Resistors on an H-Bridge

[Warhawk87]

What was the purpose of using the 1K and 10k resistors in this H-Bridge circuit?  Honestly if i was building it i would completely leave those out, which apparently wouldnt be good.... Why?

http://www.mcmanis.com/chuck/robotics/tutorial/h-bridge/bjt-circuit.html

[futmacl]

The purpose of 1k resistors in this circuit is to limit the base current; if you do not understand why this is necessary with BJTs, I strongly suggest pausing to read a bit about transistor operation first. You would not need these 1k resistors with MOSFETs, but they are necessary here. Since these are power BJTs, and the unnamed phototransistor may not have a particularly high current capability, there is *some* chance that skipping this resistor will not cause the circuit to catch fire; but you don't want to bank on this.

The 10k resistors are just pull-ups / pull-downs to control the sensitivity of the phototransistors by requiring a certain current to be sourced before the base voltage increases goes above a threshold, making this largely independent of the characteristics of the driven transistors. Leaving them out in this circuit may make it too sensitive, but will have no other side effects. Leaving them out in a MOSFET circuit would make it misbehave, because when the phototransistor is off, residual charges will have nowhere to go.

[amando96]

I'm pretty sure all transistors as "switches" need a resistor, or will burn something(?) sooner or later, buy a h-bridge in a chip, like the sn754410ne.

[futmacl]

I'm pretty sure all transistors as "switches" need a resistor, or will burn something(?) sooner or later, buy a h-bridge in a chip, like the sn754410ne.

Most MOSFETs certainly do not need a current-limiting resistor on the gate, at least with low-frequency signals (in an ideal MOSFET, gate current is zero with DC signal applied; but at high frequencies, gate capacitance becomes a factor).

[Warhawk87]

I thought resistors just limited voltage, and in a series circuit the current remained constant at  total circuit voltage divided by total circuit resistance... And if thats true would u be able to throw the resistor anywhere in the circuit? Or it needs to be where it is?

And i completely lost you on the 10k part... But that part isnt that important so i think ill forget it for now...

[billhowl]

Read this
http://www.mcmanis.com/chuck/robotics/tutorial/h-bridge/bjt-impl.html

All the resistors are needed for the transistors to function, without it the transistors will not function.

[Soeren]

Hi,

You would not need these 1k resistors with MOSFETs,

Not 1k, but you need something in the vicinity of 10 Ohm to 100 Ohm, if a well mannered circuit is the goal.

The 10k resistors are just pull-ups / pull-downs to control the sensitivity of the phototransistors by requiring a certain current to be sourced before the base voltage increases goes above a threshold,

The 10k base-emitter resistors is there solely to keep the base in check when the transistor in the opto coupler is closed - not to control the sensitivity of the opto.

[Warhawk87]

I read that link, and it said that the 1K resistor was limiting the current coming OUT of the base... Why would current come out of the base? I thought you applied current TO the base!!

[futmacl]

You would not need these 1k resistors with MOSFETs,

Not 1k, but you need something in the vicinity of 10 Ohm to 100 Ohm, if a well mannered circuit is the goal.

Why? For DC signals, the gate can be essentially considered an unconnected wire. Spare for its inherent capacitance, why would you need a resistor?

The 10k base-emitter resistors is there solely to keep the base in check when the transistor in the opto coupler is closed - not to control the sensitivity of the opto.

I thought that in BJTs, the base will not keep a residual charge and will drain all by itself (MOSFET would need them, though)? As to the sensitivity of the circuit: the optotransistor needs to conduct a higher current for the threshold voltage to be exceeded with this resistor in place, no?

[waltr]

I read that link, and it said that the 1K resistor was limiting the current coming OUT of the base... Why would current come out of the base? I thought you applied current TO the base!!

It all in how you reference the direction of current flow. Even though we may think the current flows from Pos (+) to Neg (-), thanks to Ben Franklin, in reality electrons, which are negatively charged, flow from the Neg terminal to the Pos terminal of a voltage source.
So, in an NPN transistor the current flow really is from the emitter to the base or current flows out of the transistor's base.

In electronics the current flow direction can be arbitrarily defined as long as one remains consistent with the assignment of the 'sign'. However, in Solid State Physics current flow is dictated by electron flow.

[Warhawk87]

Hm thats interesting... But i still dont understand why the resistor limits current... I thought it just created a voltage drop which would limit voltage not current... The only drop in current i can see is a total circuit current drop due to the total circuit resistance going up...

[futmacl]

Hm thats interesting... But i still dont understand why the resistor limits current... I thought it just created a voltage drop which would limit voltage not current... The only drop in current i can see is a total circuit current drop due to the total circuit resistance going up...

There are two ways of looking at it, per Ohm's law. Current through a resistor is equal to voltage applied, divided by resistance. Consequently, voltage drop across a resistor is equal to current supplied, times resistance.

[waltr]

Hm thats interesting... But i still dont understand why the resistor limits current... I thought it just created a voltage drop which would limit voltage not current... The only drop in current i can see is a total circuit current drop due to the total circuit resistance going up...

Its that the voltage drop is the constant. The base to emitter of an NPN BiPolar is a diode (simple model) and is always near 0.6V delta when biased ON. If the emitter is tied to ground and you connect the base to say a 5V source through a resistor then the voltage drop across the resistor is 5-0.6 = 4.4V.
But, BiPolar transistors are current devices, base current controls the collector current. So, one needs to set the base current with the base resistor. In this example lets assume that the Beta = 100 and you wish to pass 1Amp through the collector. Therefore you need 1A divided by 100 = 10mA of base current. The voltage drop across the resistor will be 4.4V so with Ohm's law (4.4V/.01A) the resistor value is 440 Ohm.
Now you wish the pass 2Amps through the collector. The voltage drop across the base resistor is still 4.4V but you now need 20mA of base current. This require a 220 Ohm resistor. You can see the the base resistor sets or 'limits' the current flowing through the base and the voltage drop across the resistor is the same regardless of the current.

[Warhawk87]

Ok i think after everyone pounding it into my head im starting to see the picture... The voltage drop being the constant in the equation helped a lot....Im taking an electronics class in fall so maybe it will become clearer to me after that..... Thanks for your help everyone!