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| N-channel Mosfets in a charge pump, replacing diodes |
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| TheBaconWizard:
Hi, I want to build a Dickson style charge-pump with several characteristics: 1) It is being driven by a clock signal but I would like to use higher voltages than those of the clock. 2) I would like to keep voltage drop and overall resistance as low as possible I am proposing to try the following, but would appreciate it if anyone can spot issues before I start breadboarding and letting magic-smoke out. Description: Clock signal drives a pair of BJTs such that a current at higher voltage flows into the caps of the charge-pump. Instead of diodes, I am electing to use N-channel mosfets with the body diode blocking reverse flow. The Gate is pinned to an acceptable voltage by means of a resistor-zener network. Only 2 stages are shown. I am planning to run this at 5Mhz. Many thanks :) |
| Zero999:
--- Quote from: TheBaconWizard on June 22, 2018, 07:56:52 am ---Hi, I want to build a Dickson style charge-pump with several characteristics: 1) It is being driven by a clock signal but I would like to use higher voltages than those of the clock. 2) I would like to keep voltage drop and overall resistance as low as possible I am proposing to try the following, but would appreciate it if anyone can spot issues before I start breadboarding and letting magic-smoke out. Description: Clock signal drives a pair of BJTs such that a current at higher voltage flows into the caps of the charge-pump. Instead of diodes, I am electing to use N-channel mosfets with the body diode blocking reverse flow. The Gate is pinned to an acceptable voltage by means of a resistor-zener network. Only 2 stages are shown. I am planning to run this at 5Mhz. Many thanks :) --- End quote --- That circuit won't work. The PNP transistor is backwards and the NPN transistor has no current limiting base resistor. A MOSFET driver could be used to level shift the clock signal to +Vcc. What's the value of +Vcc? A Dickson charge pump also requires two pulse trains, in anti-phase with one another. https://en.wikipedia.org/wiki/Voltage_multiplier#Dickson_charge_pump |
| TheBaconWizard:
Hi Hero999, Yeah, the PNP was a slip-up, I've corrected it thanks. As my clock signal happens to be both well within limits but also operates the NPN in saturation mode, no resistor is needed. With regards to needed 2 clock signals, I believe not, owing to the following tutorial from Dave: https://youtu.be/I4ED_8cuVTU Any issues with how the N-chan mosfets are utilized? |
| Zero999:
--- Quote from: TheBaconWizard on June 22, 2018, 08:37:36 am ---Hi Hero999, Yeah, the PNP was a slip-up, I've corrected it thanks. As my clock signal happens to be both well within limits but also operates the NPN in saturation mode, no resistor is needed. --- End quote --- It still won't work. Why do you believe the transistor being saturated means it doesn't require a base resistor? The only reason why it wouldn't need one is if the output resistance of the clock source is high enough to limit the current to a safe level. Now you have the PNP transistor smoking too. Even if you added base resistors, the circuit is still no good. The multiplier circuit needs a push-pull output for the clock signal. A PNP transistor connected to +Vcc, can only pull up, not down, so it can only charge the capacitors. --- Quote ---With regards to needed 2 clock signals, I believe not, owing to the following tutorial from Dave: https://youtu.be/I4ED_8cuVTU Any issues with how the N-chan mosfets are utilized? --- End quote --- I haven't seen that video and can't watch it at the moment, but if only one clock signal is used, you'll only get half the multiplication factor. |
| TheBaconWizard:
--- Quote from: Hero999 on June 22, 2018, 08:56:33 am ---Why do you believe the transistor being saturated means it doesn't require a base resistor? --- End quote --- I don't. As a current source, I believe then clock is within what the base can take AND still operates the BJT as a switch, however it's no trouble to throw a resistor between them. I'm perfectly happy to "only" charge the caps, since the load will discharge them. However, if using a 2nd clock signal would improve performance that's certainly something I want to look at doing. |
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