# Design a circuit with IRLZ34N N-channel MOSFET

Thread Starter

#### j13

Joined Oct 15, 2022
4
I have to design the circuit assuming the existence of +3.3V power supply ( more on the image

-I am attaching my approach to the problem but I don't think its correct.

related questions:
-What is the role of pull up resistor in terms of sourcing and sinking the current?
-Should I implement two IRLZ34N MOSFET to achieve source/sink characteristics or should I attach pull up resistors?

#note - I am not using IRLZ34N MOSFET I'm using a BJT

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#### dl324

Joined Mar 30, 2015
17,122
#note - I am not using IRLZ34N MOSFET I'm using a BJT
That's going to get you a failing grade for not being able to follow instructions.

Thread Starter

#### j13

Joined Oct 15, 2022
4
Hey, I know its an homework assignment. I don't know how to approach this problem. Here's a different approach

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#### Irving

Joined Jan 30, 2016
4,047
Welcome to AAC!

Why are you using a BJT? Achieving a Vce[sat] of 0.1v @ Ic = 10mA for a BJT is hard, whereas achieving an Rds[on] for the MOSFET of <0.1/10mA or 10ohm. Looking at the datasheet for the given MOSFET a gate voltage of 4v will achieve this (60mOhm!).

Another key reason for not using a BJT is that you cannot calculate Ib the way you did. At saturation the gain of a BJT is about 10 so if Ic = 10mA, Ib needs to be around 1mA, twice what you are allowed.

So it's clear a MOSFET is required; not using one would be a fail.

Otherwise your approach is essentially correct. The role of the load resistors is to simulate a following circuit, eg the input of some device.
Looking out from the collector of the BJT/drain of the MOSFET it will see a resistance to the V+ that needs pulling down (Rpu) at 10mA with a Vout of <=0.1v and one to Ground (Rpd) that will pull down 1mA when the transistor is off and Vout needs to be >= 3.1v.

There is a key piece of info missing. What is your supply voltage Vs? (incidentally your battery symbol is reversed; the long side is +ve).

Starting with the latter requirement the values of Rpd & Rpu are:

Rpd >= 3.1/1mA >= 3.1k
Rpu <= (Vs - 3.1)/1mA <= ?ohm

The former requirement gives
Rpu <= (Vs-0.1)/10mA.

Using a MOSFET fully turned on at this low current effectively reduces this to:
Rpu = Vs/10mA

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#### dl324

Joined Mar 30, 2015
17,122
I don't know how to approach this problem.
What part of this do you not understand?

#### Papabravo

Joined Feb 24, 2006
21,306
I would start by looking for a MOSFET circuit.

#### WBahn

Joined Mar 31, 2012
30,294
What is this circuit supposed to do? Is it supposed to be a linear amplifier? Is it a digital inverter? What is the Vout vs Vin characteristic supposed to look like?

Thread Starter

#### j13

Joined Oct 15, 2022
4
Welcome to AAC!

Why are you using a BJT? Achieving a Vce[sat] of 0.1v @ Ic = 10mA for a BJT is hard, whereas achieving an Rds[on] for the MOSFET of <0.1/10mA or 10ohm. Looking at the datasheet for the given MOSFET a gate voltage of 4v will achieve this (60mOhm!).

Another key reason for not using a BJT is that you cannot calculate Ib the way you did. At saturation the gain of a BJT is about 10 so if Ic = 10mA, Ib needs to be around 1mA, twice what you are allowed.

So it's clear a MOSFET is required; not using one would be a fail.

Otherwise your approach is essentially correct. The role of the load resistors is to simulate a following circuit, eg the input of some device.
Looking out from the collector of the BJT/drain of the MOSFET it will see a resistance to the V+ that needs pulling down (Rpu) at 10mA with a Vout of <=0.1v and one to Ground (Rpd) that will pull down 1mA when the transistor is off and Vout needs to be >= 3.1v.

There is a key piece of info missing. What is your supply voltage Vs? (incidentally your battery symbol is reversed; the long side is +ve).

Starting with the latter requirement the values of Rpd & Rpu are:

Rpd >= 3.1/1mA >= 3.1k
Rpu <= (Vs - 3.1)/1mA <= ?ohm

The former requirement gives
Rpu <= (Vs-0.1)/10mA.

Using a MOSFET fully turned on at this low current effectively reduces this to:
Rpu = Vs/10mA
Thanks for replying
Vs is 0-5volt
Please see the attach pdf for my solution

-What is the pull up and pulldown resistor really doing to can you please explain in steps? I've been reading a lot online but its not really sinking in.

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Thread Starter

#### j13

Joined Oct 15, 2022
4
What part of this do you not understand?
View attachment 278561
Hey,

-what does it mean Vout capable of sourcing and sinking current?
-what is "external device" output? is it the load?
-what does it mean " external device with an output no more than 0.1 V"? ( I can't visualize this)
-sourcing 1mA with output less than 3.1 V not sure the about terminology here

Hope I explained the best I don't understand about question here

thanks

#### BobTPH

Joined Jun 5, 2013
9,264
@Irving:

Your 490Ω pullup drops 0.49V when sourcing 1mA, so the output voltage is 3.3-0.49= 2.8V.

If the circuit must source 1mA at 3.1V with a 3.3V supply, it is easy to calculate the max pullup resistance based on the max voltage drop. Purposely not giving the equation or result since this is homework help.

Bob

#### Irving

Joined Jan 30, 2016
4,047
@Irving:

Your 490Ω pullup drops 0.49V when sourcing 1mA, so the output voltage is 3.3-0.49= 2.8V.

If the circuit must source 1mA at 3.1V with a 3.3V supply, it is easy to calculate the max pullup resistance based on the max voltage drop. Purposely not giving the equation or result since this is homework help.

Bob
Supply volts is 5v, not 3.3v, as per post #8

Also the 10mA pull down requirement is incompatible with a 3.3v supply and a 3.1v/1mA sink.

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#### BobTPH

Joined Jun 5, 2013
9,264
Post #8 clearly shows the pull-up connected to 3.3V. The 5V is the input signal.

#### Irving

Joined Jan 30, 2016
4,047
Post #8 clearly shows the pull-up connected to 3.3V. The 5V is the input signal.
My bad, I was refering to the text where the TS says Vs =5v, I didn't look at the diagram.

The TS is clearly confused. Bit miffed my post explaining how the concept of sink & source is applied was deleted. I'll repost with less content...

#### Irving

Joined Jan 30, 2016
4,047
Hey,

-what does it mean Vout capable of sourcing and sinking current?
-what is "external device" output? is it the load?
-what does it mean " external device with an output no more than 0.1 V"? ( I can't visualize this)
-sourcing 1mA with output less than 3.1 V not sure the about terminology here

Hope I explained the best I don't understand about question here

thanks
Ok, your circuit is an inverter - a logic 1 @5v results in a logic 0 (<=0.1v) at the output and a logic 0 @ 0v results in a logic1 >= 3.1v at the output, which is the input to some "device":
When driven with a logic 1 the device sinks 1mA and requires the input to be >=3.1v (your circuit is sourcing the 1mA through Rpu)
When driven with a logic 0 the device requires the input to be <=0.1v (your circuit is sinking current through Rpu)

For various reasons (noise immunity being a common one) the user requires the pulldown to be strong and has specified it as a 10mA pull-down current.

Consider your circuit, but using a MOSFET:

Based on those design requirements we can ascertain the input impedance of the device (Rin). When the transistor is off Vout must rise to at least 3.1v. Rpu and Rin form a potential divider so Vout is a %age of the supply voltage.

When the transistor is on, current flows from drain to source and the requirement is that Vout < 0.1v at a current of 10mA. Vout in this circumstance is the result of a potential divider between Rpu and the MOSFET's Rds[on] parameter.

These concepts are valid whether the transistor is a MOSFET or a BJT, however meeting the 0.1v requirement will be very difficult with a BJT.

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