What does it mean when it says that d1 and d2 are forward biased so capacitor c3 starts charging with double the combined voltage of the power supply and capacitor c2?
"By giving DC voltage supply to Vin, Capacitor C2 starts charging through the buffer circuit created by the four NOT gate of the IC, C2 charge till the peak of the input voltage. Now the Capacitor C2 behaves as a second power source of Vin (3-15v). As shown in circuit diagram the D1 and D2 are forward biased so capacitor C3 starts charging with the double or combined voltage of the supply and capacitor C2. Hence, C3 charge with the combined value of voltage which is nearly the twice of Vin. Now we can get double voltage across capacitor C3 as output."
When the outputs of the gates are low, D1 is forward biased and D2 is reversed biased. When the outputs of the gates are high, D1 is reversed biased, and D2 is forward biased.
If the voltage across C3 is lower than Vin (in the beginning), both D1 and D2 will be forward biased, and C3 will charge up to Vin - 1.4V.
Southpark:
Does this circuit require negative supply voltage? ie. positive and negative rails.
No.
The four parallelled gates are essentially a switch that connect one side of C2 to ground or to Vin.
The two other gates are the oscillator that flip that switch at a high frequency.
When the 'switch' has connected C2 to ground, C2 is charged to the voltage of Vin through D1.
C3 is also charged to Vin through D1 and D2.
Next, the 'switch' connects the left side of C2 to Vin.
C2 is still charged, so the D2 side is now Vin plus the charge voltage of C2.
That higher voltage is only allowed to flow through D2 to C3 (D1 blocks that higher voltage).
C3 contains now a higher voltage. This 'pumping' process is repeated.
Leo..
@tjones9163, do you seriously want to make that circuit ?
The two gates on the left with C1 and R1 is the oscillator. You can use an Arduino for that as well. The tone() function will set a frequency on a output pin.
As PieterP wrote, the output voltage is 2 * Vin - 1.4. If you use 5V, then the output is 8.6V. The current output for these circuits is low. It is half (in theory) of what the driver (the four gates) can deliver.
C2 is a large capacitor with 220µF. When the frequency is higher, then C2 can perhaps be 100nF.
The four parallelled gates are essentially a switch that connect one side of C2 to ground or to Vin.
The two other gates are the oscillator that flip that switch at a high frequency.
When the 'switch' has connected C2 to ground, C2 is charged to the voltage of Vin through D1.
C3 is also charged to Vin through D1 and D2.
Next, the 'switch' connects the left side of C2 to Vin.
C2 is still charged, so the D2 side is now Vin plus the charge voltage of C2.
That higher voltage is only allowed to flow through D2 to C3 (D1 blocks that higher voltage).
C3 contains now a higher voltage. This 'pumping' process is repeated.
Leo..
Thanks for the explanation, I'm still trying to wrap my head around this. Is this statement on the right track?
In simple terms, the two gates on the left create an oscillation which connects the parallel gates to ground and voltage in, acting like a switch. When the gates are connected to ground the left side of the capacitor is negative and the right side of the capacitor is positive and charged to voltage in through D1. When the ocillation swithces the output of the gates are connected to voltage in, and now the left side of the capacitor is positive,( is that when c2 discharges, when the oscillation swithches???)Then c3 gets the voltage charge from the voltage in, and the discharge from the capactitor??
Is this semi correct?
thank you for your time and great explanation!
The four parallelled gates are essentially a switch that connect one side of C2 to ground or to Vin.
The two other gates are the oscillator that flip that switch at a high frequency.
When the 'switch' has connected C2 to ground, C2 is charged to the voltage of Vin through D1.
C3 is also charged to Vin through D1 and D2.
Next, the 'switch' connects the left side of C2 to Vin.
C2 is still charged, so the D2 side is now Vin plus the charge voltage of C2.
That higher voltage is only allowed to flow through D2 to C3 (D1 blocks that higher voltage).
C3 contains now a higher voltage. This 'pumping' process is repeated.
Leo..
The trick to understanding this is that:-
When end A of a capacitor is at 0V and end B is at 5V the capacitor is charged. Then if end A is suddenly switched to be at 5V then end B has to go to 10V because instantaneously the charge on the capacitor stays the same.
