Current direction using transistors

Thread Starter

MrsssSu

Joined Sep 28, 2021
40
1635079895607.png1635079975404.png

Hi readers, I am trying to simulate current flow in both direction of the load by using 4 relays or transistors. For example, in this circuit, the load will have left current flow when 2 of the switch or transistors are on and 2 of the switch are off. The opposite way applies too if 2 of the other switch are on and other 2 are off.

I have lt spice schematic above. Any ideas, people :) ?

Please help improve my understanding by attaching a lt spice schematic. Above schematic seems not working
 

Irving

Joined Jan 30, 2016
2,308
1. Your LTSpice circuit doesn't reflect the example below it - the battery V11 is wired incorrectly. The +end of V11 goes to the drains of M2 & M3. The -end of V11 goes to the sources of M1 & M4 and GND. Note that the example below is upside down, the + side is towards you not at the top.
2.. The MOSFETs you've selected need more than 1 volt to turn them on. Try changing V7 - V10 to 10v

Hope that helps...
 

Ramussons

Joined May 3, 2013
1,098
The schematic is wrong.
Please take the photo as a reference.
The voltage source +ve should be common to M2 and M3
and the -ve to M4 and M5.
In the present schematic, M4 and M5 have no connection to main power source V11.

Connect M2 and M3 drain to V11 +ve.
Connect M4 and M5 source to V11 -ve.
 

Alec_t

Joined Sep 17, 2013
12,309
Here you go.
BridgeSim.jpg
Things to note:
1) To turn a MOSFET on it is the voltage beween gate and source which must exceecd the threshold voltage Vthr specified in the part's datasheet.
2) Vthr is the gate-source voltage at which the part only just begins to turn on, passing a few microamps. When the MOSFET is being used as a switch it must be turned fully on to prevent heating. That requires a gate-source voltage of about 10V usually, or >3V for parts specified as 'logic level'.
 

Attachments

Thread Starter

MrsssSu

Joined Sep 28, 2021
40
Here you go.
View attachment 250965
Things to note:
1) To turn a MOSFET on it is the voltage beween gate and source which must exceecd the threshold voltage Vthr specified in the part's datasheet.
2) Vthr is the gate-source voltage at which the part only just begins to turn on, passing a few microamps. When the MOSFET is being used as a switch it must be turned fully on to prevent heating. That requires a gate-source voltage of about 10V usually, or >3V for parts specified as 'logic level'.
Hi, thank you for correcting me !! Hope you have a nice day!!
 

Thread Starter

MrsssSu

Joined Sep 28, 2021
40
Here you go.
View attachment 250965
Things to note:
1) To turn a MOSFET on it is the voltage beween gate and source which must exceecd the threshold voltage Vthr specified in the part's datasheet.
2) Vthr is the gate-source voltage at which the part only just begins to turn on, passing a few microamps. When the MOSFET is being used as a switch it must be turned fully on to prevent heating. That requires a gate-source voltage of about 10V usually, or >3V for parts specified as 'logic level'.
Hi, I have a question regarding the voltage 10V that you used to turn on mosfet. For the top 10V, it does not have ground at negative terminal whereas for the bottom 10V, it has ground at neagtaive terminal. So, these 2 cannot come from the same voltage source right? How am I suppose to turn both of this 10V simultaneously since it only be achieved by a same voltage source? I can only think that you will need 2 batteries to used as 10V source to turn on mosfet. Thank you :)
 

Irving

Joined Jan 30, 2016
2,308
You need an isolated supply or a 'boost' supply which is built into the driver chip. Here is an example (one half). D1 and C1 generate a 15v supply relative to VS to drive the upper gate.


1635089415648.png

Or for smaller low-voltage/medium-power bridge circuits you can use P-channel MOSFETs at the top and a level-shifter. Here is one such example:

1635089112582.png
 

Thread Starter

MrsssSu

Joined Sep 28, 2021
40
You need an isolated supply or a 'boost' supply which is built into the driver chip. Here is an example (one half). D1 and C1 generate a 15v supply relative to VS to drive the upper gate.


View attachment 250978

Or for smaller low-voltage/medium-power bridge circuits you can use P-channel MOSFETs at the top and a level-shifter. Here is one such example:

View attachment 250977
Hi sir, thank you for your reply :)
1635090672246.png
What about if i build another seperate voltage divider circuit and feed into the gate of the mosfet? Is that a good and simple idea? Actually, I am planning to get the square wave pulses from NE 555 timer to turn the current left or right at a frequency (can turn a motor clockwise and anticlockwise periodically).
I simulated it in LT spice and it works perfectly fine :)
 

Irving

Joined Jan 30, 2016
2,308
That only works statically and for one 'pair'. It won't work dynamically across two pairs of MOSFETs. When turning on a MOSFET dynamically a large current pulse is needed to charge the intrinsic gate capacitance; a resistor network will be so slow your MOSFET will die quickly from overheating, if it turns on at all before its time to turn it off again... (and then turning off will be slow and it'll die from shoot-through!). A properly designed full-bridge is much more complex than the simplistic video you've watched...

Here's a dynamic simulation:
1635091980574.png

The turn-on pulse width is always slightly shorter than the turn off time to avoid upper and lower MOSFETs on the same side being on at the same time, else a large current flows through both - shoot through - which kills them pretty quickly.

You also need to drive the gate of M2 independently from M3 because there are other operational states where M2 and M3 are not both ON. For example, to apply motor braking you need to turn M3 and M4 on and M1 and M2 off.
 

Irving

Joined Jan 30, 2016
2,308
If you use P-MOSFETs for the two two transistors, then you don't need a voltage above 10V.
Grounding their gates will turn them on.
As per the second diagram in post #9. For low to medium power requirements, IMHO, its often the simplest/easiest approach (other than a dedicated IC).
 
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