A real working driver is needed for signal from 74AC14 to voltage multiplier - Page 1
Forum section: electronics / general

author: Rhouland
author_rank: Newbie
timestamp: September 27, 2025, 11:47:57 am
content: A real working driver is needed for signal from 74AC14 to voltage multiplier via driver.Supply 5 Volts from 278R05. on one element 74AC14 create generator 583Khz.Next 5 triggers connect in paralel for more current value.output signal need connect to driver (2 or 4 mosfets) on IRF7319 (P-channel and N-channel in one package).driver connect to voltage multiplier x6 VCC. Next connect to stabilizer L7812, and need output about 50mA current.need driver circuit with IRF7319.Not working driver sample I attach:this me circuit now, need simple working driver:

author: Kleinstein
author_rank: Super Contributor
timestamp: September 27, 2025, 12:54:12 pm
content: The describtion is a bit confusing. For 50 mA at 12 V one would not really need a very powerfull driver. The AC14, especially 4 in parallel should be powerfull enough to directly drive the gates of MOSFETs as driver. For finally 12 V one would not need a x6 mulitplier, more like x 3 or maybe x4.The other point is that for a bit more power a normal SMPS with magnetics is often better. A big problem with capacitor type multipliers is the large peak current and thus quite some ripple currents. So I would more look for a boost converter and maybe a LDO (not so much the old L7812 with a large drop out) behind.

author: Rhouland
author_rank: Newbie
timestamp: September 27, 2025, 01:07:03 pm
content: this circuit supply only 5Volts. 5 paralel inverters on 74ac14 not enough current for voltage multipier, need a driver.

author: mtwieg
author_rank: Frequent Contributor
timestamp: September 27, 2025, 02:15:25 pm
content: I think you forgot to explain what is "not working" about this circuit. Are you saying the 74AC14 are not strong enough to drive the IRF7319, or the IRF7319 is not strong enough to drive the multiplier...?

author: Rhouland
author_rank: Newbie
timestamp: September 27, 2025, 02:37:09 pm
content: this circuit - only sample. I don't have diver circuit. Need working driver. I having ic 74ac14 with supply 5 volts. and standard voltage multipier x6 with load stabilizer 7812. but current not enough. need driver, I have mosfets IRF7319. need draw circuit working driver.

author: TimFox
author_rank: Super Contributor
timestamp: September 27, 2025, 03:30:46 pm
content: Since the threshold gate voltage on the IRF7319 is only 1 V, you can connect the two devices in that package together as a simple CMOS inverter that can drive amperes into the load from a 5 V supply.

author: Rhouland
author_rank: Newbie
timestamp: September 27, 2025, 03:43:25 pm
content: now circuit:

author: Rhouland
author_rank: Newbie
timestamp: September 27, 2025, 04:15:09 pm
content: how connected mosfets? need ending draw me circuit.

author: TimFox
author_rank: Super Contributor
timestamp: September 28, 2025, 03:01:14 am
content: No time tonight for drawing.Two MOSFETs: connect two FET gates together and drive from CMOS output.Connect two drains together for high-current output.PMOS source to +5 V, NMOS source to ground.

author: Rhouland
author_rank: Newbie
timestamp: September 28, 2025, 07:46:01 am
content: Quote from: TimFox on September 28, 2025, 03:01:14 amif connect gates to one point - mosfets damages. not right.this circuits - not work:

author: mtwieg
author_rank: Frequent Contributor
timestamp: September 28, 2025, 02:03:47 pm
content: Quote from: Rhouland on September 27, 2025, 02:37:09 pmThe "driver" in your first schematic looked reasonable (though the 10K connecting to AO3416 should be from gate to source, not gate to drain). The resistor/diode networks should prevent cross conduction.

author: DavidAlfa
author_rank: Super Contributor
timestamp: September 28, 2025, 04:28:20 pm
content: Nothing makes sense...if you want help, take your time to properly show what you're after.But, your main issue is cross-conduction.Transistors don't turn off instantly, if you drive P+N fets with the save drive signal, they shoot-through (Cross-conduct) and burn.To prevent this, you need separate signals, adding a dead time, where both transistors are off, to let everything settle down before enabling the other.A charge pump need push-pull, and you need two separate signals with 180º offset.So, you need two stages of transistors, then add any number of multiplying stages.I've exaggerated the dead time effect by adding an external resistor not visible in the picture, showing how it would float durign the dead time..As you see, the on/off transition isn't instant.You can greatly simplify this with a Class-B amplifier using BJT transistors, they won't be as efficient, by they can't shoot-through as their emitters are joined.

author: TimFox
author_rank: Super Contributor
timestamp: September 28, 2025, 06:00:31 pm
content: I built and tested this circuit for a variable-voltage drive (varying Vdd to the circuit block) that I used to drive a few 100 mA.The two drain resistors were added to obtain approximately the same output resistance in the two states.This is essentially the same circuit used in 4000-series CMOS (except for the added drain resistors).There will be cross-conduction when the gate voltages are near the middle of the range, but none at full positive or zero drain voltage (wrt ground).The load was 25 ohms through a 220 uF aluminum capacitor, and the input is an appropriate 400 Hz square wave for that application.