Transistor convert pure square wave signal into shark fin signal

Question

I have a square wave generator that puts out roughly 200 kilohertz of pure square wave signal. When I give this square wave which is I am sure it is a square to the base of a transistor (NPN) the output turns into something that I would call shark fin wave. The rise has a curve but the fall is perfectly perpendicular. What am I doing wrong?

---

@Burglups "Welcome. A schematic could ..........." Thank you friend for your warm welcome. Here is the schematic of my entire project. It is going to be a soldering Iron . Thanks for taking time to reply to my post

@periblepsis "Hrmann, You have said what  entirely different? I can't tell......" Thanks friend for cool and nice sketches and clarifying points I really learned a lot from your sketches and commentary that was added.  I'll definitely benefit a lot from these lessons in the next steps of my project.

@Justme "What is the purpose of buffering the NE555...." The purpose of my original circuit is not only buffing the output but also reversing the duty cycle with off-cycle. I have this nice little pulse module N555 (see photos) and it allows me to adjust the frequency as well as the duty cycle. in my project though I need a lower duty cycle because I am working with higher voltage. Unfortunately when I try to reduce the duty cycle the model output becomes really unstable and one of the potentiometers gets a little warm. From the other side I have pretty much no limitation and I can increase the duty cycle to something like 90%. My idea was that in my original circuit, the R1 was gonna provide the voltage for the gate of ES170 MOSFET and when the transistor switches, it should ground the gate of the MOSFET and thus turning it off. The turn off part is working fine but turning on is kind of gradual (shark fin) which is gonna heat up the mosfet and probably kill the MOSFET. My project is building Weller soldering gun. Here are the materials that I have, the simplified diagram, and here are my findings so far:

• Lowering the duty cycle, since the input to the primary winding is relatively high (110v) is actually the solution (by my understanding) since the ferrite core reaches saturation fast and no need for longer duty cycle. -It somewhat works but the heat produced by the secondary winding is not even close to what I need
• the D1 (reverse induction protection) gets hot. Even 5A diod gets too hot, especially when there is no load on the secondary winding.
• The interesting part was that as soon as I short the secondary winding, the linear light dims a bit when I expectet more current go thru the circuit, -the primary winding is roughly 100 turn, but I have no way or knowledge to calculate the precise loops so I go but gut feeling

@ Michal Podmanický ..... for 200kHz I would use some fast switching signal transistor. Thanks friend! I will give it a shot and will let you know how it changed the results. . Cheers!!! @ D.A.S.'s  Friend Thanks!! But too technical for me! I appreciate if you could elaborate a bit or suggest some books etc, to help me understand those vccrccr.... stuff enter image description here

Answer 1

There is a trick to overdrive the base during the transitions (edges) with series capacitor, but not to drive hard (not use low base resistor) during levels to avoid saturation.

Putting the transistor out of hard saturation takes a lot of time , i.e. it slows down the speed/edges.

Btw, for 200kHz I would use some fast switching signal transistor.

Answer 2

There is nothing wrong, the shark fin wave shape is to be expected from that circuit.

The transistor base is driven hard with quite a lot of current into saturation when turning it on, so this makes fast falling edge.

But the transistor is slow to turn off from saturation, and the only thing pulling the output high is just a resistor, so expect to see an RC charging waveform.

The transistor is quite useless though. The NE555 outputs drivers are quite strong and it is a push pull output, capable of likely driving a better square wave directly to your load than the transistor.

Answer 3

All transistors have capacitance and probe capacitance doesn't help. While the Vbe is low the rise time of the initial slope matches RC and the slew rate reduces exponentially with voltage across the resistor. dVc/dt = Ic/C = V/RC.

Since the fall slew rate has a very low resistance that results in the rapid switch shorting out the Vce capacitance with a low T=RC. You won't be able to match Rce =Vce(sat)/Ic calculated in the daa sheet of a few ohms or less as a pull up for symmetrical square wave unless you reduce the supply voltage to prevent overheating parts but 50 Ohms pullup to 3.3V is 200 mW which on a 1/4 W resistor means that the Temp rise is 200/250 80% of the rated 125'C temp. rise.

A better solution is just use the totem pole BJT output and expect a 2V drop from Vcc for the high level or use the CMOS version depending on your speed expectations.

Answer 4

A 555 timer already has a fairly capable 2-quadrant output. One that is much better than your single bipolar with a resistor for its output. Partly for that reason, it wasn't clear to me where you are headed until you wrote in comments:

I am looking to get the same square signal but boosted to something like 15v to open the gate of a MOSFET fully. the 9volt VCC can be increased, up to 15volts.

That helped a lot.

If you don't mind that the output is inverted from the input, then the following behavioral circuit provides 2-quadrant output which can run easily at \$200\:\text{kHz}\$ with commonly found bipolars and will allow you to set \$V_{_\text{PP}}=15\:\text{V}\$ while keeping \$V_{_\text{CC}}=9\:\text{V}\$ (and lower, too, if this is a small battery that gradually declines in output voltage.)

^{simulate this circuit – Schematic created using CircuitLab}

Michal Podmanický's answer reminded me about the speed-ups. So this circuit includes them for both the high and low sides (\$R_7+C_1\$ on one side and \$R_8+C_2\$ on the other.)

\$Q_3\$ operates in (to my mind) cascode mode. \$Q_1\$ and \$Q_2\$ provide the 2-quadrant output drive.

\$R_1\$ and \$R_2\$ provide some modest management of short-circuit/shoot-through and emitter degeneration. (You may remove them and just use a wire if it works for you.)

Here's the part values for \$V_{_\text{PP}}=15\:\text{V}\$ and \$V_{_\text{CC}}=9\:\text{V}\$ with an output compliance of \$100\:\text{mA}\$:

^{simulate this circuit}

Here's the LTspice run for it, using two runs: (1) essentially unloaded; and (2) fully loaded:

I just picked the usual bipolars. Dissipation, even with some shoot-through, should be under \$200\:\text{mW}\$. Tolerable.

It appears to work reasonably well at \$200\:\text{kHz}\$.