Need Help!- psu problems low current??

[dolomitedave]

Hi,

i have built myself a PSU, control and dual H-Bridge driver for my stepper X 1 so far, connected it all together and it works but the stepper does not seem to draw the current i was expecting it to and there is little torque compared to me testing this stepper on a bench PSU. please see attached schematic for the H-Bridge and control. i am stepping it from a 555 timer for now just to test it.

The PSU consists of toroidal transformer 100VA (only using 1 tap 18V AC) - bridge rectifier - smoothing cap - LM7812 for logic and LM317 with a 1R resistor between adj and output pins to make it a constant current supply of 1.25A

I was under the impression as the motors are rated at 1.7A they would still work. I know with no load they will use less but when its under load it's still the same low. it only draws 150mA under load.

could someone point me in the right direction with this please, i know there are commercial products that will do what i want but i want the pleasure and experience of building my self.

Thanks in advance for your help.

[john_100]

looking at your circuit both coils are on all the time
the direction of the current is reversed in each coil on alternate steps

so the current in each coils is 1.25/2 = 0.625A

this current may be reduced if the coil and transistor on resistance is more
than 24v/0.625A = 38 ohm

looking closer at your circuit the upper transistors in the bridge
may not be turned fully on

you need to add an extra circuit to shift the logic level from 5v or 18v(TTL or CMOSlogic levels) to what ever the constant current circuit output voltage is

as it is ,if the voltage out of the constant current supply is more than the logic 1 level

the emitters of Q5 ,8 , 13, 16 will be about logic 1 level -1.2v !
ie for CMOS logic run of 12v the emitters will be 12v - 1.2v = +10.8v max

John

[dolomitedave]

John,

Thanks for your help but i am a little confused i was using the image attached as a guide (this is from the datasheet for my steppers) which shows both windings on at the same time or have missunderstood you. also i may have misslead you but i have 2 supplies 1 at fixed 12V for logic the other at fixed current for the steppers only.

Dave

[john_100]

Hi Dave,

your circuit is as I expected , with CMOS logic levels (I just chose a different supply voltage)

If you look at the darlington transistors (Q1 +Q5 ,Q4+Q8, Q9+Q13,Q12+Q16) in the upper half of the H bridge circuit they will act as emitter followers

the diagram shows 2 possible mods, using either opto isolators or 4 pnp transistors

with the simple full step circuit the current from a single constant current supply is equally divided between both coils

with a half step drive to the coils would alternate between all and half the supply current unless each bridge had its own constant current supply

have alook at this reply to another post - half /full step circuit :-





http://www.cnczone.com/forums/genera...ar_driver.html

you can use a 4066 to switch between full and half step instead of the links shown

John

PS as the step frequency is increased ,the inductance of the motor winding will have more of an effect ,
reducing the current more than expected ,if the upper half of the bridge is not being switched hard on

PPS if you drive the circuit directly from the TTL levels (eg printer port)
you may have to reduce the logic supply to 5v
if I remember correctly 0 to 30% v supply = logic 0 ,60 to 100 % vs = logic 1

[dolomitedave]

John,

Thanks for your help, It wasn't until i was explaining this to a friend, i had the light bulb moment. i see where i went wrong i guess it's an easy mistake to make.

in your 2nd drawing the pnp one, you have an extra npn transistor with 2 diodes from base to gnd. when i first saw it it looked like your inverting the signal which it doesn't need to. would you mind explaining this a little further for me. i think i know what your doing here but don't fully understand it.

sorry for my lack of knowledge, i am still learning

Thanks again

Dave

[john_100]

Hi Dave ,

if you have not all ready done so

connect the two motor windings directly to your constant current psu

if the current is now 1.25A note the voltage across the windings
13v or less will explain the problem

if the current is less than expected
the ic may need two capacitors adding (22- 100uF) if not all ready fitted
to decouple the input and output of the IC

John

[dolomitedave]

John,

Sorry i should have explained before i have replaced the 4 upper npn's of the Darlington pair with pnp's as advised by you. I am now getting the correct current through the motor windings. but the motors are not stepping now which is why i was sent the last post questioning what the extra transistor was for and 2 diodes.

Dave

[john_100]

Hi Dave

thats great , just what we want

you posted the last reply whileI was adding the capacitors (in red) to my diagram

the npn transistor , emitter resistor and diodes provide a switched constant current to switch on the pnp transistor

this fixes the base current to pnp transistor over a wide range of supply
voltage to H bridges


now the upper transistors are switching on
is the logic counting when clocked ( stepped) ?

and are is the outputs driving the correct two transistors?

if its the correct two transistors the motor should be locked

if they where both on the same side of the bridge
the motor will be free to rotate by hand
as they are short circuiting the supply (its good its current limited!)

John



ps I'll have another look at the logic circuit

[dolomitedave]

John,

I know its stepping correctly as i also have LED's on the logic output from when i was testing with no hbridge. i did check that the right transistors are being switched on at the correct time but maybe i need to double check this, this was the first thing i checked when it wasn't working.

Thanks again for your help and time on this.

[john_100]

Hi Dave ,

just had another look at the logic circuit again

the JK flipflops ( U1a and U1b 4027) the set and reset pins 4 ,7 ,9 and 13 should be tied to logic 0 ie pin 8 the negative supply

John


oops posts crossed again

[dolomitedave]

John,

sorry my fault again the diagram i posted was an old one when i made the board i couldn't get it working then realised i hadn't tied these pins low so modified the board and not the drawing. there are possibly other mistakes in there as well.

Dave

[john_100]

Hi Dave


just another diagram of the revised H bridge

the 2 diodes limit the voltage to the base of lower
npn transistors to 1.2v
a 600 ohm emitter will result in a collector current of 1mA
(600 ohm x 1mA = 0.6V , ie 1.2v less the base emitter voltage 0.6v)

this 1mA via a resistor of 4k7 to 10k switches on the pnp transistor
and npn power transistor combination

the 1K resistors shown in red can be added if the transistor don't turn
completely off when they warm up
as a result the 10K resistors connected to the base connections
will have to be reduced to 8K2 (or 6k8)

the resistor values may need changing depending on the transistors used


John

PS what have you used to build the test circuit
Vero board ?