I have a load that has varying resistance with external pressure. In order to measure the change, I am trying to measure the voltage on the load, when constant current is applied. And the circuit in the attachment is known to supply constant current using a transistor and a zener diode. But I have no idea what resistor/ transistor/ diode I need. I do not know if the circuit will burn them out.
The current on the load has to be approximately 6 mA, I think. And the supply voltage might be 5V or 3.3V. If impossible, the supply voltage can be changed. If needed, the transistor and zener diode I purchased are KTC3198, and 1N5231B-5.1V, respectively. The minimum current zener diode needs is 0.25 mA. Please comment, if any further information needed

 

Welcome to AAC!

You can't make a current source using a 5.1V zener with a 5V (or 3.3V) voltage source.

What is the resistance range for the load?

 

i'm sorry. the supply voltage will be 10V

 

the resistance will be in range of 6-400? (one digit ohm ~ three digits ohm)

 

Welcome to AAC!

You can't make a current source using a 5.1V zener with a 5V (or 3.3V) voltage source.

What is the resistance range for the load?

i'm sorry. the supply voltage will be 10V.
the resistance will be in range of 6-400? (one digit ohm ~ three digits ohm)

 

You should operate the zener well beyond the knee. Closer to 20mA than 0.25mA.

Current is set by R2.
\( \small R=\frac{V}{I}=\frac {5.1V-0.7V}{6mA}=733\Omega \)
Use 750 ohms or use a parallel combination to get 733.

 

750 ohms in parallel with 33k will give 733 ohms.

 

I have a load that has varying resistance with external pressure. In order to measure the change, I am trying to measure the voltage on the load, when constant current is applied. And the circuit in the attachment is known to supply constant current using a transistor and a zener diode. But I have no idea what resistor/ transistor/ diode I need. I do not know if the circuit will burn them out.
The current on the load has to be approximately 6 mA, I think. And the supply voltage might be 5V or 3.3V. If impossible, the supply voltage can be changed. If needed, the transistor and zener diode I purchased are KTC3198, and 1N5231B-5.1V, respectively. The minimum current zener diode needs is 0.25 mA. Please comment, if any further information needed

How precise do you need the current to be?

Notice that I'm not asking about accuracy -- since you said it needed to be approximately 6 mA, I'm assuming that anything in that general ballpark will be suitable.

I'm asking about precision or, in this context, stability. If it's, say, 6 mA at 400 Ω, how close to 6 mA does it need to still be at 6 Ω?

That circuit is not going to have a very high output resistance, but you are operating it at a sufficiently low current that it might be good enough -- but you do need to be able to articulate how good constitutes good enough.

There are other circuits that can significantly increase the output resistance of the source.

 

Here's the LTspice simulation of the circuit.
I didn't have models for the devices you bought but the results shouldn't be that different with your devices.
Notice that the current varies less than 0.05% as sensor resistance varies from 6Ω to 400Ω

 

ZENER DIODE TESTING The testing of zener diodes requires a variable dc power supply. A typical test circuit can be constructed, as shown in figure 4-19. In this circuit, the variable power supply is used to adjust the input voltage to a suitable value for the zener diode being tested. Resistor R1 limits the current through the diode. With the zener diode connected as shown in figure 4-19, no current will flow until the voltage across the diode is equal to the zener voltage. If the diode is connected in the opposite direction, current will flow at a low voltage, usually less than 1 volt. Current flow at a low voltage in both directions indicates that the zener diode is defective.

I have a load that has varying resistance with external pressure. In order to measure the change, I am trying to measure the voltage on the load, when constant current is applied. And the circuit in the attachment is known to supply constant current using a transistor and a zener diode. But I have no idea what resistor/ transistor/ diode I need. I do not know if the circuit will burn them out.

The current on the load has to be approximately 6 mA, I think. And the supply voltage might be 5V or 3.3V. If impossible, the supply voltage can be changed. If needed, the transistor and zener diode I purchased are KTC3198, and 1N5231B-5.1V, respectively. The minimum current zener diode needs is 0.25 mA. Please comment, if any further information needed

 

ZENER DIODE TESTING

The TS is not testing Zeners.

 

Hello,

Here is an approximation to the current in the collector assuming there is enough overhead voltage from Vcc:
Ic=(10*Vd-7)*(B-1)*(B+1)/(10*B^2*R1)

where:
Ic is the collector current,
Vd is the zener voltage,
R1 is as in post #9 the emitter resistor,
B is the transistor Beta.

Fooling around with that equation tells us the minimum Beta should be around 20 which should be easy to obtain. This is based on the values in post #9. Also if the Beta is 20 or above the collector current is relatively insensitive to a change in that Beta.

 

ZENER DIODE TESTING The testing of zener diodes requires a variable dc power supply. A typical test circuit can be constructed, as shown in figure 4-19. In this circuit, the variable power supply is used to adjust the input voltage to a suitable value for the zener diode being tested. Resistor R1 limits the current through the diode. With the zener diode connected as shown in figure 4-19, no current will flow until the voltage across the diode is equal to the zener voltage. If the diode is connected in the opposite direction, current will flow at a low voltage, usually less than 1 volt. Current flow at a low voltage in both directions indicates that the zener diode is defective.

Even if the TS were testing zener diodes, what use is a post referring to figures that aren't included and component references in schematics that aren't included.

 

Hello,

Here is an approximation to the current in the collector assuming there is enough overhead voltage from Vcc:
Ic=(10*Vd-7)*(B-1)*(B+1)/(10*B^2*R1)

where:
Ic is the collector current,
Vd is the zener voltage,
R1 is as in post #9 the emitter resistor,
B is the transistor Beta.

Fooling around with that equation tells us the minimum Beta should be around 20 which should be easy to obtain. This is based on the values in post #9. Also if the Beta is 20 or above the collector current is relatively insensitive to a change in that Beta.

I think the equation would cause less head scratching if it weren't forced to have only integers. As it is, it's a head-scratcher for a while about why the zener diode voltage is multiplied by ten and then seven is subtracted from it.

How about simply

Ic = (Vd - 0.7 V)/R1 * (B² - 1)/B²

 

Andy why not simply Ic = (Vd - 0.7 V)/R1?
The (B² - 1)/B² term is basically one for any practical beta. At extremely low B=5 the error is 4% when you just use 1 instead of calculated 0.96. You will get grater error from the accuracy and temperature dependance of zener voltage and temperature dependence of Vbe.

 

Andy why not simply Ic = (Vd - 0.7 V)/R1?
The (B² - 1)/B² term is basically one for any practical beta. At extremely low B=5 the error is 4% when you just use 1 instead of calculated 0.96. You will get grater error from the accuracy and temperature dependance of zener voltage and temperature dependence of Vbe.

Agreed (even the actual Vbe compared to the assumed 0.7 V will have a larger impact), but I think MrAl was specifically making a point about β and that's hard to do if you don't include those terms.

 

Fooling around with that equation tells us the minimum Beta should be around 20 which should be easy to obtain.

Typical beta for the transistor the OP mentioned is about 100 at 10mA and it's affect can be ignored.


Zener diode and base-emitter voltages will have more effect.

 

Over what temperature range must your circuit operate?

 

I think the equation would cause less head scratching if it weren't forced to have only integers. As it is, it's a head-scratcher for a while about why the zener diode voltage is multiplied by ten and then seven is subtracted from it.

How about simply

Ic = (Vd - 0.7 V)/R1 * (B² - 1)/B²

Hi,

Yes i dont always reduce the solution to the simplest possible form and using floats.
It's good to show it that way too so we can immediately spot the base emitter voltage approximation too.
So we end up with:
Ic = (Vd - Vbe)/R1 * (B² - 1)/B²

and the multiplying effect of Beta becomes more apparent too.

 

And just in case you need the load to be ground referenced: