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variable voltage source with enough current

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Stewie911

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variable voltage source

Hi

I need to build a variable voltage source that can deliver up to 1A current.
The voltage should be able to vary between 0V and 5V and it will be driven with a DAC.
The DAC can output up to 5V, but can't give a lot of current, so it will be connected to an OP-AMP.

I've heard of a design using a transistor and an OP-AMP, but it's not the one where you drive the transisor with the OP-AMP output.

Can someone please help me out?

PS: I'm working with high precision
 

After a quick google search with keywords 'switch', 'mode', 'power', 'supply', 'schematics', I found the following page :

**broken link removed**


This is the circuit you are looking for :
**broken link removed**

Here, 10K pot is used to adjust output voltage level. If you want a fixed output voltage leve, just connect the inverting input of the opamp to the output of the circuit, and non-inverting input to the zener for referece level.


Not that you don't need triple transistors to form a darlington, just do with the clasical two-transistor way.
 

Thanks for the circuit. Was looking for current sources on google. Never thought about a powersupply. I'm using a 12V battery, so I'll just have to adjust a bit.
 

Hi,

If on the low voltage side you can manage with 1.25V, you can get a nicely protected adjustable voltage output just by direcltly driving the Vadj. pin of LM317T regulator from DAC.

Regards,
Laktornics
 

Was thinking about the LM317, but I'm not sure if that voltage will work. I need the voltage to vary from the threshold voltage of any MOSFET to 5V. A general MOSFET's Vtn is about 1.7V, but I'm still trying to find out if some of them are lower than 1.25V.

What if I put two diodes in series with the output of the LM317? They will have a voltage drop of about 1.4V.
 

If you connect a diode to drop voltage, don't forget to connect a 10k resistor with it. If you don't put a resistor and get very very low current there will be no voltage drop on the diode.

You say you want to work with MOSFETs. MOSFETs drain a very little amount of current. You should be aware of this issue.
 

I want to characterize any kind of MOSFET. It's because of power MOSFETS that I need the high current.
 

If you guaranty that you will always drain high current then there is no need for that resistor. But as an precaution for the worst case you should better connect a resistor there. Why not, its small and cheap.
 

The problem is that the current will go as low as 100uA, and possibly lower. I'm plotting the current vs. Vds of the MOSFET. Voltage source connected to the drain.

I just saw that the LM317 has a quiescent current of about 1.3mA, so I don't think my design will work with the LM317.
 

Hi,
I still do not understand why you can not use two diodes and a bleeder resistor of say 2.2k to ground after the diodes, which will provide the necessary quiescent current for LM317? You may also need to put a diode from Vadj. pin of 317 to its Vin to take care of the situation of applying the DAC voltage in the absence of external Vin to LM317. Use a current limiting resistor at the output of DAC and connect to Vadj. Remember DAC should be of +ve output Unipolar type.

Regards,
Laktronics
 

laktronics said:
I still do not understand why you can not use two diodes and a bleeder resistor of say 2.2k to ground after the diodes, which will provide the necessary quiescent current for LM317?
Click to expand...

You are right. Thank you for correcting my mistake. The bleeding resistor must be connecter between the output of the diodes and the ground, not parallel with the diodes.
 

Hi,
Sorry, again I made a mistake, you will need a bleeder resistor of about 1k before the diodes to ground and use another resistor of 10k to ground after the diodes to provide the ground biasing for the MOSFET.
Regards,
Laktronics
 

Why do you need the 1k resistor before the diodes? Even if you need it, isn't 1k a too low value, wouldn't it be a waste of energy?
 

Hi,
Because he wanted a biasing current of 1.3mA and with the diodes there, the resistor after diodes will not be able to sink any current at the low output of 1.25V. As regards waste of power at max 5V output, it will be 25mw, but I do agree that is still a wasted power.

Regards,
Laktronics
 

Ok, with the resistors to ground, they would take care of the quiescent current and now I can get current in the range of 100uA's through my MOSFET.

I think the bleeding resistor should be small, so that it can take all the quiescent current if the output voltage of the LM317 is low.

The power wasted is ok, as I'm using 12V 1.2AH lead acid battery's and the circuit will only be on for a few milliseconds
 

Add a power switch to your design. Turn off the power when you are not using the circuit, or it will kill your battery in time.
 

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