FETs switching in resistors.

[JMG]

FETs switching resistors into a circuit.

Hello,

I have a small circuit based around an LM3409 LED driver. I need to control the analog dimming using 2 inputs. Inputs are 0 or 5v. I will be using logic level FETs. The LM3409's analog input pin is a 1.24V 50uA supply that can either be varied using an external analog voltage, or reduced by tying it to ground through resistors.

Attached is a schematic that uses a PFET to tie the analog in pin to ground when there is no inputs, and 2 NFETS that switch in resistors depending on which input is turned on. This gives 4 light output options. Off, low, mid and high.

The 2 diodes stop the other input from turning on due to the PFET being common between the 2 inputs. Obviously as the input is turned on, the PFET is turned off, thus disconnecting the analog pin from ground and allowing current to flow through one or both resistors.

The input to the LM3409 is through R5 at the top right of the schematic.

Does this look like a good solution? Any better ideas? I am missing something obvious here?

Thanks for your help.

James.

 

[alexan_e]

Note that mosfets don't turn off completely, they have a leakage current even when they are off which will probably be close to your 50uA rating (maybe even more) so I think you will have problems with a solution like the one described in your post

Alex

 

[JMG]

I don't think leakage through the nfets would be an issue as when they are off the pfet is on short circuiting their drains to ground.

When the pfet is off the nfets are on and the resistance of the pfet is megaohms, thus having little to no effect on the resistors connected to the nfets.

 

[alexan_e]

I'm not talking about Q4, you said you want to control the current using Q2 Q3 so when you have

Q2=on Q3=off
or
Q2=off Q3=on
or
Q2=0ff Q3=off

the leakage will be a problem.
For the four states (00,01,10,11) only one involves the pmos Q4 shorting the inputs

Alex

 

[JMG]

Ahh, yes, I see your point.

I guess I'm out of ideas. I'm not sure how to create a fast acting switch that completely isolates one resistor when its corresponding input is low. Opto is too expensive and too many components for the space I have. Relay is too slow and to big.

Hmm...

 

[BradtheRad]

If you're controlling current flow into a device, perhaps you can find a way to use a single transistor to do the job? Rather than switch resistors in and out.

A transistor because it's current operated. (Whereas mosfets are voltage operated.)

A 5V signal can energize a transistor bias. You would install a resistor inline to tailor adjust the bias current.

The proper resistor values will give you desired current flow through the transistor C-E legs.

Use a separate bias resistor for each incoming control pulse. Isolate each incoming control pulse with its own steering diode.

When all control pulses are ON, you'll get the combined bias currents, for maximum flow through the transistor. You must adjust all values so that you don't exceed spec ratings of neighboring components.

====================

Edited to add:

Since you're connecting the IC to ground through various resistances...

It may be feasible simply to use your control pulses alone to do the job. Choose resistor values to achieve desired operation.

Your incoming control would need to be able to sink current. If it doesn't then you'll need to invert operation.