in this video we're gonna talk about voltage multipliers a circuit that can increase the voltage of the input by some multiple so in this example what we have is a half wave of voltage doubler we have an AC signal at the input and we're gonna get a DC signal at the output we have two capacitors which we'll call c1 and c2 and we have two diodes d1 and d2 and the resistor is the load of resistor now let's talk about how this circuit works this circuit can increase the voltage so let's say the voltage of the input is 10 the voltage of the output if you don't take into account the voltage drop of the diodes will be approximately 20 volts however there's a trade off this circuit doesn't increase the energy of the system if you increase the voltage the current will decrease so at a hundred percent efficiency let's say the current was 10 amps the maximum current that you can get at the output will be 5 amps such that the power transferred is a hundred watts power is voltage times current so whenever a circuit increases the voltage without applying any input energy to the circuit the current is going to decrease because we don't have any external power sources that's going to increase the voltage the voltage is being increased by passive elements not by any active element so as the voltage goes up the current goes down now let's talk about how the circuit increases the voltage and to do that let's briefly review batteries let's say if we have two batteries connected in series let's say the first one is a 10 volt battery and the second one is a 25 volt battery what is the output voltage across those two batteries and now what's going to happen if we switch the polarity of the 10 volt battery what will be the output now so feel free to pause the video and think about that for a moment whenever you connect the negative terminal of one battery to the positive terminal of the other battery the voltages add up so the total voltage across these two batteries will be 35 now for the other one the two negative terminals are facing each other so these batteries they oppose each other this one is the stronger one so it's gonna win out the total voltage across these two batteries will be 15 volts it's 25 minus 10 so it's important to understand that whenever you have two batteries or two elements with a voltage across it if you have the positive side of one element attached to the negative side of the other element the voltage will add up it's going to increase and that's going to happen in this circuit during the negative half cycle of the AC signal the bottom will be positive the top part will be negative so conventional current will flow from the positive portion of the AC of signal through d1 because the d1 is going to be forward biased so d1 is on and it's gonna flow back to the negative terminal through c1 now really doesn't flow through c1 but what's really happening is c1 is being charged by the negative half cycle of the sine wave now this current won't be able to flow through c2 because the two will be in Reverse by small so that diode is off during the negative half cycle conventional current won't flow this way now keep in mind conventional current and the actual flow of electrons are opposite so in this circuit electrons are actually flown in this direction but many textbooks use the direction of conventional current so we're gonna go with that in this video but just remember electrons actually flow in the opposite direction so that's it for the first half cycle that is the negative cycle so once c1 has been charged this is now going to be positive and this is going to be negative now let's say that this sine wave has a peak voltage of 12 volts so that's the maximum voltage and let's say we're using a germanium diode with a voltage drop of 0.3 volts so 12 minus 0.3 that means that the maximum voltage across C 1 will be eleven point seven it's 12 minus the voltage drop of the diode so now that C one has been charged let's see what's going to happen during the positive half cycle of the sine wave so during the positive half cycle the polarity across the 12 volt AC sine wave is going to change the positive side will now be at the top and the negative sign will be at the bottom so notice the polarities of the AC signal and c1 positive to negative so like those two batteries in series the voltages are now additive so let's call this ground let's say this is at zero this will now be at 12 and at this point the potential will be twelve plus eleven point seven which means it's going to be twenty three point seven volts so that's the sum of the voltages of the AC power source and c1 during the positive half cycle so now let's talk about the direction of the current at this point so current is going to flow away from the positive terminal and current will also flow from c1 as well away from its positive terminal it won't be able to flow through d1 because d1 is in reverse bias mode so that diode is off now it will flow through d2 so d2 is on and that's going to give us a voltage drop of 0.3 volts so this is at 0 this is at 23.7 this will now be at 23 point four and then current is going to flow through c2 charging it so see two charges during the positive half cycle of the sine wave and it's going to charged to a potential of twenty three point four volts now some of the current will flow through the load resistor but the load of resistor has to have a high resistance such that it doesn't quickly discharge c2 so you want to have you want to make sure that c2 its capacitance it's high enough such that the RC constant with c2 and a load resistor it doesn't quickly discharge c2 if RL is too low this voltage may not be sustained but if RL is high where the rate at which the circuit charges c2 is greater than the rate at which c2 discharges through RL then this voltage will be maintained at twenty three point four so you Nate you need to make sure that you use the right capacitance value based on the load of the existence of the external circuit so that c2 can maintain its voltage at twenty three point four if you find that the voltage is not being maintained at twenty three point four just increase the capacitance of c2 until it is and also make sure you choose a capacitor with the appropriate maximum voltage rating you don't want to use a 16 volt electrolytic capacitor for c2 because it's gonna be damaged by the high potential across it so you may want to use a 48 volt capacitor or a capacitor that can handle 100 volts something that's higher than twenty three point four so you need to make sure you choose the appropriate elements so now that c2 has been charged to the twenty three point four volts that energy is now stored in a capacitor d2 will prevent c2 from discharging back into the circuit because it will be in reverse bias mode so c2 can only discharge through the load resistor and so that's how the circuit works that's how the voltage can be increased so for this particular example we had an input voltage of 12 volts of alternating current and we will now have an an output voltage of 23 point 4 volts DC so the output voltage is equal to twice the peak voltage minus 2 times the voltage drop of the diodes assuming that they have the same voltage drop so that's how the half wave a voltage doubler circuit works now let's talk about the signal at the output so let me just take a minute and delete a few things so at the input we have a sine wave at 12 volts and at the output we're gonna have a DC signal at 23 point 4 volts but it's not a perfect straight line we're gonna have a ripple voltage so it's gonna go up and down because the capacitor is being charged and discharged slightly now the amount of ripple voltage depends on the capacitance of c2 and the amount of current that the external circuit draws from c2 which can be controlled by the loading resistance so to decrease the ripple voltage what you can do is increase the capacitance of c2 or you can increase the resistance of the you can increase the load resistance of the external circuit so that less current is drawn from c2 so those are some factors in which you can reduce the ripple voltage so that you can get a smooth output DC voltage and that's basically it for this particular circuit