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20v PWM motor control suggestions

20v PWM motor control suggestions

hi, i am currently working on a project that involves driving a DC brushed motor rated for 20V and around 6A (measured running with battery). i needed to run the motor off the mains as well as control the speed of the motor using PWM.

i have searched online for a while, and the closest thing i could find was which is another persons project that needed a 20v motor but only 2A. his PWM circuit worked for him, so i tried using it with a couple of changes, i am using a 20V wall wart to give power to the motor, and a 12V wall wart to power the PWM portion, i could not find the transistor he was using, i only have an MJE3055t on hand so i slapped that one on in hopes of it working.

i had a few problems at first that burned a 555 and a resistor, but i managed to find the faults and fixed them. the circuit now runs small DC motors and i seem to have control of the speed range, however with the big 20V motor connected the speed range is greatly reduced and the transistor heats up quite quickly. i am pretty sure i am using the wrong transistor for the job, so if anyone can suggest one that would work for this application, that would be great, any other advice would also be greatly appreciated. thank you

Replies

ScottWang

If you want to get a 20V dc for the motor, you may consider the Vac decrease to 16 Vac, Vout_dc = 16 Vac*1.414 = 22.6 Vdc.

And the circuit you may change to this way.

tsan

DC current gain of MJE3055T is low and it would be better to use darlington transistor or mosfet. DC current gain of MJE3055T is 20-100 at 4 A. 220 ohm base resistor is too big and MJE3055T can't turn fully on. That's why the speed range is reduced as too big part part of the voltage is across the transistor. Your MJE3055T can work if it's current gain happens to be 30 or more. NE555 can supply 200 mA max. If base resistor is selected so that base current is close to 200mA transistor losses are smaller and controlled speed range is increased. 200 mA x 30 = 6 A.

2STW100 is an example of darlington transistor (available at least on Mouser). Current gain is 500 minimum at 10 A.

ScottWang

To get an IRF540N is better than 3055.

IRF540N -- 33A, 100V, 0.040 Ohm, N-Channel, Power MOSFET.

Original Poster

so smaller resistor or change to IRF540? maybe ill try a 10ohm 5 watt since the 2 watt im using still gets kinda warm

+Scottwang i am not using 20VAC, the first pic is of the original where the original author was using a rectifier.

ScottWang

If you want to using the bjt as switch, you should calculate the current as :

Ic_3055 = 2A,

Ib_3055 = 2A/10 = 0.2A = 200 mA.

Although the NE555 can offer 200 mA, but for the power dissipation, so you can't use it to the full current as 200 mA, and you need to adding one stage npn as 2SC1384(Ic=1A) or some other bjt has higher Ic current to drive the Ib_3055, the 2SC1384 and 2N3055 can be connected as darlington pair.

If you want to using mosfet to drive the the fan, then you can adding a 100 Ω~1K resistor from Output(pin 3) of 555 to the Vgs of mosfet, and the s connecting to GND, and the d connecting to the fan.

Dodgydave

Ditch the 220R resistor and replace the transistor with a mosfet like IRF540, BUZ11

Darrell Teegarden

I simulated your circuit in SystemVision and it indeed has a constant 50% duty cycle for all pot settings. Should the timer, though, be configured so that the pot adjusts the duty cycle, so that the motor speed will change with the pot?

ScottWang

Oh, No, I posted the wrong circuit, that circuit is a 50%/50% duty cycles pulse generator, it was different from pwm controller, when you adjust the pot that it only can be adjust the frequency of pulse.

Pwm controller.

NE555 Pulse Generator and PWM controller circuit, it can be choose the function by switches.

Original Poster

Change the transistor. If you only have 3055 you can try changing resistor but better to buy irf540.

crutschow

Yes, use an N-MOSFET.

Easier to drive and has lower ON voltage for better efficiency.

Darrell Teegarden

Here's a SystemVision circuit configured for variable duty cycle, controlled by a potentiometer

Darrell Teegarden
ScottWang

The pin 8 should be a Vcc, why you connected to a pulse?

Darrell Teegarden

This simply models the initial turn-on transient for the system. Vcc starts at zero and quickly ramps up to the full 5 volts. The pulse "duration" is for the entire simulation.

It can be useful to start everything "off" when simulating circuits that are fundamentally oscillators.

ScottWang

You using the SystemVision to simulating the circuit online, but the method seems different with LTspice, the members here almost using the LTspice, this SystemVision is a new software to me.

Why not using a switch to 5V Vcc to replacing the the 5V pulse, does it has any good advantage for the simulation?

Darrell Teegarden