Voltage ramp

254

05:28:43

When the npn base has positive voltage, the transistor is switched on and the upper current mirror pushes its current down to ground. When the npn base voltage drops to 0V, the npn turns off, so the current mirror can charge the capacitor to 6V in 10ms with a linear voltage ramp. Then, when the npn base voltage raises again the capacitor voltage drops to 0V. 
Try to switch the collector of the npn on the 680k resistor and compare the linear current charging with the classic exponential RC envelope.

published 10 years ago

thebugger

10 years ago

Make the base resistor 105Mohm.

halvin

10 years ago

Actually, the voltage across the cap at discharge will never be zero (after the first discharge cycle) . The voltage across the cap will be determined by the pn junctions in the transistor .

Mellotron89

10 years ago

I think that when the npn is working as a closed switch, the cap is connected between the npn collector, which is now a short circuit between the current generator and ground, and the ground itself; so the voltage across the capacitor can be only 0 because it's connected to the ground with both its poles... It would be different if the cap was connected to the base-emitter junction of the transistor.

UdorriUdorri

10 years ago

There is a saturation voltage from emitter to collector, it might be as small as .2 volts or .1, but it is not a perfect short, better described as a near short circuit. Capacitor voltage wouldn't drop lower than that.

UdorriUdorri

10 years ago

Another way to think about this is if the voltage across the ce juntion was 0 it means that ZERO power is consumed across the junction. That would basically make it an ideal switch. Maybe times have changed, but I had a professor that preached that there is no such thing as an ideal switch. In a practical application though .1 or .2 volts is close enough to ground

Mellotron89

10 years ago

Thanks for your kind explanation!

UdorriUdorri

10 years ago

No problem, cool circuit. If you scale the time frame you can see that the voltage is actually 18.8 mV which probably is possible.

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