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HomeElectronicsEliminating the diode voltage drop in a cost pump circuit

Eliminating the diode voltage drop in a cost pump circuit


These days microcontrollers and microprocessors are powered with a DC voltage of three.3V (even 2.5V, or 1.8V, and many others.). So, the well-known ahead voltage drop of diodes actually turns into a problem for a lot of digital circuits such because the AC/DC bridge, HF detector, and cost pump to call a couple of. Let’s speak about a traditional cost pump circuit: In a nutshell, it’s composed of two capacitors and two diodes, and the circuit is meant to double its enter voltage. In apply, we get hold of Vout=2*Vin–2*Vd (the place Vd=0.4V for Schottky diodes); so, as an illustration, if Vin=3.3V, we get hold of 5.8V as an alternative of 6.6V (a lack of about 12%). Since MOSFET expertise has tremendously enhanced, its value “collapses”, so it’s attention-grabbing to make use of MOSFETs as an alternative of diodes with a purpose to do away with the ahead voltage drop.

Wow the engineering world together with your distinctive design: Design Concepts Submission Information

The next circuit (Determine 1) makes use of two equivalent P-MOSFETs T1 and T2, the place the essential level is to decide on them with a really low RDS(ON) characteristic, to extend the general effectivity, e.g., IRF9310, IRF7410, AOD403, and many others. N-MOSFETs T3 and T4 are basic function low energy (e.g., BS170, 2N7000, and many others.) which drive T2 and T1 respectively.

Determine 1 Cost pump circuit with MOSFETs as an alternative of diodes.

Here’s a transient description of how this circuit works:

  • At energy ON, pin RA5 is configured as output LOW. Capacitor C1 begins to cost via the physique diode of T1 (similar factor for capacitor C2, via the physique diodes of T1 and T2). MOSFET T4 turns ON and drives T1 ON additionally, which displays RDS(ON) (therefore, capacitor C1 costs near Vdd however capacitor C2 costs to Vdd-0.6V). In the meantime, MOSFET driver T3 is OFF which retains T2 OFF additionally.
  • When pin RA5 outputs a HIGH stage, MOSFET T3 goes ON and drives T2 ON, then capacitor C2 costs as much as a worth very near 2* Vdd. In the meantime, T4 goes OFF, and drives T1 OFF (which retains C1 charged to Vdd and holds the drain potential worth of T2 to 2* Vdd).

The PWM sign from pin RA5 will need to have an obligation cycle of fifty%, however the frequency shouldn’t be essential—some are within the vary of kHz to tens of kHz (the upper the frequency, the decrease the capacitors C1 and C2). The values of resistors R1 and R2 are additionally not essential—about a whole bunch of kΩ if we choose to acquire a really low quiescent present, or about tens of kΩ if we choose to lower the swap time commutation for T1 and T2.

Hichem Benabadji is a freelancer and has a Grasp’s diploma in industrial and automated computing. 

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