Recent content by jrive

I think the issue is with the material of the core. I did a quick calculation for the Resr using the following relation: $$\frac{u''}{u'}=\frac{Resr}{2*\pi*f*L}$$ using data from the MN60 ferrite material datasheet, u''=1800, u'=6000, f=200kHz, and the corresponding L as shown in the Ls-Rs...

No, @sophicentaur...I completely understand and appreciate all the questions and inputs.-- thank you. Here are two of the plots I generated with data from the LRC meter, same coil, with and without the ferrite. (Ls-Rs, Ls-Q). I'm also including the coil and the ferrite for reference...

I have an image I can share, but don't know how (the image icon wants a URL, so I can't navigate to my folder to get image)...It shows the Ls-Q of the inductor with and without the ferrite...the self resonance of the coils is around 840kHz.

Thanks for engaging in the discussion...I find it useful to discuss these things since almost always one of these questions leads to an answer (or at least deeper understanding on my part!). I don't think I need to do the Rs to Rp conversion. The meter is giving me Rs, and that is indeed how...

The RLC meter? It's a Keysight e4980a Precision LCR meter, 20-2MHz range. If it is Flux 3D, it is a 3 dimensional finite element analysis tool to model and simulate electromagnetics. Similar to Ansys.

Thanks Thanks for your response. The resistance (real part) is measuring 8.4kohms, not the reactance. The meter measures Ls-Rs (L and R in series model). The rise in R is kills the Q that I need (and have at lower frequencies). Yes, I believe the resistance is increasing due to the skin and...

Yes., that's 54ohms in series (Ls-Rs setting)...

edit: The key thing I am after is why the ferrite might be responsible for the increase in resistance I measure on the coil. Hello, i have a coil designed to be 2.2mH and54 ohms. When an MN60 ferrite (T-ish core) is used with it, the inductance rises to 6.1mH as expected, and the resistance...

Never mind, i figured it out...

Sorry, my latex is rusty and i can't figure out how to make it work. So, here it is directly Ipd/(R1*C1*s^2). In time domain, this results in t*Ipd/(R1*C1). My problem is I can't seem to get there via partial fraction expansion...

Hello, When evaluating the step response of a circuit, the resulting Laplace representation is: $\frac{I_{pd}}{s^2 C1 R1}$ If I look this up on a table of Laplace Transforms, this results in $\frac{I_{pd}*t}{C1 R1}$. However, I'm struggling to solve this via partial fraction expansion--is...

Yep, thanks @Baluncore. My concern was about the lack of voltage on the gate after that time. During the ON time of the top bjt, the gate of the FET has a steady voltage ( PWM -vbe (of the top BJT). During the ON time of the bottom BJT, the FET does not have a steady voltage driving the...

I respectfully disagree with this statement. When the PWM output goes low, there is no vbe drop across the lower bjt. The only current path during this time is the discharge of the gate capacitance of the FET...what happens after it discharges? Isn't the FET gate floating at this time until...

How does a low turn on the lower bjt? where is the current flowing from? the base is 0, and the emitter is floating (the upper transistor is off, and there is no current path (except from the gate capacitance of the MOSFET and leakage currents). In the best case, the Rgate you mention (which...

Is it that this topology assumes the drive (the pwm output) goes above and below ground?