author_rank: Newbie
timestamp: February 13, 2021, 12:05:19 am
content: I'm working on a design based on an existing schematic, and everything makes sense to me until I start considering the bandwidth of some amplifiers/buffers implemented with BJTs.Hopefully attached is the image of the relevant part of the schematic.The transistor is a BC850CL, acting as a video output driver at VGA rate (approx. 25MHz pixel clock). The output signal on the right side goes straight to a 75Ω load.VCC is 5V and DAC inputs are 3.3V.The way I currently understand it is:1. The hFEfor this transistor is specified as between 400 and 800, so when operating in this arrangement I should be getting a current gain in this range (simulations agree).2. The current gain-bandwidth-product or fTis specified as 100MHz min. I assume the typ/max values won't be much higher (am I wrong?)3. Taking current gain as 400 and gain bandwidth product as 100MHz, I calculate a bandwidth of only 250kHz, about 50-100 times lower than needed for this signal!So as far as I'm concerned, this choice of transistor should not work in this circuit. But the thing is, it does. The schematic I'm looking at is from a device that was produced (in relatively small quantities) and managed perfectly clean looking VGA output, no indication of low bandwidth.So, what's happening here? As someone who primarily does digital stuff I've only recently wrapped my head around BJT operation and now I'm doubting myself. Is my understanding flawed? Do I suck at reading datasheets? Or is there something weirder going on?

author: bob91343
author_rank: Super Contributor
timestamp: February 13, 2021, 01:02:30 am
content: It's only an emitter follower. The bandwidth you are talking about is probably for common emitter operation.

author: penfold
author_rank: Frequent Contributor
timestamp: February 13, 2021, 01:32:47 am
content: The transition frequency (ft) is based on voltage gain. The current-gain parameters (beta, hFE etc) are not (for the analysis you need to do here) dependant on frequency. In your emitter follower, the voltage gain is considered to be 1 (since the AC voltage between base and collector is the same as the input voltage). If you look up about the hybrid-pi or t-model of the BJT it might help explain this conceptually and remember theres a base-resistance resistance.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 13, 2021, 01:42:46 am
content: How are you calculating bandwidth? With a source impedance of some hundreds ohms, and output impedance of some hundreds also (a bit lower), h_fe ~ 2 would seem to be adequate, which puts BW in the 50MHz range, or lower if limited by Ccb or other factors.Tim

author: ksk
author_rank: Newbie
timestamp: February 13, 2021, 02:01:30 am
content: Thanks all, looks like my very limited understanding didn't get me that far after all.Quote from: bob91343 on February 13, 2021, 01:02:30 amI clearly have a lot to learn!Quote from: penfold on February 13, 2021, 01:32:47 amOf course, I completely missed the fact that a lot of stuff would still be in terms of voltage despite BJTs being "current operated devices". The datasheet referring to it as "current-gain-bandwidth product" probably contributed, but now I realize current is probably a term in the product and not referring to "current gain". I also looked at the wrong points in my simulation and completely missed that the voltage gain was 1!Quote from: T3sl4co1l on February 13, 2021, 01:42:46 amClearly the wrong way.Since I don't need exact numbers, I just wanted to figure out "why it works", I guess I've got my answer for now. I'll keep reading up on this stuff, models etc. and try not to jump ahead like I have.

author: gnuarm
author_rank: Super Contributor
timestamp: February 13, 2021, 02:46:25 am
content: I'm surprised you aren't asking about the emitter resistors. Why the two and why 100 ohms when driving a 75 ohms cable? Inquiring minds want to know!I know part of the answer, but it's not completely clear to me.The two resistors form a Thevenin equivalent driver at half the voltage of the emitter and with a series resistor equivalent to the two 100 ohms in parallel, so 50 ohms.Why give up half the voltage using this circuit rather than just using a 1K ohm resistor to ground and driving the cable from the emitter? If that produces too much drive, the input to the transistor could lowered to set the scale factor correctly. I know in RF work the output of a transmitter is not matched to the impedance of the cable. What is important is to match the cable with a terminating impedance, but the source, not so much.I'm thinking the 50 ohm equivalent resistance protects the transistor in case of a short. But that can be mitigated with a resistor in the collector leg which drops enough voltage so when the load impedance drops significantly below 75 ohms, the collector voltage is too low to let more current pass.So why is it important to drive the cable with 50 ohms and why 50 ohms instead of 75?

author: Alex Eisenhut
author_rank: Super Contributor
timestamp: February 13, 2021, 03:04:24 am
content: If you have the time and inclination, read the General Electric Transistor Manual, like the 1964 edition. I'm sure you can find a scanned copy, and printed copies abound here and there.

author: penfold
author_rank: Frequent Contributor
timestamp: February 13, 2021, 04:12:53 am
content: Quote from: gnuarm on February 13, 2021, 02:46:25 amWell... I did notice that, but I assumed (a dangerous thing I know) that it had been designed around keeping the emitter current in the low 10's mA for the hFE and fT sweet spot - for a base voltage in the 2-4V range it should be about right... the two 100ohms would also then put "unloaded" output level in the 0.65V to 1.65V range and a 75 ohm load brings it to 0.3V to 1V range... so I figured it was correct for VGA signalling and my mental-ladder-network arithmetic is off tonight.