# How to lower 5v to 1.6v and 0.8v?

## Main Question or Discussion Point

Hey guys, ive been stuck on this for the past few hours and im about to lose it, what i am using is a usb interface that outputs 5v on each pin and when i activate one pin i need it to output 1.6v and when i activate ithe other i need it to output 0.8v, i attempted this using a voltage divider and got around 0.83v which was great but i couldnt get 1.6v at the same time, any help with this would be fantastic.

Thanks Scott

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How mich current do you need in each case? Can you use a resistor divider, or do you need to regulate it?

i dont belive the current matters, its actually for a analog stick, 0 volts is backwards, 0.8 is centered and 1.6v is forwards, i need to be able to change between the three, i cant figure it out. Thanks for the reply

If current doesn't matter you could use a 1k resistor in series with a 1k pot and adjust the pot for whatever voltage you need. It would give you a range between 0 and 2.5 volts.

Hey thanks for that, it has to be automatic though, so when i select pin one in hyperterminal it outputs 1.6v and pin two outputs 0.8v although they are both 5v by defualt

So everything has to go to one input on your analog stick?

If I'm understanding you correctly you can do it with two diodes, one 1k resistor, and one 1k pot. Connect the anode of one of the diodes to one of the USB outputs. Connect the anode of the other diode to the other USB output. Connect the 1k resistor between the cathodes of the two diodes. Connect one end of the pot to ground. Connect the other end to the cathode of the diode which is connected to the USB output you want to use for the 1.6 volt. Connect the middle tap of the pot to your analog stick.

You can adjust the pot for 0.8 volts or 1.6 volts, depending on which USB output you have activated. You only have to adjust the pot once. Having both USB outputs off will give you 0 volts.

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Hey TurtleMeister thanks alot for the reply, yes both outputs must go to the same connection im sorry if im being silly here or mabye not explaining it correctly, i dont understand how the circuit you gave me would change between 0.8v and 1.6v, with only adjusting once, wouldnt i have to adjust it each time between voltages? sorry if im being silly, i assumed i would need difffrent valued resistors or something. Thanks again

No, you only have to adjust once. The reason it works is that 0.8 is half the value of 1.6. And since the resistor and the pot have the same value, when you switch to the 0.8 USB output it will cut the current in half. So whatever voltage you have the pot set at, it will be halved also. Just use ohms law and you will see that it works. I've already calculated that you will need to set the pot at 320 ohms (as measured from tap to ground). With the 1.6 output active you will have a total resistance of 1k, so the current will be 5/1000=.005, and the voltage at the tap will be .005*320=1.6. When you swith to the 0.8 output you will have a total resistance of 2k, so the current will be 5/2000=.0025, and the voltage at the tap will be .0025*320=0.8.

Of course you could also use a 680 ohm resistor in series with a 320 ohm resistor in place of the pot and get the same results. I should also mention that this circuit assumes that the current draw from your analog stick will be negligible and that your USB interface can source 5ma. And of course the USB interface ground and the analog ground must be connected together.

ahh thats excellent, its kind of over my head though, should really stick to programming lol, i belive i have a few 680ohm and 320 ohm resistors althought i am unsure how to connect them, do i attatch the 680ohm to one 5v output and the 320ohm to the other 5v output and connect them both to the anloug stick?, the analoug stick wont be a problem with current draw, it was at first so i just removed it and am attatching directly onto the y axis point on the board, thanks again i appreciate your help.

On second thought, I would not recommend using the resistors in place of the pot. The reason is that I did not account for the voltage drop across the diodes, which would be somewhere around 0.7 volts and may vary depending on the diode. Because of this I cannot predict exactly what value resistors you would need to get the voltages you want. However, this would not be a problem using the pot because you can adjust it using a voltmeter. Make sure both diodes are of the same type.

ahh thats great, i wont be able to get to an electronics store for a few days and id love to get this complteted today, just now i am using a 3300 ohm resistor and a 660 ohm resistor which is giving me 0.8v i assumed if i connected both 5v pins anf activated one it would be 5v and give me 0.8v and if i activated both pins it would be 10v and give me 1.6v although the pins are both active and 5v each when i combine them im only getting 5v, im sure thats really basic stuff but it seems wrong lol

Yeah, connecting the two 5v together will still only give you 5v. By the way, it's not a good idea to connect two outputs together like that without using diodes. In some cases it could damage the outputs when one is high (5v) and the other is low (0v). That's the reason I used the diodes in my circuit. However, the interface could already have the diodes internally, but I had no way of knowing that. To be safe you should use the diodes.

ahh ok, thanks for all your help, guess ill have to go up to maplin :P

i found a few POT's although i dont have diodes, i figured id give it a bash anyway, i made a voltage divider using a 430 ohm resistor and a 680 ohm i then connected the POT, one output gave me 0.8v and the other was 1.8v which is great, the 0.8v has to be as close as but the 1.6v is a minumum up to about 1.8-2v, but when i connnected the same setup again to do the other axis the values all changed, i belive its because they are using the same ground perhaps?

Sorry, I'm not following that. You can change the resistance values in my circuit as long as you keep the pot value and the resistor value the same. 2k resistor and 2k pot, 5k resistor and 5k pot etc... However, I would recommend staying with the 1k values because if there is any appreciable current sink from your analog stick the lower values will work better.

sorry i explained it poorly, what my ideal solution would be is to create a voltage divider on each pin one that lowers 5v to 0.8v and one that lowers it to 1.6v, i can do that and it works great, i enable pin one and get 0.8v, i enable pin two and get 1.6v when they are seperate although because they are going to the same place the y axis, it seems that the resistance changes once i join them together although only one pin is active at a time, do you know any way around this?

Thanks again

Edit: ahhh i was just looking up the parts u suggested getting and relised that a diode stops current from flowing in a certian dirrection, i was able to pull four from an old control pad and it works fantasticlly, thanks soo much for your help.

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Congratulations. Glad you got it working.

Hey TurtleMeister sorry to bring this up again, the control pad was freaking out when i applied 1.9v that must be too high a voltage this this type of controller, i tried the setup you suggested today and i cant get it to work for some reason, i have a diode on each usb port, a 1k resistor between them and a POT connected to one of the outputs,when i set output one to 0.8v using the POT it works great, although when i change to out put two it dosent double to 1.6v it goes up to 0.92v, sorry for the trouble.

Are you using a 1k pot? If all components are correct then the most likely cause of this is that your analog device is sinking too much current for this circuit. You said in your earlier post that the current did not matter so I designed the circuit based on that. If the device requires current then I will need more information about the analog device.

You can use your voltmeter to determine if the analog device is sinking current or not. Check the voltage with the analog device connected, and then check it with the analog device disconnected. If the voltage changes then it is sinking current.

ahh im sorry, i was using a 10k POT, the voltage is the same when connected or disconnected from the controller when it is powered, although when the controller is disconnected from the console it drops to 0.72, if i were to get some POT's could i just connect a POT to each output and adjust each one?

vk6kro
Sound like there is voltage coming OUT of the controller.

You could try to measure if there is a voltage coming out of the controller when it is connected to the console. This is very important and it would upset the voltages supplied by Turtlemeister's very clever circuit.
Incidentally, you can't substitute parts that are 10 times the suggested value (or omit diodes) and then be surprised that it doesn't work.

You also need to measure the resistance into the controller where the voltage is being supplied to. If this is low enough (like less than 2000 ohms ) it will have an effect on the voltages you get.

If this is for an analog joystick, why do you need 0.8 volts and 1.6 volts from a USB port?

The pot must be 1k when using a 1k resistor. I realize you may not have the correct components on hand, but using trial and error is not a good way to go about doing this. It would serve you well if you would take some time to learn ohms law. It's not that difficult and you could have fun designing circuits like this on your own.
if i were to get some POT's could i just connect a POT to each output and adjust each one?
Yes, but the components values would probably be different.

Circuit with individual pot adjustments. This circuit assumes that both USB outputs will not be active at the same time. If that happens, the output to the controller will be higher than 1.6 volts. However, this circuit is not as critical on component values. If you don't have a 5k pot, you can substitute a 10k pot. The reason I selected a 5k pot for the 1.6 output was so that it's set-point at 1.6 volts would be around mid range.

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