Why the output voltage amplitude is reducing as I increase the frequency - Page 1
Forum section: electronics / beginners

author: Aakash
author_rank: Contributor
timestamp: November 30, 2021, 08:41:34 pm
content: Can anyone let me know why the amplitude at Vo is reducing as I increase the frequency of the input signal generator. My expectation is to create a 1.7MHz Square waveform with 30V amplitude.As soon as the frequency is increased over few kHz, I see that amplitude of 30V starts decreasing.I'm using 2N3904 NPN Transistor and it is saying it has a bandwidth of 300MHz. Please help, what's wrong with this.

author: Benta
author_rank: Super Contributor
timestamp: November 30, 2021, 08:51:14 pm
content: It has a bandwidth of 300 MHz in linear operation.But you're switching it, so the relevant parameters are rise/fall/storage time, and they are significant. Period time at 1.7 MHz is ~590 ns. Work it out from there.A solution could be a two/three-diode Baker clamp.

author: Zero999
author_rank: Super Contributor
timestamp: November 30, 2021, 09:18:03 pm
content: It has a gain bandwidth product of 300MHz, which is the point when the current gain falls to 1. This doesn't necessarily mean it can't be used as an voltage amplifier above 300MHz. A BJT can still behave has a voltage amplifier, even if its hFEis below 1.The voltage gain of a circuit attached to the original post, will start to fall at a much lower frequency. If the transistor is saturated, it will take a finite amount of time for it to turn off and the Miller effect will cause a roll-off at a much lower frequency.

author: CaptDon
author_rank: Super Contributor
timestamp: November 30, 2021, 09:39:20 pm
content: 33k is RIDICULOUSLY high in value for anything above a couple of KHZ!!!! You need to come way down to 1K or even less to get the response you need. Where did you get that resistor value from??? Also, a square wave is a sine wave with all harmonics present, which in this case means if you want to see a square wave at 2Mhz you better have a circuit flat to about 20+ Mhz. Where do people come up with this stuff?

author: Zero999
author_rank: Super Contributor
timestamp: November 30, 2021, 10:00:56 pm
content: Swap the base and collector resistors round. Then it won't be driving it so hard into saturation. A 47pF capacitor in parallel with the base resistor will help to speed up the turn-off time.

author: David Hess
author_rank: Super Contributor
timestamp: December 01, 2021, 12:14:17 am
content: You have two problems:At 1.7 MHz the period is 588 nanoseconds so the roughly 200 nanosecond storage time becomes a problem and steps should be taken to keep the transistor out of saturation, like using Baker clamping. A small signal Schottky diode between the base and collector can handle this.The input, output, and reverse transfer capacitance are too high to yield the performance you need with a 33 kilohm load and 1 kilohm source. A collector current of 10 to 20 milliamps is more suitable so a maximum collector load resistance of 3 kilohms is more suitable. If that is not good enough, then a more complex circuit is required, like a differential pair or cascode or something more sophisticated.Note that you can get a lot more performance out of your example by putting about 22 picofarads cross the series base resistance, and this is a common feature in discrete bipolar switching circuits.

author: Aakash
author_rank: Contributor
timestamp: December 01, 2021, 03:11:57 pm
content: Thanks for your input, I chose 33k because earlier I had 1k which was burned out of more power. Hence I increased. I see that 1k works but can you let me know why we should not choose high resistor value here? appreciate your inputs.

author: Zero999
author_rank: Super Contributor
timestamp: December 01, 2021, 04:43:34 pm
content: Quote from: Aakash on December 01, 2021, 03:11:57 pmThe transistor has a parasitic capacitance, between the collector and emitter, which has to charge, via the collector resistor, when it switches off. It takes time to charge the capacitance and the lower value, the resistor, the quicker it will charge.Assuming the transistor gets turned on continuously, the power can be calculated using Ohm's law:P = V2/RV = 30VR = 1000RP = 302/1000 = 900/1000 = 0.9WHow big is your resistor? If it's only rated to ¼WP = 0.25WV = 10VRearrange for RP = V2/PP = 900/0.25 = 3600RSo try a 3k6, 3k9, 4k3, or 4k7 resistor.

author: Aakash
author_rank: Contributor
timestamp: December 01, 2021, 05:18:48 pm
content: You are right, I had used 1/4 W resistor.I appreciate your inputs