How to calculate Frequency from a capacitor and a resistor? - Electronic Circuits Projects Diagrams Free

andy257 · 14th January 2004, 06:11 PM

Hi Guys

I posted earlier about my project of building an inverter. Basically what i want to know is:

how do i calculte the frequency which the circuit runs at?
i have been told its to do with the resistors and capacitors.

what is the formula? i assume its one of the RC circuit formulas but i have done a few rough calculations and i am not getting the result i need.

I want my inverter to run at 50Hz like the UK mains power supply.

In the diagram does it mean each side should be set for 25Hz or each side 50Hz?

cheers guys, almost there i think

andy

crust · 14th January 2004, 07:57 PM

The period of oscillation is about 1.4*R*C. R is the 160 ohm and C is 68uF. Your period is about 57Hz for this circuit. If you wanted it to be closer to 50Hz, increase R to about 180 ohms or so.

*EDIT* The number should have been 1.4

andy257 · 14th January 2004, 08:21 PM

thank you so much, just what i wanted.

where the hell does the 1.6 come into it though? no wonder i was getting the wrong answer

cheers

andy

ljcox · 14th January 2004, 10:09 PM

There is a problem with your circuit. The reverse breakdown voltage of the Base/Emitter junction of the transistors will be about 5 volt. You need diodes (pointing downwards) in series with the emitters to prevent this.

I can't explain how the formula is derived in this forum since my computer skills don't extend to attaching diagrams etc, so if you would like to know, send me an email and I'll send you a diagram plus the maths. It is relatively easy.

Len

nanlen@bigpond.com[/quote]

crust · 14th January 2004, 11:15 PM

The total off time of the circuit (the sum of the off times for each q) is 0.693*R1*C1 + 0.693*R2*C2. In this case, your R1=R2, C1=C2, so it is about 1.4*R1*C1.

The charging time of the

The 0.693 is the amount of time to get 1/2 full charge given by ln(1/2)*R*C.

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14th January 2004, 06:11 PM	(permalink)
andy257	How to calculate Frequency from a capacitor and a resistor? Hi Guys I posted earlier about my project of building an inverter. Basically what i want to know is: how do i calculte the frequency which the circuit runs at? i have been told its to do with the resistors and capacitors. what is the formula? i assume its one of the RC circuit formulas but i have done a few rough calculations and i am not getting the result i need. I want my inverter to run at 50Hz like the UK mains power supply. In the diagram does it mean each side should be set for 25Hz or each side 50Hz? cheers guys, almost there i think andy

14th January 2004, 07:57 PM	(permalink)
crust	The period of oscillation is about 1.4RC. R is the 160 ohm and C is 68uF. Your period is about 57Hz for this circuit. If you wanted it to be closer to 50Hz, increase R to about 180 ohms or so. EDIT The number should have been 1.4

14th January 2004, 08:21 PM	(permalink)
andy257	thank you so much, just what i wanted. where the hell does the 1.6 come into it though? no wonder i was getting the wrong answer cheers andy

14th January 2004, 10:09 PM	(permalink)
ljcox	Andy There is a problem with your circuit. The reverse breakdown voltage of the Base/Emitter junction of the transistors will be about 5 volt. You need diodes (pointing downwards) in series with the emitters to prevent this. I can't explain how the formula is derived in this forum since my computer skills don't extend to attaching diagrams etc, so if you would like to know, send me an email and I'll send you a diagram plus the maths. It is relatively easy. Len nanlen@bigpond.com[/quote]

14th January 2004, 11:15 PM	(permalink)
crust	The total off time of the circuit (the sum of the off times for each q) is 0.693R1C1 + 0.693R2C2. In this case, your R1=R2, C1=C2, so it is about 1.4R1C1. The charging time of the The 0.693 is the amount of time to get 1/2 full charge given by ln(1/2)RC.