Recent content by ntetlow

Yes, I see it now, just a matter of shifting things round, I wondered if that would do it. Thanks for your help. Neil

V1 is 1 volt. it is used to measure the thevenin resistance. the screenshot shows the current The current is 151.51mA, dividing 1 by 151.51mA gives a resistance of 6.6 ohms.

Sorry, just realised that i've sent a different version of the ltspice, instead of 20ohms as above, please read 9 ohms(2 +3+4) making the thevenin resistance 6.6ohms

Thanks for the reply. Screenshot attached. Neil

Attached is an ltcpice net file. The thevenin resistance betweed the 3ohm and 6ohm resistor is 7.6 ohms. This works out as 3ohms plus the parallel value of 20||6 (4.6ohms). My problem is - Why is this the thevenin resistance, could someone please explain?

I mean that R1 is in parallel to R2 and RC is in parallel to RE whereas before R1 was in series with R2 and RC was in series with RE.

Hello, attached are two screenshots showing a common emitter and the same with source voltage removes to find the input impedance. How is it that resistors R1 and RC can be placed in parallel to work out the input and output impedances? Examples i have seen only show straightforward...

Attached is a screenshot showing my workings out so far. I cannot get the correct answer from these, however.

Attached is the ltspice schematic as per the netlist. I think it will be easier understanding if II is replaced by say SINE (0 1 1000).

attached is screenshot of the spice netlist for the schematics in the URL Under "ideal integrator" you will see RIN (sorry, I've called it R1). It is the resistance under the number 1 in the first screenshot that is not shown.

it's how you get to the equation V2/II= -K/(1+KSC) where I'm confused. Analysis gives II = V/R1 + (V1 - V2),S,C where V is the voltage over R1. Also V2 = V1.K. In order to get the right equation from the above I'd have to eliminate R1 which I cant.

Hello, attached is a screenshot showing a schematic of an ideal miller integrator. Can anyone describe how the formula for V2/II is arrived at, can't work it out thru nodal analysis myself. Thanks in advance.

Audi_amp_bjt_1 seems to have got rid of the spikes (thank you) but why are the base, collector and emitter currents of Q3 so ugly, is this normal? Also, the current shapes of Q4 and Q5 don't look the best either but are beautifully righted in Q6 and Q7, is this too normal? May I also draw your...

Thank you for looking and for your info though I'm going to have to look up 'dead band' and 'step' to try and understand what you mean. I got the schematic from http://www.ecircuitcenter.com/Circuits_Audio_Amp/Advanced_Amplifier/Advanced_Audio_Amplifier.htm which I would expect would work...

ok. thanks