I'm trying to drive a 10ohm load with 20Volts using a MOSFET. I want the load to be grounded, so My Vs would be around 20volts, so I need my gate voltage to be at least 25Volts or more. So I got a ISL6801 High side mosfet driver, hooked up everything as depicted in the datasheet.
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But I'm not getting a output voltage that is 25Volts, instead below
20Volts. What am I doing wrong? I hooked up a function generator with 5Vpp with frequency changing from
1Hz to 1kHz. I'm using Vcc of 5Volts, which is connected to VB via diode. 22uF capacitor between VB and VS.
i'm not sure if I'm not understanding the circuit very well, or if there is something that I didn't do. Help!
It's a "bootstrap" part. That means that when the output is low, a capacitor is charged, and when it switches high, the "bottom" of the capacitor is carried up...
If you try to hold the output high for a long time, the capacitor will discharge (e.g., through leakage of the diode used to charge the capacitor. The part is intended for use in something like a half-bridge circuit that switches frequently--fast enough to keep the cap charged.
Does that help?
Cheers, Tom
kyuj> I'm trying to drive a 10ohm load with 20Volts using a MOSFET.
Tom is right. I haven't checked out the linked datasheet, but if you want to keep the output high for any length of time, you'll need to use a combination of a bootstrap capacitor and something along the lines of a charge-pump or other supply that goes about 10V above the source voltage. The bootstrap is needed for quickly charging the MOSFET's gate capacitance and the charge-pump will then supply enough current to keep the output high. As a reference/learning tool, check out the datasheet for the Intersil/Harris HIP4080/HIP4081 or similar. Even if you don't need these chips, the datasheets are very informative on this subject.
sorry, that component is design to be pulsed or gated for a short time. what your seeing is the effects of Fet biasing after the cap internally has charged/discharged. have you thought of about a Pmos or reconfiging your app ?
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Real Programmers Do things like this.
http://webpages.charter.net/jamie_5
I'd be cautious about exceeding the GS max voltage, with a 20V supply; and though it's not difficult to do a level shift (e.g. with an NPN with base tied to the logic supply and emitter driven from the logic source through a resistor), it's not quite as straightforward as "high side MOSFET gate driver with logic-level input," at least if you're not accustomed to thinking in terms of transistor amplifier circuits. Depends too on how fast it must act.
Another alternative is an optoisolator of the sort that has photovoltaic output and can drive a low current to a few volts. By itself, that's not going to drive a FET gate very fast, but in conjunction with a storage cap and something like the "bootstrap high side driver" (now fed high-side gate drive power from the opto output), it should work. It's a takeoff on the theme mentioned by other posters of a simple switching supply to get the gate voltage.
In this application, since the current is fairly low, a P-channel MOSFET or PNP transistor isn't too bad an idea. Usually, as high-side drivers, P-channel MOSFETs are avoided because the on resistance, (Rds on) is higher than in an N-channel device, resulting in a higher power dissipation due to the higher voltage drop across the device. Also, in high current apps, transistors tend to have higher dissipation than MOSFETs and so are usually avoided. In this case, even if you used a P-channel MOSFET with an 'Rds on' of 0.1 ohm, you'd only drop about 100mV, resulting in a device power dissipation of 200mW. You shouldn't even need a heatsink for this. (I'm assuming that your load is resistive and not inductive) What type of load are you driving?
I need to use N-channel mosfet, because actual application uses 9Amps of current, and 60volts. I was using 20Volts just to test to see how the high side mosfet driver works.
I have 8 loads, which I will be multiplexing with 80ms period, meaning each load gets turned on for 10ms, every 80ms. (about 12Hz) I'm using PIC to do this.
My Question:
Where do I connect the charge pump? Do I connect it to the bootstrap power supply, which is connects to the diode? All the diagrams I saw seemed like I should connect the diode to the Vcc of the gate driver, which is 5Volts.
What do I use for the charge pump? Does the charge pump need the voltage that I will be driving the gate with? So does this mean I need an extra power supply with 70volts?
So I basically need to drive the gate to 70volts using a 5 volt output from PIC.
My actual application is more complicated than what I've described. For people who are interested, here it is.
------------------------------------------------------------------------------------------------ My actual application uses 9 Amps to drive multiple resistive loads(artificial muscle actuator) that are connected in series via multiple of IRF8901 MOSFETS, which are surface mount N-mosfets. I have 8 of these, that I will be driving at the same time. So I use PIC to program (Please don't ask why I need to connect the loads with switches. Believe me, I need to.) The total load requires about 60Volts to get 9Amps. So the actual problem requires different gate voltages for each MOSFETS that drive the loads, since it cannot be over 20volts. For this, I decided to use a zener diode at every step where I need to get Vgs between 10Volts and 20Volts. I'm using zener diode because of space limitations. I don't have room to use high side mosfet driver for each MOSFETs.
If all the loads are in series as you have shown, just WHY do you need multiple MOSFETs? If you think you are going to have isolation between the loads when the MOSFETs are off, you are most likely fooling yourself because of the parasitic diode between source and drain in the MOSFETs.
If you indeed are turning each series string of loads (and there are eight of them) on for 10 milliseconds out of each 80 milliseconds, and if the loads pull down to "ground" when off, then the bootstrapped driver you first posted about should work just fine. You only need to use a storage capacitor large enough to hold its charge against whatever leakage currents you have, for only 10 milliseconds. I gather the supply current for the output stage is on the order of a milliamp. Since I=C*dv/dt, to limit the voltage sag in 10 milliseconds to, say, 1 volt, and assuming 1mA, you'd need about 10uF of capacitance. That's the capacitor between Vb and Vs in the data sheet diagram for the part you suggested.
With that arrangement, you should not need an isolated supply; the little SO-8 driver plus the bootstrap capacitor and the diode from the bootstrap supply (presumably about 10 volts) should be enough, and you can drive the input nicely from logic levels. That keeps everything simple, no?
Just DO NOT expect that it will hold the FET ON for much longer than 10 milliseconds if you use 10uF for the bootstrap capacitor. Yeah, it may hold it for 100 milliseconds, but not for seconds. For testing you could use an even larger bootstrap capacitor (which should be fine with a voltage rating twice the bootstrap supply voltage--twice is just a safety margin), and have longer "hold" times.
***The "Typical Application Block Diagram" that I saw in the ISL6801 datasheet does not connect the bootstrap supply to Vcc, it is a separate supply here, connected to pin 8, and the diode is connected to that. Take another look at the datasheet. This is probably irrelevant anyway, since, as Tom points out, if you're only turning the load on for 10mS, you can probably do without a charge-pump type supply.
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*** I must confess that I didn't look at your diagram until now. Are these really connected as depicted, with no ground path for each? A clearer diagram would help.
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