Nice, thanks...

Now I have a few theoretical questions to answer that I hope you can help me with...

1 - Could we use a BJT instead of a MOSFET and get the same or identical results????
I would say No.

2 - What should we do to invert the Vd waveform? (We could use a simulation here to prove it)
You could arrange that the capacitor charges up while the switch is closed or instead of an 'enhancement' FET use a 'depletion' FET

3 - Could we use this circuit to control a DC engine/motor of 12V and 2A??? Explain your answer. If not, what should we change in the circuit to be able to control that engine/motor???
Check the Ids and Pdis rating of the BS170.

4 - How can we use this circuit to measure the capacitor value??? (assuming the value requested is the capacitance).

Use the formula for a capacitor discharge rate, you know the Rg = 1 meg and C is currently a 10uF... what do you calculate the discharge time to be for that combination.?
Measure the time to discharge an 'unknown' capacitor value..and use the same formula, with the 'new' discharge time.

Thanks

hi,
Lets see your revised answers.
E

 

Nice, thanks...

Now I have a few theoretical questions to answer that I hope you can help me with...

1 - Could we use a BJT instead of a MOSFET and get the same or identical results????

No, we couldn't use a BJT because as the Collector current at a BJT is controlled by the Base current, and in our circuit this current, so called Gate current, is so small that we would need a huge Hfe in a BJT to get the same results. -> Is the answer better now?

2 - What should we do to invert the Vd waveform? (We could use a simulation here to prove it)

We could use a depletion FET as this one would invert the circuit response! Another way could be to find a way, changing the circuit setup, to charge the capacitor while the switch was "OFF".

3 - Could we use this circuit to control a DC engine/motor of 12V and 2A??? Explain your answer. If not, what should we change in the circuit to be able to control that engine/motor???

Is Pdis the maximum power dissipation???? If so, I can see in the datasheet 830mW.

4 - How can we use this circuit to measure the capacitor value??? (assuming the value requested is the capacitance).

The discharge time would be ζ = R*C = 1E6*10E-6 = 10ms. To completely discharge would be 10ms times 5 which is 50ms.

Then we would need to measure it's discharge time and calculate C value back from tau's formula.

But in this case, the capacitor charges from the 1K resistor which is in series with the capacitor and I discharges by the 1Meg resistor. Is the discharge time formula the same as the charge time formula? I mean, when the capacitor charges from the 1K resistor, both, capacitor and resistor are in series, but when it discharges, they in a parallel setup, so I'm asking if the formulas are the same for series and parallel setups!
Thanks

I've answered in blue...

Thanks
Psy

 

I could use some help here! Hope anyone can!

I have now completed the answer to the 2nd question by changing the circuit as the one attached!

I need help to answer the 3rd question about the engine/motor!

 

For the other question I would answer like:

As the BS170 can only "deliver" 500mA of Id max peak, this circuit would not be able to control the engine/motor.
To do so, we could change the MOSFET and use another one with enough ID current to fulfill the engine/motor demands. With this change we would also need to do some arrangments so that the new circuit wopuld match the same response as with the BS170. We would need, for instance, to adjust the Vt by increasing/decreasing the C value or the Rg2 value.

I need some help here, please!

 

hi,
That looks OK to me.
You could consider using FET with a higher Vgs turn on voltage.

E

 

Morning... Eric, I need some help with my last circuit...

The goal was to invert the circuit response! To be honest, someone helped me designing that circuit. The fact is that I'm struggling to explain the changes.

The plot says me that at t=2s the pulse is OFF and the capacitor starts charging.
Then, when the Vg >= Vt the current starts flowing thru the MOSFET. At the end of the simulation, next to t=35s, the capacitor charge is close to it's max (10V from the Voltaqe source on the right).

Is this correct????

 

hi,
Before the switch is operated, the FET is conducting due the 1meg resistor to +10V, so both ends of the 10uF and the Gate are at +10v.

Operating the switch connects the Gate and the lower end of the 10uF to 0V , so the 10uF charges upto 10V and the FET is not conducting as the Vgs is below the turn on threshold voltage of the FET.

When the switch opens the 10uF starts to discharge thru the 1meg, the FET will not be conducting until the voltage across the 10uF falls to the Vgs threshold of the FET, at which point it will conduct.
The voltage on the 10uF will continue to decrease until it is fully discharged via the 1meg.

E

 

hi,
Before the switch is operated, the FET is conducting due the 1meg resistor to +10V, so both ends of the 10uF and the Gate are at +10v.

So, before we operate the switch, what is going on at the capacitor? Is it a short, is it like it is not there, is it charged to 10V or what?

Operating the switch connects the Gate and the lower end of the 10uF to 0V , so the 10uF charges upto 10V and the FET is not conducting as the Vgs is below the turn on threshold voltage of the FET.

The pulse I'm giving to the capacitor is enough to charge it up to it's max, right? I have calculated the time constant before, and it was 10ms so the capacitor will charge up to it's max in 5 times the constant time, which means that a pulse lasting 55ms should be more than enough to charge it up at 100%, right?

So shouldn't we see the Vg in the plot going up to 10V in that 55ms that the pulse lasts?

When the switch opens the 10uF starts to discharge thru the 1meg, the FET will not be conducting until the voltage across the 10uF falls to the Vgs threshold of the FET, at which point it will conduct.
The voltage on the 10uF will continue to decrease until it is fully discharged via the 1meg.
E

At this point, don't we have our circuit at the starting point where both the C and the Vg are connected to 10V???

 

So, before we operate the switch, what is going on at the capacitor? Is it a short, is it like it is not there, is it charged to 10V or what?
so both ends of the 10uF and the Gate are at +10v.

The pulse I'm giving to the capacitor is enough to charge it up to it's max, right? I have calculated the time constant before, and it was 10ms so the capacitor will charge up to it's max in 5 times the constant time, which means that a pulse lasting 55ms should be more than enough to charge it up at 100%, right?
Yes.
So shouldn't we see the Vg in the plot going up to 10V in that 55ms that the pulse lasts?
We can see if you expand the time scale.

At this point, don't we have our circuit at the starting point where both the C and the Vg are connected to 10V???

At the end of the charge pulse, the discharge time starts.

Two seconds after starting the discharge the Vgs threshold is reached and the FET conducts, the 10uF finally discharges to zero, ie: +10V at both ends.

 

Ok, I need to move to the next step that it an hard one, at least for me...

We were asked to build a DC engine speed and motion direction controller using a DC voltage source of 9V to control motion direction (0V spins to the right, 9V spins to the left) and a PWL voltage source to control speed.

I'm only trying to simulate the motion direction control with switches but in fact I don't have idea what resistor values I should use. I'm also using leds to check the direction of current flow but I can't get what I'm looking for!

I'm following this schematics but I can't make it work properly...

I'm using 4 switches, combined 2 by 2 in diagonal in the MOSFETS H bridge, 2 of them delayed by half period of the first so that we can see the correct LEDs light up at each half-cycle!

 

hi,
Redo the sim using 2 Pulse Voltage sources, one driving the top left and lower right FET's
and the other driving lower left and upper right FET's.

The switches are not necessary.

Increase the value of the resistors in series with the LED's.

E

 

hi,
Redo the sim using 2 Pulse Voltage sources, one driving the top left and lower right FET's
and the other driving lower left and upper right FET's.

The switches are not necessary.

Increase the value of the resistors in series with the LED's.

E

But to redo the sim to only2 Pulse Voltage sources I need to connect the Vpl_1 to Vpl_3 and the same to the other 2, right?


Edited;

I've tried this but no good!

 

But to redo the sim to only2 Pulse Voltage sources I need to connect the Vpl_1 to Vpl_3 and the same to the other 2, right?


Edited;

I've tried this but no good!

hi,
The two upper FETS are PMOS FETs' ,,, check from the datasheet the polarity of the Vgs that is required to turn ON these PMOSFET.

E

 

Morning...

Ok, I think I have the motion direction controlled...

Now I need to know how to control the speed. Can you help, Sir Eric?

 

hi Psy,
Look at this option, a practical circuit would require additional components.
Its NOT complete!.
E

EDIT:
Woops! Corrected an error in my timing.!... now OK.

 

Morning...

Ok, I think I have the motion direction controlled...

Now I need to know how to control the speed. Can you help, Sir Eric?

hi Psy,
In order to control the speed of a motor, the 'driver' either a PIC or MCU would have to use PWM, Pulse Width Modulation of the V1/V2 input signals.

E

EDIT:
BTW: have you measured the MOSFET current thru the FETs on your circuit... its BIG.!!!
You are shorting the FET's

 

I was trying to follow a schematic from the web. http://www.bristolwatch.com/ele/h_bridge.htm

Figure 4.

BTW, when I try to open you asc, it says it can't find the 2N7002...

 

I was trying to follow a schematic from the web. http://www.bristolwatch.com/ele/h_bridge.htm

Figure 4.

BTW, when I try to open you asc, it says it can't find the 2N7002...

OK, Use any suitable N MOSFET in the LTS library.

NOTE: My EDIT on Post #35

 

But what is wrong with my schematics? Are the MOSFETs also shorted in the web schematics I'm following?

Wasn't it possible to build this simulation with only 4 MOSFETs?

 

But what is wrong with my schematics? Are the MOSFETs also shorted in the web schematics I'm following?

Wasn't it possible to build this simulation with only 4 MOSFETs?

This clip is from your schematic link... note its says 'notted' for A A' B B' which either means a inverter stages or 4 drive sources,!

To fully turn on a Power FET a Vgs of ~10V is required, this is not possible with a direct drive from a MCU.

You can use FET's with a lower Vgs threshold that can be driven directly from a MCU.