Multiple R/C servo control

Another shameless plug: Look at my servo controller (which is an open design)

Regards, Andras Tantos

The below servo chip has up to 1us resolution if desired. I did some testing with a standard servo and with ~190 deg of rotation, I could only get ~425 descrete steps in the 190 deg rotation. It appears that a regular cheap servo will get little improvement from higher than 8 bit resolution at ~255 steps.

In my servo setup I use a 12v supply and a 5v regualtor with a diode on the ground lead. This jacks the output voltage of the regulator up to 5.7v to supply the servo (makes for pretty quick servo response). I tap off of this output thru another doide to supply the servo chip. This reduces the voltage to 5v. It also supplys some isolation for voltage dips due to the servo moving. A crude schematic at the bottom of the below page.

"Si Ballenger" wrote

Thanks for the link. Could you explain a bit more what function the diode is performing in your schematics? My knowledge in EE is very lame, so go slow :) What values are you using for the diodes?

Cheers

Padu

The diodes are the cheap small generic 1N914 type that have a .7v voltage drop across them. Used on the gound pin of the voltage regulator, they cause the voltage regulator output to increase by .7v, for an output of 5.7v instead of 5.0v. Used on the regulator output, it causes the 5.7v output to be reduced back to 5.0v for use by the servo chip. It also acts like an electrical check valve keeping the power stored in the down stream capacitor from trying to back feed the servo motors. The large capacitor is connected in parallel with the voltage input to the chip to act as a power backup if the starting current of the servos causes a big voltage dip on the regulator output. The capacitor temporarly powers the chip during the voltage dip. Voltage dips on the power to the servo chips can cause erratic operation of the chips.

"Si Ballenger" wrote

Yes, interesting. After I posted I went and try attaching a diode the the GND lead of a 3.3V voltage regulator. The voltage jumped from 3.29V to 3.97. The diode has a "1N4" and "14*" marked on it (where * stands for "I can't read it"). When I measure it using my dmm, it reads 580mV.

What I knew about diodes is that voltage will go through anode to cathode, but won't go the other way. What is the theory behind the voltade increase when you plug it to the ground lead of the vreg?

I'm just guessing, but I would think it changes a reference voltage in the regulator from 5.0v to 5.7v, and this is the voltage the regulator tries to maintain.

The regulator tries to maintain 5.0V between its ground pin and its output pin - it has no other external references (it does have an internal reference, so it can keep the ground and output pins 5.0V apart, but this reference is unchanged by the external diode). The ground pin of the regulator is 0.7V above circuit ground thanks to the diode, and the output pin is 5.0V above the ground pin, so the output pin is 5.7V above circuit ground.

Tim

"Tim Auton"

Nice explanation, thanks a lot. Now, at the risk of start being annoying, there's one final thing that is still unclear in my mind. I see that if I install a diode forward biased (is it the right way to say?) on a 5V line, there will be a small voltage drop on the other side of the diode. So why connecting it to the GND lead makes the voltage to actually raise?

Cheers

Padu

...

A 5V regulator maintains a 5V difference between its output leads, call them Vreg and Vgnd. The forward-biased diode maintains a 0.7V drop between its input anode and output cathode.

Consider this circuit diagram: Vin | Reg -> Vreg | (Vgnd) Diode | (0V) Earth Ground

The regulator tries to maintain Vreg-Vgnd=5V. The diode maintains Vgnd-0V=0.7V. Thus Vreg = 5V+Vgnd = 5V+0.7V = 5.7V.

Later, Daniel

You may like to have a look at

David - just pass> "Wayne Lundberg"

Also, have a look at the Silvertone site

David - try> Hi folks,