This is my first post here! I am designing (actually adapting an existing design) a regulator for an automobile generator (not alternator!). THe control circuitry results in a PWM signal. The original design needed low-side switching. For my generator, I only have access to the high side, so I need to switch the high side.

I am attaching a JPG of my current design. I have tested it in Linear Tech's SWCad, and it seems OK to me, but I know nothing. I thought I would run it by the experts here. I am not averse to adding some components to improve reliability. I am open to completely scrapping this design and using something else!!!

The original design used a push-pull pair of transistors driving an N-channel MOSFET. I was not too sure why the push-pull was advantageous, but I needed to reverse the logic on the first stage output to accomodate the P-channel MOSFET. I thought that using an N-Channel first stage would work well.

Any thoughts/suggestions?

-Tony

 

The BS170 Mosfet needs a gate voltage of 10V to turn on. You have a voltage divider that is not wanted at its gate. None will turn on when the input pulses are only 1.8V without the voltage divider. Only a few in a bucket-full will turn on when the gate voltage is 5V.

Push-pull transistors are used to quickly charge the high gate capacitance of Mosfets. Get rid of the 100nF capacitor at the gate of the P-channel Mosfet because it slows down its switching so that it will spend a lot of time as a slow linear resistor and getting hot.

The 1k resistor will spend a long time discharging the gate capacitance of the P-channel Mosfet and the Mosfet will get hot if the switching frequency is high.
Use a lower resistor value to speed the switching and keep the Mosfet cooler.

 

Thank yo so much for your reply. I appreciate your time and effort. I guess there is something very basic about MOSFETS that I don't understand. I have been a little concerned that there is something I am missing when I see stuff about the need for charge pumps, etc.

When I looked at the datasheet for the ON Semiconductor BS170, I see that see that the Gate Threshold Voltage is listed as 2 volts nominally. I thought that this was the gate voltage (source = 0 volts) that would get the device to start conducting.

I used the voltage divider to get the low end of the PWM square wave (100Hz) comfortably below 2v.

What specifications should I look at in the datasheed to determine the turn-on voltage of a MOSFET? Incidentally mt LTSpice program seems to think that an N-channel MOS pretty similar (same Vto) seems to switch on at 2 volts....

But maybe I should use continue to use the push-pull transistors to feed the P-channel FET, to improve the ability to charge the gate compared to an N-MOS? Is there a way to re-orient the push-pull transistors to reverse the logic on the signal, or do I need to insert an NPN transistor ahead of the push-pull?

Here is a circuit with the original push-pull transistors with an NPN inverter ahead of it. Is this superior? Any other suggestions? I changed the power MOSFET device to one that is in my simulator. But I intend to use the IRF5210PbF if it is OK.

-Tony

 

Tony,

A push-pull arrangement is a better configuration because it can equally sink and source large amounts of current. If you treat the gate to source junction as a capacitance, you should be able to understand that you need a high amount of current to charge and discharge this capacitive junction. The less time in the linear region between ground and Vthres, the better. If you spend too much time in the device's linear region, you will have great power losses.

Don't worry about limiting voltages to this junction, just as long as it is below the maximum recommended Vgs.

Steve

 

Tony,

A push-pull arrangement is a better configuration because it can equally sink and source large amounts of current. If you treat the gate to source junction as a capacitance, you should be able to understand that you need a high amount of current to charge and discharge this capacitive junction. The less time in the linear region between ground and Vthres, the better. If you spend too much time in the device's linear region, you will have great power losses.

Don't worry about limiting voltages to this junction, just as long as it is below the maximum recommended Vgs.

Steve

OK, It is easy to use the push-pull transistors. When I ran the spice simulation as I drew it in the previous schematic, I found that the fact that the input to the first NPN transistor never got down to zero meant that the voltage going to the push-pull transistors did not vary much. It looks as if I need to reverse the logic on the Op-Amp that is creating the signal I am using. Then I can feed that directly to the push-pull transistors.

This solution ought to work pretty well.

-Tony

 

OK, I tested the circuit in my Spice simulator. It works well with an N-MOS driving the power P-MOS.

It also works well with the N-MOS driving the push-pull transistors which then drive the P-MOS.

WHen I looked at the switching of the P-MOS, the rise time was as close to identical as I could perceive, within the limits of the simulator.

It looks as if I can use just the simple N and P MOSFET combo and avoid the push-pull transistors.



This leaves one last question: am I misunderstanding the meaning of the Gate Threshold Voltage? Isn't this the voltage (above Supply) that it takes to switch a MOSFET on? In the cas if a BS-170, this is specified as 2.0V. Doesn't this mean that when the gate exceeds 2V it will begin to conduct?

-Tony

 

The threshold is where it just starts to conduct. It is measured relative to the voltage on the source. The FET isn't on until the gate is 10 volts above the source.

 

The threshold is where it just starts to conduct. It is measured relative to the voltage on the source. The FET isn't on until the gate is 10 volts above the source.

Ah, I understand. "Starting to be On" vs "Fully On".

Is the 10V value universal, or is it particular to this device? Where in the specifications do I find the "full on" gate voltage?

-Tony

 

Look for the ON RESISTANCE figure. It is always given for the device turned full on, and Vgs is stated.

 

OK, I see the Rds(on) which is typ. 1.2 Ohms.

And it specifies that Vgs = 10V. OK, Now I know where the 10V comes from.

Thanks!

-Tony